Differential Equations

1. The highest derivative present is the second derivative $y''$, so the order is 2.
2. The function $y$ and its derivatives $y'$ and $y''$ all appear to the first power and are not multiplied by each other.
3. The term $f(t)$ depends only on the independent variable $t$, which does not affect linearity.

Answer: Order 2, Linear

1. First, compute the derivative of the proposed solution: $y' = -5 e^{-5t}$.
2. Substitute $y$ and $y'$ into the equation: $(-5 e^{-5t}) + 5(e^{-5t})$.
3. Simplify the expression: $-5 e^{-5t} + 5 e^{-5t} = 0$.
4. Since the result is 0, the equation is satisfied.

Answer: Yes, it is a solution.

1. The highest derivative is $y'$, so the order is 1.
2. Rewrite the equation as $y y' = t$.
3. Because the unknown function $y$ is multiplied by its derivative $y'$, the equation is non-linear.

Answer: Order 1, Non-linear

1. Identify the highest derivative: the second derivative $y''$ is present, so the order is 2.
2. Check for linearity: the term $y' y$ is a product of the unknown function and its derivative.
3. Since a product of $y$ and its derivatives is present, the equation is non-linear.

Answer: Order 2, Non-linear

1. Compute the derivative of the candidate function: $y' = \frac{d}{dt}(e^{3t}) = 3e^{3t}$.
2. Substitute $y$ and $y'$ into the ODE: $3e^{3t} - 3(e^{3t})$.
3. Simplify the expression: $3e^{3t} - 3e^{3t} = 0$.
4. Since the result is $0 = 0$, the function is a solution.

Answer: Yes, it is a solution

1. Identify the highest derivative: $y'''$ is the third derivative, so the order is 3.
2. Check for linearity: $y'''$, $y''$, $y'$, and $y$ all appear to the first power.
3. Check coefficients: $1$, $t^2$, $-e^t$, and $\cos(t)$ depend only on the independent variable $t$.
4. Since all conditions for linearity are met, the equation is linear.

Answer: Order 3, Linear

1. Compute the derivative: $y' = 15e^{3t}$.
2. Substitute $y$ and $y'$ into the equation: $15e^{3t} - 3(5e^{3t})$.
3. Simplify: $15e^{3t} - 15e^{3t} = 0$.
4. The identity $0 = 0$ holds.

Answer: It is a solution.

1. Substitute $t = 0$ and $y = 10$ into the general solution: $10 = Ce^{0/2}$.
2. Since $e^0 = 1$, we have $10 = C(1)$.
3. Thus, $C = 10$.
4. Substitute $C$ back into the general solution.

Answer: $y = 10e^{t/2}$

1. Compute the derivative: $y' = 2t$.
2. Substitute into the ODE: $2t = 2t$. This is an identity, so it is a solution.
3. To find the initial condition, evaluate $y$ at $t = 0$: $y(0) = 0^2 + 1 = 1$.
4. The initial condition is $y(0) = 1$.

Answer: Yes; $y(0) = 1$

1. Set $y' = 0$, which gives $y(y - 3) = 0$.
2. The equilibrium solutions are $y = 0$ and $y = 3$.
3. For $y > 3$, $y'$ is positive, so solutions move away from $y = 3$.
4. For $0 < y < 3$, $y'$ is negative, so solutions move away from $y = 3$ and toward $y = 0$.
5. For $y < 0$, $y'$ is positive, so solutions move toward $y = 0$.
6. Therefore, $y = 3$ is unstable and $y = 0$ is stable.

Answer: $y = 0$ (stable), $y = 3$ (unstable)

1. At $(0, 2)$, the slope is $0 - 2 = -2$, so the curve starts by decreasing.
2. As $t$ increases and $y$ decreases, the slope $t - y$ becomes less negative.
3. We check the line $y = t - 1$. Its slope is $1$.
4. Plugging $y = t - 1$ into the ODE gives $t - (t - 1) = 1$.
5. Since the slopes match, $y = t - 1$ is a solution.
6. The solution starting at $(0, 2)$ will be attracted to the line $y = t - 1$.

Answer: The solution decreases and asymptotically approaches the line $y = t - 1$.

1. The equation is not autonomous because the slope depends on $t$.
2. At $(2, 1)$, the slope is $y' = \frac{2^2}{1} = 4$.
3. To find concavity, we differentiate $y' = t^2 y^{-1}$ with respect to $t$ using the product and chain rules: $y'' = 2t y^{-1} + t^2 (-y^{-2} y')$.
4. Substitute $t = 2, y = 1, y' = 4$: $y'' = 2(2)(1)^{-1} + (2^2)(-(1)^{-2})(4)$.
5. Compute: $y'' = 4 - 16 = -12$.
6. Since $y'' < 0$, the solution is concave down at this point.

Answer: Not autonomous; slope is $4$; concave down.

1. Identify $f(t, y) = y^2 + t$. This is a polynomial, so it is continuous everywhere, including $(0, 0)$.
2. Compute the partial derivative with respect to $y$: $\frac{\partial f}{\partial y} = 2y$.
3. Check continuity of the partial derivative: $2y$ is a polynomial and is continuous everywhere, including $(0, 0)$.
4. Since both $f$ and $\frac{\partial f}{\partial y}$ are continuous, a unique solution is guaranteed.

Answer: Yes, because both $f(t, y) = y^2 + t$ and $\frac{\partial f}{\partial y} = 2y$ are continuous at $(0, 0)$.

1. Substitute $y(t) = 0$ into the ODE: $y' = 0$ and $2\sqrt{|0|} = 0$. Since $0 = 0$, it is a solution.
2. Identify $f(t, y) = 2\sqrt{|y|}$.
3. Compute the partial derivative: $\frac{\partial f}{\partial y} = \frac{1}{\sqrt{|y|}}$ (for $y \neq 0$).
4. Evaluate at the initial point $y = 0$: the expression $\frac{1}{\sqrt{0}}$ is undefined (approaches infinity).
5. The uniqueness condition is not satisfied.

Answer: The uniqueness condition is not satisfied because $\frac{\partial f}{\partial y}$ is undefined at $y=0$.

1. Identify $f(t, y) = \frac{y}{t-1}$.
2. Check continuity of $f$ at the point $(1, 2)$.
3. The denominator $t-1$ becomes $1-1 = 0$ at $t=1$.
4. Since the function $f$ is undefined at $t=1$, it is not continuous at the initial point.
5. Therefore, the existence of a solution is not guaranteed by the Picard theorem.

Answer: No, because $f(t, y)$ is not continuous at $t=1$.

1. Separate the variables: $\frac{1}{y} dy = t^2 dt$.
2. Integrate both sides: $\int \frac{1}{y} dy = \int t^2 dt$.
3. Compute integrals: $\ln|y| = \frac{1}{3}t^3 + C$.
4. Exponentiate: $|y| = e^{\frac{1}{3}t^3 + C} = e^C e^{\frac{1}{3}t^3}$.
5. Simplify constant: $y(t) = A e^{\frac{1}{3}t^3}$.

Answer: $y(t) = A e^{\frac{1}{3}t^3}$

1. Separate variables: $\frac{1}{y^2} dy = 2t dt$.
2. Integrate: $\int y^{-2} dy = \int 2t dt \implies -\frac{1}{y} = t^2 + C$.
3. Apply initial condition $y(0)=1$: $-\frac{1}{1} = 0^2 + C \implies C = -1$.
4. Substitute $C$ back: $-\frac{1}{y} = t^2 - 1$.
5. Solve for $y$: $\frac{1}{y} = 1 - t^2 \implies y = \frac{1}{1-t^2}$.

Answer: $y(t) = \frac{1}{1-t^2}$

1. Check for equilibrium: $h(y) = y = 0 \implies y(t) = 0$ is a solution.
2. Separate variables: $\frac{1}{y} dy = \cos(t) dt$.
3. Integrate: $\ln|y| = \sin(t) + C$.
4. Exponentiate: $y = A e^{\sin(t)}$.

Answer: $y(t) = A e^{\sin(t)}$ and $y(t) = 0$

1. Identify $P(t) = 2$ and $Q(t) = e^{-t}$.
2. Compute the integrating factor $\mu(t) = e^{\int 2 dt} = e^{2t}$.
3. Multiply the ODE by $e^{2t}$ to get $\frac{d}{dt}(e^{2t}y) = e^{2t}e^{-t} = e^t$.
4. Integrate both sides: $e^{2t}y = e^t + C$.
5. Solve for $y$: $y = e^{-t} + Ce^{-2t}$.
6. Use $y(0) = 1$: $1 = 1 + C \implies C = 0$.
7. The particular solution is $y = e^{-t}$.

Answer: $y = e^{-t}$

1. Identify $P(t) = -1/t$ and $Q(t) = t^2$.
2. Compute $\mu(t) = e^{\int -1/t dt} = e^{-\ln t} = e^{\ln(1/t)} = 1/t$.
3. Multiply the ODE by $1/t$: $\frac{d}{dt}(\frac{1}{t}y) = \frac{1}{t}t^2 = t$.
4. Integrate both sides: $\frac{1}{t}y = \frac{1}{2}t^2 + C$.
5. Solve for $y$: $y = \frac{1}{2}t^3 + Ct$.

Answer: $y = \frac{1}{2}t^3 + Ct$

1. Identify $P(t) = t$ and $Q(t) = t e^{-t^2/2}$.
2. Compute $\mu(t) = e^{\int t dt} = e^{t^2/2}$.
3. Multiply the ODE by $e^{t^2/2}$: $\frac{d}{dt}(e^{t^2/2}y) = e^{t^2/2} t e^{-t^2/2} = t$.
4. Integrate both sides: $e^{t^2/2}y = \frac{1}{2}t^2 + C$.
5. Solve for $y$: $y = (\frac{1}{2}t^2 + C)e^{-t^2/2}$.

Answer: $y = (\frac{1}{2}t^2 + C)e^{-t^2/2}$

1. Identify $M = 3t^2 + 2ty$ and $N = t^2$.
2. Check exactness: $\frac{\partial M}{\partial y} = 2t$ and $\frac{\partial N}{\partial t} = 2t$. It is exact.
3. Integrate $M$ with respect to $t$: $\psi = \int (3t^2 + 2ty) dt = t^3 + t^2 y + g(y)$.
4. Differentiate $\psi$ with respect to $y$: $\frac{\partial \psi}{\partial y} = t^2 + g'(y)$.
5. Set equal to $N$: $t^2 + g'(y) = t^2 \implies g'(y) = 0 \implies g(y) = C_1$.
6. The general solution is $t^3 + t^2 y = C$.

Answer: t^3 + t^2 y = C

1. Check exactness: $\frac{\partial M}{\partial y} = 1$ and $\frac{\partial N}{\partial t} = 1$. Exact.
2. Integrate $M$: $\psi = \int (2t + y) dt = t^2 + ty + g(y)$.
3. Differentiate $\psi$ wrt $y$: $t + g'(y) = t + 2y \implies g'(y) = 2y \implies g(y) = y^2$.
4. General solution: $t^2 + ty + y^2 = C$.
5. Apply $(1, 1)$: $1^2 + (1)(1) + 1^2 = 3$, so $C = 3$.

Answer: t^2 + ty + y^2 = 3

1. Check exactness: $\frac{\partial M}{\partial y} = \cos(t) e^y$ and $\frac{\partial N}{\partial t} = \cos(t) e^y$. Exact.
2. Integrate $M$: $\psi = \int \cos(t) e^y dt = \sin(t) e^y + g(y)$.
3. Differentiate $\psi$ wrt $y$: $\sin(t) e^y + g'(y) = \sin(t) e^y \implies g'(y) = 0$.
4. General solution: $\sin(t) e^y = C$.

Answer: \sin(t) e^y = C

1. Identify $n = 3$, so use substitution $v = y^{1-3} = y^{-2}$.
2. Compute $v' = -2y^{-3}y'$.
3. Multiply the ODE by $-2y^{-3}$ to get $-2y^{-3}y' - \frac{2}{t}y^{-2} = -2t$.
4. Substitute $v$ and $v'$ to get $v' - \frac{2}{t}v = -2t$.
5. Use integrating factor $\mu(t) = e^{\int -2/t dt} = t^{-2}$.
6. Solve $t^{-2}v = \int -2t(t^{-2}) dt = -2\ln|t| + C$.
7. Find $v = t^2(-2\ln|t| + C)$ and substitute back $y = \pm 1/\sqrt{v}$.

Answer: $y = \pm \frac{1}{\sqrt{t^2(C - 2\ln|t|)}}$

1. Rewrite as $y' = \frac{y}{t} + \frac{t}{y}$, which is $f(v) = v + 1/v$ where $v = y/t$.
2. Substitute $y' = v + tv'$, so $v + tv' = v + 1/v$.
3. Simplify to $tv' = 1/v$.
4. Separate variables: $v dv = \frac{1}{t} dt$.
5. Integrate: $\frac{1}{2}v^2 = \ln|t| + C$.
6. Substitute $v = y/t$: $\frac{y^2}{2t^2} = \ln|t| + C$.

Answer: $y^2 = 2t^2(\ln|t| + C)$

1. This is a Bernoulli equation with $n = 2$. Use $v = y^{1-2} = y^{-1}$.
2. Then $v' = -y^{-2}y'$.
3. Multiply ODE by $-y^{-2}$: $-y^{-2}y' = -y^{-2}(y/t) - t^2$.
4. Substitute: $v' = -v/t - t^2$, or $v' + \frac{1}{t}v = -t^2$.
5. Integrating factor $\mu(t) = e^{\int 1/t dt} = t$.
6. Solve $tv = \int -t^3 dt = -\frac{1}{4}t^4 + C$.
7. Find $v = -\frac{1}{4}t^3 + C/t$.
8. Substitute $y = 1/v$: $y = \frac{1}{C/t - t^3/4}$.
9. Use $y(1) = 1$: $1 = \frac{1}{C - 1/4} \implies C = 5/4$.

Answer: $y = \frac{4t}{5 - t^4}$

1. Identify the equation as first order linear with $P(t) = 2$ and $Q(t) = 4$.
2. Compute the integrating factor $\mu(t) = e^{\int 2 dt} = e^{2t}$.
3. Multiply the ODE by $\mu(t)$ to get $(e^{2t}y)' = 4e^{2t}$.
4. Integrate both sides: $e^{2t}y = 2e^{2t} + C$.
5. Solve for $y$: $y = 2 + Ce^{-2t}$.
6. Apply $y(0) = 3$: $3 = 2 + C \implies C = 1$.

Answer: $y(t) = 2 + e^{-2t}$

1. Identify the equation as homogeneous since the right side depends only on $y/t$.
2. Substitute $v = y/t$, so $y = vt$ and $y' = v + tv'$.
3. Rewrite the ODE: $v + tv' = v^2$.
4. Separate variables: $tv' = v^2 - v \implies \frac{dv}{v(v-1)} = \frac{dt}{t}$.
5. Integrate using partial fractions: $\ln|v-1| - \ln|v| = \ln|t| + C$.
6. Simplify: $\frac{v-1}{v} = Ct \implies 1 - \frac{1}{v} = Ct$.
7. Substitute back $v = y/t$: $1 - \frac{t}{y} = Ct$.

Answer: $y(t) = \frac{t}{1 - Ct}$

1. Identify as Bernoulli with $n = 2$.
2. Substitute $v = y^{1-2} = y^{-1}$, so $v' = -y^{-2}y'$.
3. Divide the original ODE by $y^2$: $y^{-2}y' + \frac{1}{t}y^{-1} = t$.
4. Substitute $v$: $-v' + \frac{1}{t}v = t \implies v' - \frac{1}{t}v = -t$.
5. Find integrating factor $\mu(t) = e^{\int -1/t dt} = e^{-\ln t} = 1/t$.
6. Multiply: $(v/t)' = -1$.
7. Integrate: $v/t = -t + C \implies v = -t^2 + Ct$.
8. Substitute back $y = 1/v$.

Answer: $y(t) = \frac{1}{Ct - t^2}$

1. Step 1: Translate the rule to an ODE. The rate of change is $dT/dt$. The difference is $(T - 20)$. Since it is cooling, the rate is negative: $dT/dt = -k(T - 20)$.
2. Step 2: Solve the separable ODE. $\int \frac{1}{T-20} dT = \int -k dt$, which gives $\ln(T-20) = -kt + C$, or $T(t) = 20 + Ce^{-kt}$.
3. Step 3: Use initial condition $T(0) = 100$. $100 = 20 + C e^0$, so $C = 80$. The solution is $T(t) = 20 + 80e^{-kt}$.
4. Step 4: Use $T(10) = 60$ to find $k$. $60 = 20 + 80e^{-10k} \implies 40 = 80e^{-10k} \implies 0.5 = e^{-10k}$. Thus $-10k = \ln(0.5)$, so $k = \frac{\ln(2)}{10} \approx 0.0693$.

Answer: $T(t) = 20 + 80e^{-0.0693t}$

1. Step 1: Write the ODE. Growth is $0.1P$ and harvest is $-500$. So $dP/dt = 0.1P - 500$.
2. Step 2: Solve the linear ODE. This is separable: $dP/(0.1P - 500) = dt$. Integrating gives $10 \ln(0.1P - 500) = t + C$, or $0.1P - 500 = Ae^{0.1t}$.
3. Step 3: Use $P(0) = 2000$. $0.1(2000) - 500 = A e^0 \implies 200 - 500 = A \implies A = -300$.
4. Step 4: Solve for $P$. $0.1P = 500 - 300e^{0.1t} \implies P(t) = 5000 - 3000e^{0.1t}$.
5. Step 5: Analyze long term behavior. As $t \to \infty$, $e^{0.1t} \to \infty$, so $P(t) \to -\infty$. The population will hit zero and go extinct.

Answer: $P(t) = 5000 - 3000e^{0.1t}$; Yes, it goes extinct.

1. Step 1: Write the ODE. Rate in is $0.1 \text{ kg/L} \times 5 \text{ L/min} = 0.5 \text{ kg/min}$. Rate out is $(y/500) \text{ kg/L} \times 5 \text{ L/min} = 0.01y \text{ kg/min}$.
2. Step 2: The ODE is $dy/dt = 0.5 - 0.01y$.
3. Step 3: Solve the ODE. $\int \frac{1}{0.5 - 0.01y} dy = \int dt \implies -100 \ln(0.5 - 0.01y) = t + C$.
4. Step 4: $0.5 - 0.01y = Ae^{-0.01t} \implies 0.01y = 0.5 - Ae^{-0.01t} \implies y(t) = 50 - 100Ae^{-0.01t}$.
5. Step 5: Use $y(0) = 20$. $20 = 50 - 100A \implies 100A = 30$. So $y(t) = 50 - 30e^{-0.01t}$.
6. Step 6: Steady state is $\lim_{t \to \infty} (50 - 30e^{-0.01t}) = 50$ kg.

Answer: $y(t) = 50 - 30e^{-0.01t}$; Steady state is $50$ kg.

1. Identify the ratio $y/y_0 = 0.40$.
2. Calculate the decay constant $k = \ln(0.5)/5730 \approx -0.00012097$.
3. Set up the equation $0.40 = e^{kt}$.
4. Solve for $t$ by taking the natural log: $\ln(0.40) = kt$.
5. Compute $t = \ln(0.40)/(-0.00012097) \approx 7575$ years.

Answer: 7575 \text{ years}

1. Use the doubling time to find $k$: $2y_0 = y_0 e^{12k} \implies k = \ln(2)/12$.
2. Set up the equation for tripling: $3y_0 = y_0 e^{kt}$.
3. Simplify to $3 = e^{(\ln(2)/12)t}$.
4. Take the natural log: $\ln(3) = (\ln(2)/12)t$.
5. Solve for $t = 12 \ln(3)/\ln(2) \approx 19.02$ days.

Answer: 19.02 \text{ days}

1. Find $k = \ln(0.5)/150 \approx -0.004621$.
2. Set up the equation $50 = 500 e^{kt}$.
3. Divide by $500$: $0.1 = e^{kt}$.
4. Take the natural log: $\ln(0.1) = kt$.
5. Solve for $t = \ln(0.1)/-0.004621 \approx 498.29$ years.

Answer: 498.29 \text{ years}

1. Use the model $T(t) = 20 + (100 - 20)e^{-kt} = 20 + 80e^{-kt}$.
2. Use $T(1) = 80$ to find $k$: $80 = 20 + 80e^{-k} \implies 60 = 80e^{-k} \implies e^{-k} = 0.75$.
3. Find $k = -\ln(0.75) \approx 0.2877$.
4. Calculate $T(2) = 20 + 80(e^{-k})^2 = 20 + 80(0.75)^2$.
5. Compute $T(2) = 20 + 80(0.5625) = 20 + 45 = 65$.

Answer: 65^{\circ}C

1. Model: $T(t) = 22 + (180 - 22)e^{-kt} = 22 + 158e^{-kt}$.
2. Find $k$ using $T(10) = 150$: $150 = 22 + 158e^{-10k} \implies 128 = 158e^{-10k} \implies e^{-10k} = 128/158 \approx 0.8101$.
3. $-10k = \ln(0.8101) \implies k \approx 0.02106$.
4. Set $T(t) = 100$: $100 = 22 + 158e^{-0.02106t} \implies 78 = 158e^{-0.02106t}$.
5. Divide by $158$: $e^{-0.02106t} \approx 0.4937$.
6. Take log: $-0.02106t = \ln(0.4937) \approx -0.7058 \implies t \approx 33.5$.

Answer: \approx 33.5 \text{ minutes}

1. At $t=0$, $T(0) = T_s + 50e^0 = T_s + 50$.
2. Given $T(0) = 40$, we have $40 = T_s + 50 \implies T_s = -10^{\circ}C$.
3. The model is $T(t) = -10 + 50e^{-0.05t}$.
4. For $t=20$: $T(20) = -10 + 50e^{-0.05(20)} = -10 + 50e^{-1}$.
5. Using $e^{-1} \approx 0.3679$, $T(20) = -10 + 50(0.3679) = -10 + 18.395 = 8.395$.

Answer: T_s = -10^{\circ}C, T(20) \approx 8.4^{\circ}C

1. Identify parameters: $V = 500$, $r_{in} = 5$, $c_{in} = 1$, $r_{out} = 5$.
2. Set up the ODE: $x' = (5 \times 1) - (5 \times x/500)$, which simplifies to $x' = 5 - x/100$.
3. Solve using integrating factor $e^{t/100}$: $x(t) = 500 + Ce^{-t/100}$.
4. Apply initial condition $x(0) = 10$: $10 = 500 + C \implies C = -490$.
5. Final solution: $x(t) = 500 - 490e^{-t/100}$.

Answer: x(t) = 500 - 490e^{-t/100}

1. Set up ODE: $x' = (3 \times 0.5) - (3 \times x/200) = 1.5 - 0.015x$.
2. Solve the ODE: $x(t) = 100(1 - e^{-0.015t})$.
3. Set $x(t) = 50$: $50 = 100(1 - e^{-0.015t}) \implies 0.5 = 1 - e^{-0.015t}$.
4. Solve for $t$: $e^{-0.015t} = 0.5 \implies -0.015t = \ln(0.5)$.
5. Calculate $t$: $t = \ln(0.5) / -0.015 \approx 46.21$ minutes.

Answer: t \approx 46.21 \text{ minutes}

1. Volume function: $V(t) = 100 + (5-3)t = 100 + 2t$.
2. Set up ODE: $x' = (5 \times 2) - (3 \times x/(100 + 2t)) = 10 - 3x/(100 + 2t)$.
3. Integrating factor: $\int \frac{3}{100+2t} dt = \frac{3}{2} \ln(100+2t)$, so $\mu(t) = (100+2t)^{1.5}$.
4. Integrate: $x(100+2t)^{1.5} = \int 10(100+2t)^{1.5} dt = \frac{10}{2 \times 2.5} (100+2t)^{2.5} + C = 2(100+2t)^{2.5} + C$.
5. General solution: $x(t) = 2(100+2t) + C(100+2t)^{-1.5}$.
6. Initial condition $x(0)=0$: $0 = 200 + C(100)^{-1.5} \implies C = -200 \times 1000 = -200,000$.
7. At $t=10$: $x(10) = 2(120) - 200,000(120)^{-1.5} \approx 240 - 152.1 = 87.9$ g.

Answer: x(10) \approx 87.9 \text{ g}

1. Identify constants: $K = 5000$, $r = 0.2$, $y_0 = 500$.
2. Solve for $C$: $500 = 5000 / (1 + C) \implies 1 + C = 10 \implies C = 9$.
3. Write the solution: $y(t) = 5000 / (1 + 9 e^{-0.2t})$.
4. Calculate $y(5) = 5000 / (1 + 9 e^{-0.2 \cdot 5}) = 5000 / (1 + 9 e^{-1}) \approx 5000 / (1 + 3.31) \approx 1160$.

Answer: $y(5) \approx 1160$

1. Since $y(0) = 2500$ is greater than $K = 2000$, the term $(1 - y/K)$ is negative, so $y'$ is negative and the population decreases.
2. Solve for $C$: $2500 = 2000 / (1 + C) \implies 1 + C = 0.8 \implies C = -0.2$.
3. Use the formula $y(t) = 2000 / (1 - 0.2 e^{-0.05t})$.
4. For $t = 10$: $y(10) = 2000 / (1 - 0.2 e^{-0.5}) \approx 2000 / (1 - 0.2 \cdot 0.6065) \approx 2000 / 0.8787 \approx 2275$.

Answer: Decreases; $y(10) \approx 2275$

1. Use $y(t) = 1000 / (1 + C e^{-rt})$.
2. At $t=1$: $100 = 1000 / (1 + C e^{-r}) \implies 1 + C e^{-r} = 10 \implies C e^{-r} = 9$.
3. At $t=2$: $400 = 1000 / (1 + C e^{-2r}) \implies 1 + C e^{-2r} = 2.5 \implies C e^{-2r} = 1.5$.
4. Divide the two equations: $(C e^{-r}) / (C e^{-2r}) = 9 / 1.5 \implies e^r = 6$.
5. Solve for $r$: $r = \ln(6) \approx 1.79$.

Answer: $r = \ln(6) \approx 1.79$

1. Set $0.5y (1 - y/800) - 50 = 0$.
2. Multiply by $2$ to simplify: $y (1 - y/800) - 100 = 0$.
3. Expand: $y - y^2/800 - 100 = 0$.
4. Multiply by $-800$: $y^2 - 800y + 80000 = 0$.
5. Use quadratic formula: $y = [800 \pm \sqrt{640000 - 320000}] / 2 = [800 \pm \sqrt{320000}] / 2$.
6. Calculate: $y = 400 \pm 200\sqrt{2}$, so $y \approx 682.8$ and $y \approx 117.2$.

Answer: Stable equilibrium at $y \approx 682.8$, collapse threshold at $y \approx 117.2$.

1. Use the formula $H_{max} = rK/4$.
2. Substitute values: $H_{max} = (0.2 \cdot 5000) / 4 = 1000 / 4 = 250$.
3. Compare $H = 260$ to $H_{max} = 250$.
4. Since $260 > 250$, the harvest rate exceeds the maximum growth rate of the population.

Answer: $H_{max} = 250$; if $H = 260$, the population will collapse regardless of the initial size.

1. Set $y' = 0$ at $y = 400$: $1(400)(1 - 400/1000) - H = 0$.
2. Calculate: $400(1 - 0.4) - H = 400(0.6) - H = 240 - H = 0$.
3. Solve for $H$: $H = 240$.
4. Check stability: The growth parabola $y(1-y/1000)$ peaks at $y=500$. Since $400 < 500$, this is the lower equilibrium point.

Answer: $H = 240$; the equilibrium is unstable (it is the collapse threshold).

1. Calculate $k = c/m = 0.0002 / 0.001 = 0.2$ s$^{-1}$.
2. Calculate $v_{term} = g/k = 9.8 / 0.2 = 49$ m/s.
3. Use the formula $v(t) = v_{term}(1 - e^{-kt})$ for an object dropped from rest.
4. Substitute the values: $v(t) = 49(1 - e^{-0.2t})$.

Answer: v_{term} = 49 \text{ m/s}, v(t) = 49(1 - e^{-0.2t})

1. The velocity for an object dropped from rest is $v(t) = v_{term}(1 - e^{-kt})$.
2. Set $v(10) = 0.9 v_{term}$, so $0.9 v_{term} = v_{term}(1 - e^{-10k})$.
3. Divide by $v_{term}$ to get $0.9 = 1 - e^{-10k}$, which simplifies to $e^{-10k} = 0.1$.
4. Take the natural logarithm of both sides: $-10k = \ln(0.1)$.
5. Solve for $k$: $k = -\ln(0.1)/10 \approx 2.303/10 = 0.2303$ s$^{-1}$.

Answer: k \approx 0.2303 \text{ s}^{-1}

1. The general solution to $v' = g - kv$ is $v(t) = v_{term} + C e^{-kt}$.
2. We are given $v(0) = 2 v_{term}$.
3. Substitute $t=0$ into the general solution: $2 v_{term} = v_{term} + C e^{0}$.
4. Solve for $C$: $C = 2 v_{term} - v_{term} = v_{term}$.
5. Substitute $C$ back into the general solution: $v(t) = v_{term} + v_{term} e^{-kt}$.
6. Factor out $v_{term}$ to get $v(t) = v_{term}(1 + e^{-kt})$.

Answer: v(t) = v_{term}(1 + e^{-kt})

1. Calculate $\tau = RC = 200 \times 0.01 = 2\text{ s}$.
2. Calculate $Q_{max} = CV = 0.01 \times 12 = 0.12\text{ C}$.
3. Use the charging formula $q(t) = 0.12(1 - e^{-t/2})$.
4. Substitute $t = 3$: $q(3) = 0.12(1 - e^{-1.5}) \approx 0.12(1 - 0.223) \approx 0.0932\text{ C}$.

Answer: $\tau = 2\text{ s}, q(3) \approx 0.0932\text{ C}$

1. Calculate $\tau = RC = 100 \times 0.05 = 5\text{ s}$.
2. Set up the discharge equation: $0.2 = 0.8 e^{-t/5}$.
3. Divide by $0.8$: $0.25 = e^{-t/5}$.
4. Take the natural log: $\ln(0.25) = -t/5$.
5. Solve for $t$: $t = -5 \ln(0.25) = 5 \ln(4) \approx 6.93\text{ s}$.

Answer: $t \approx 6.93\text{ s}$

1. Use the ratio $q(2)/q(1) = (1 - e^{-2/\tau}) / (1 - e^{-1/\tau})$.
2. Let $u = e^{-1/\tau}$. Then $0.08/0.05 = (1 - u^2) / (1 - u) = 1 + u$.
3. Solve for $u$: $1.6 = 1 + u \implies u = 0.6$.
4. Solve for $\tau$: $e^{-1/\tau} = 0.6 \implies -1/\tau = \ln(0.6) \implies \tau = -1/\ln(0.6) \approx 2.23\text{ s}$.
5. Find $C$: $C = \tau / R = 2.23 / 100 = 0.0223\text{ F}$.
6. Find $Q_{max}$: $0.05 = Q_{max}(1 - 0.6) \implies Q_{max} = 0.05 / 0.4 = 0.125\text{ C}$.
7. Find $V$: $V = Q_{max} / C = 0.125 / 0.0223 \approx 5.61\text{ V}$.

Answer: $\tau \approx 2.23\text{ s}, V \approx 5.61\text{ V}$

1. Identify constants: $B_0 = 15000$, $r = 0.06$, $P = 2000$.
2. Calculate equilibrium: $P/r = 2000 / 0.06 = 33333.33$.
3. Use formula: $B(t) = 33333.33 + (15000 - 33333.33)e^{0.06t}$.
4. Substitute $t=5$: $B(5) = 33333.33 - 18333.33 e^{0.3}$.
5. Compute: $B(5) \approx 33333.33 - 18333.33(1.3498) \approx 8586.33$.

Answer: $8,586.33$

1. Identify constants: $B_0 = 50000$, $r = 0.04$, $P = 6000$.
2. Use the payoff time formula: $T = \frac{1}{0.04} \ln \left( \frac{6000}{6000 - 0.04(50000)} \right)$.
3. Simplify denominator: $6000 - 2000 = 4000$.
4. Calculate ratio: $6000 / 4000 = 1.5$.
5. Compute $T = 25 \ln(1.5) \approx 25(0.4055) \approx 10.14$.

Answer: 10.14 years

1. Identify constants: $B_0 = 5000$, $r = 0.03$, $P = -1200$ (since it is a deposit).
2. The ODE is $B' = 0.03B + 1200$.
3. Equilibrium $P/r = -1200 / 0.03 = -40000$.
4. Use formula: $B(t) = -40000 + (5000 - (-40000))e^{0.03t} = -40000 + 45000e^{0.03t}$.
5. Substitute $t=20$: $B(20) = -40000 + 45000e^{0.6} \approx -40000 + 45000(1.8221) \approx 41994.50$.

Answer: $41,994.50$

1. Set $y^3 - y = 0$, which factors as $y(y-1)(y+1) = 0$.
2. The equilibria are $y = -1, 0, 1$.
3. For $y > 1$, $y' > 0$ (up). For $0 < y < 1$, $y' < 0$ (down). For $-1 < y < 0$, $y' > 0$ (up). For $y < -1$, $y' < 0$ (down).
4. At $y = 1$, arrows point away (source). At $y = 0$, arrows point toward (sink). At $y = -1$, arrows point away (source).

Answer: $y = 0$ is stable; $y = 1$ and $y = -1$ are unstable.

1. Find equilibria: $y(2 - y) = 0$ gives $y = 0$ and $y = 2$.
2. Check the sign for $y > 2$: if $y = 3$, $y' = 3(2 - 3) = -3$, which is negative.
3. Since the derivative is negative for $y > 2$, the solution $y(t)$ will decrease toward the equilibrium at $y = 2$.
4. Because $y = 2$ is a sink (arrows from above point down, arrows from below point up), the solution converges to 2.

Answer: $y(t) \to 2$ as $t \to \infty$.

1. Equilibria occur where $y^2 - k = 0$, so $y^2 = k$.
2. If $k > 0$, there are two distinct equilibria $y = \pm \sqrt{k}$, which are typically a sink and a source.
3. If $k < 0$, there are no real equilibria.
4. If $k = 0$, the equation is $y' = y^2$. We found in the lesson that $y = 0$ is a semi-stable equilibrium because $y^2 \ge 0$ for all $y$.

Answer: $k = 0$

1. Set $f(y) = y^3 - 4y = 0$. Factoring gives $y(y-2)(y+2) = 0$, so equilibria are $y = -2, 0, 2$.
2. Compute the derivative $f'(y) = 3y^2 - 4$.
3. At $y = -2$, $f'(-2) = 3(4) - 4 = 8 > 0$, so it is unstable.
4. At $y = 0$, $f'(0) = -4 < 0$, so it is stable.
5. At $y = 2$, $f'(2) = 3(4) - 4 = 8 > 0$, so it is unstable.

Answer: y = -2 (unstable), y = 0 (stable), y = 2 (unstable)

1. Set $y^2 - 3y + 2 = 0$. Factoring gives $(y-1)(y-2) = 0$, so equilibria are $y = 1$ and $y = 2$.
2. Compute $f'(y) = 2y - 3$.
3. At $y = 1$, $f'(1) = 2(1) - 3 = -1 < 0$, so it is stable.
4. At $y = 2$, $f'(2) = 2(2) - 3 = 1 > 0$, so it is unstable.

Answer: y = 1 (stable), y = 2 (unstable)

1. The equilibrium is at $y = 0$ since $f(0) = 0^3 = 0$.
2. Compute $f'(y) = 3y^2$. At $y = 0$, $f'(0) = 0$, so the derivative test is inconclusive.
3. Check the sign of $f(y) = y^3$. For $y > 0$, $f(y) > 0$ (flow is away from 0).
4. For $y < 0$, $f(y) < 0$ (flow is away from 0).
5. Since the flow is away from the equilibrium on both sides, it is unstable.

Answer: unstable

1. Separate the variables: $\frac{1}{y} dy = \tan(t) dt$.
2. Integrate both sides: $\ln|y| = -\ln|\cos(t)| + C$.
3. Simplify using log laws: $\ln|y| = \ln|\sec(t)| + C$.
4. Exponentiate: $y = A \sec(t)$.
5. Apply initial condition $y(0) = 5$: $5 = A \sec(0) \implies A = 5$.

Answer: $y(t) = 5 \sec(t)$

1. Rate in: $3 \text{ L/min} \times 2 \text{ g/L} = 6 \text{ g/min}$.
2. Rate out: $3 \text{ L/min} \times \frac{y}{50} \text{ g/L} = \frac{3y}{50} \text{ g/min}$.
3. ODE: $\frac{dy}{dt} = 6 - \frac{3y}{50}$.
4. Equilibrium: Set $\frac{dy}{dt} = 0 \implies 6 = \frac{3y}{50} \implies 300 = 3y \implies y = 100$.

Answer: $\frac{dy}{dt} = 6 - \frac{3y}{50}, \quad y^* = 100$

1. Equilibria are the roots of $f(y) = y(y-2)(y-4)$, which are $y=0, 2, 4$.
2. Compute $f'(y) = (y-2)(y-4) + y(y-4) + y(y-2)$.
3. At $y=0$: $f'(0) = (-2)(-4) = 8 > 0 \implies$ Unstable.
4. At $y=2$: $f'(2) = 2(2-4) = -4 < 0 \implies$ Stable.
5. At $y=4$: $f'(4) = 4(4-2) = 8 > 0 \implies$ Unstable.

Answer: $y=0 \text{ (Unstable)}, y=2 \text{ (Stable)}, y=4 \text{ (Unstable)}$

1. Identify constants: $m = 5$, $c = 3$, $k = 20$.
2. Apply Newton's second law: $m x'' = -c x' - k x + f(t)$.
3. Substitute values: $5 x'' = -3 x' - 20 x + 5\cos(2t)$.
4. Rearrange to standard form: $5 x'' + 3 x' + 20 x = 5\cos(2t)$.

Answer: 5 x'' + 3 x' + 20 x = 5\cos(2t)

1. Equation: $2 x'' = -18 x \implies x'' + 9 x = 0$.
2. First derivative: $x'(t) = 3\cos(3t)$.
3. Second derivative: $x''(t) = -9\sin(3t)$.
4. Substitute into ODE: $-9\sin(3t) + 9\sin(3t) = 0$.
5. The equation holds.

Answer: x'' + 9 x = 0; Yes, it is a solution.

1. Compare $x'' + 4 x' + 4 x = 0$ to $m x'' + c x' + k x = 0$.
2. Given $m = 1$, we see $c = 4$ and $k = 4$.
3. The system consists of a $1\text{ kg}$ mass, a damping coefficient of $4\text{ kg/s}$, and a spring constant of $4\text{ N/m}$.

Answer: m = 1\text{ kg}, c = 4\text{ kg/s}, k = 4\text{ N/m}

1. Find the derivatives: $y_1' = 2e^{2t}$ and $y_2' = -2e^{-2t}$.
2. Set up the Wronskian: $W = (e^{2t})(-2e^{-2t}) - (e^{-2t})(2e^{2t})$.
3. Simplify the terms: $W = -2(e^0) - 2(e^0) = -2 - 2 = -4$.
4. Since $W = -4$, which is never zero, the functions are linearly independent.

Answer: Linearly independent, W = -4

1. Verify $y_1$: $y_1'' = 4e^{2t}$, so $4e^{2t} - 4(e^{2t}) = 0$. Correct.
2. Verify $y_2$: $y_2'' = 4e^{-2t}$, so $4e^{-2t} - 4(e^{-2t}) = 0$. Correct.
3. Compute Wronskian: $W = (e^{2t})(-2e^{-2t}) - (e^{-2t})(2e^{2t}) = -4$.
4. Since both are solutions and $W \neq 0$, they form a fundamental set.

Answer: They form a fundamental set because both solve the ODE and $W = -4 \neq 0$.

1. Compute the Wronskian to ensure independence: $W = (\sin 3t)(-3\sin 3t) - (\cos 3t)(3\cos 3t)$.
2. Simplify: $W = -3(\sin^2 3t + \cos^2 3t) = -3(1) = -3$.
3. Since $W \neq 0$, the solutions are linearly independent.
4. The general solution is the linear combination $y(t) = c_1 \sin(3t) + c_2 \cos(3t)$.

Answer: $y(t) = c_1 \sin(3t) + c_2 \cos(3t)$

1. Write the characteristic equation: $r^2 + 7r + 10 = 0$.
2. Factor the quadratic: $(r + 2)(r + 5) = 0$.
3. Identify the roots: $r_1 = -2$ and $r_2 = -5$.
4. Construct the general solution using the linear combination of exponentials.

Answer: $y(t) = c_1 e^{-2t} + c_2 e^{-5t}$

1. Characteristic equation: $r^2 - 4r + 3 = 0$, which factors to $(r-1)(r-3)=0$.
2. General solution: $y(t) = c_1 e^t + c_2 e^{3t}$.
3. Use $y(0)=2$: $c_1 + c_2 = 2$.
4. Find $y'(t) = c_1 e^t + 3c_2 e^{3t}$ and use $y'(0)=5$: $c_1 + 3c_2 = 5$.
5. Subtract the first equation from the second: $2c_2 = 3$, so $c_2 = 1.5$.
6. Substitute back: $c_1 + 1.5 = 2$, so $c_1 = 0.5$.

Answer: $y(t) = 0.5 e^t + 1.5 e^{3t}$

1. The characteristic equation is $r^2 + kr + 4 = 0$.
2. For the roots to be distinct and real, the discriminant must be strictly positive: $b^2 - 4ac > 0$.
3. Substitute the coefficients: $k^2 - 4(1)(4) > 0$.
4. Solve the inequality: $k^2 - 16 > 0$.
5. This means $k^2 > 16$, which implies $k > 4$ or $k < -4$.

Answer: $k \text{ must be in the interval } (-\infty, -4) \cup (4, \infty)$

1. The characteristic equation is $r^2 + 5r + 6 = 0$, which factors as $(r+2)(r+3) = 0$.
2. The roots are $r_1 = -2$ and $r_2 = -3$, so the general solution is $y(t) = c_1 e^{-2t} + c_2 e^{-3t}$.
3. Using $y(0) = 2$, we get $c_1 + c_2 = 2$.
4. The derivative is $y'(t) = -2c_1 e^{-2t} - 3c_2 e^{-3t}$. Using $y'(0) = 1$, we get $-2c_1 - 3c_2 = 1$.
5. Solving the system: $2(c_1 + c_2) = 4$, so $(-2c_1 - 3c_2) + (2c_1 + 2c_2) = 1 + 4$, which gives $-c_2 = 5$, so $c_2 = -5$.
6. Then $c_1 = 2 - (-5) = 7$.

Answer: $y(t) = 7e^{-2t} - 5e^{-3t}$

1. The characteristic equation is $r^2 - 4 = 0$, so $r^2 = 4$.
2. The roots are $r = 2$ and $r = -2$.
3. The general solution is $y(t) = c_1 e^{2t} + c_2 e^{-2t}$.
4. As $t$ approaches infinity, the $e^{2t}$ term grows without bound (unless $c_1 = 0$), so the solution is unstable.

Answer: $y(t) = c_1 e^{2t} + c_2 e^{-2t}$; unstable growth

1. The characteristic equation is $r^2 + 2r = 0$, which factors as $r(r+2) = 0$.
2. The roots are $r_1 = 0$ and $r_2 = -2$.
3. The general solution is $y(t) = c_1 e^{0t} + c_2 e^{-2t} = c_1 + c_2 e^{-2t}$.
4. Using $y(0) = 1$, we get $c_1 + c_2 = 1$.
5. The derivative is $y'(t) = -2c_2 e^{-2t}$. Using $y'(0) = -1$, we get $-2c_2 = -1$, so $c_2 = 1/2$.
6. Then $c_1 = 1 - 1/2 = 1/2$.

Answer: $y(t) = \frac{1}{2} + \frac{1}{2}e^{-2t}$

1. Write the characteristic equation: $r^2 + 10r + 25 = 0$.
2. Factor the quadratic: $(r + 5)^2 = 0$.
3. Identify the repeated root: $r = -5$.
4. Apply the general form for repeated roots: $y(t) = (c_1 + c_2 t) e^{-5t}$.

Answer: $y(t) = (c_1 + c_2 t) e^{-5t}$

1. Characteristic equation: $r^2 + 4r + 4 = (r+2)^2 = 0$, so $r = -2$.
2. General solution: $y(t) = (c_1 + c_2 t) e^{-2t}$.
3. Use $y(0) = 3$: $c_1 e^0 = 3 \implies c_1 = 3$.
4. Find derivative: $y'(t) = c_2 e^{-2t} - 2(c_1 + c_2 t) e^{-2t}$.
5. Use $y'(0) = 1$: $c_2 - 2c_1 = 1 \implies c_2 - 6 = 1 \implies c_2 = 7$.
6. Final solution: $y(t) = (3 + 7t) e^{-2t}$.

Answer: $y(t) = (3 + 7t) e^{-2t}$

1. Characteristic equation: $r^2 + 2\beta r + \beta^2 = (r + \beta)^2 = 0$, so $r = -\beta$.
2. General solution: $y(t) = (c_1 + c_2 t) e^{-\beta t}$.
3. Apply $y(0) = y_0$: $c_1 e^0 = y_0 \implies c_1 = y_0$.
4. Find derivative: $y'(t) = c_2 e^{-\beta t} - \beta(c_1 + c_2 t) e^{-\beta t}$.
5. Apply $y'(0) = 0$: $c_2 - \beta c_1 = 0 \implies c_2 = \beta y_0$.
6. Substitute constants: $y(t) = (y_0 + \beta y_0 t) e^{-\beta t} = y_0(1 + \beta t) e^{-\beta t}$.

Answer: $y(t) = y_0(1 + \beta t) e^{-\beta t}$

1. Write the characteristic equation: $r^2 + 6r + 25 = 0$.
2. Use the quadratic formula: $r = \frac{-6 \pm \sqrt{36 - 100}}{2} = \frac{-6 \pm \sqrt{-64}}{2}$.
3. Simplify the roots: $r = \frac{-6 \pm 8i}{2} = -3 \pm 4i$.
4. Identify $\alpha = -3$ and $\beta = 4$.
5. Substitute into the general form: $y(t) = e^{-3t}(c_1 \cos(4t) + c_2 \sin(4t))$.

Answer: $y(t) = e^{-3t}(c_1 \cos(4t) + c_2 \sin(4t))$

1. Characteristic equation: $r^2 + 9 = 0$, so $r = \pm 3i$.
2. Here $\alpha = 0$ and $\beta = 3$, so $y(t) = c_1 \cos(3t) + c_2 \sin(3t)$.
3. Apply $y(0) = 2$: $c_1 \cos(0) + c_2 \sin(0) = 2 \implies c_1 = 2$.
4. Find the derivative: $y'(t) = -3c_1 \sin(3t) + 3c_2 \cos(3t)$.
5. Apply $y'(0) = 3$: $-3(2)(0) + 3c_2(1) = 3 \implies 3c_2 = 3 \implies c_2 = 1$.

Answer: $y(t) = 2 \cos(3t) + \sin(3t)$

1. Compute $y'(t)$: Using product rule, $y' = -e^{-t} \sin(t) + e^{-t} \cos(t)$.
2. Compute $y''(t)$: $y'' = (e^{-t} \sin(t) - e^{-t} \cos(t)) + (-e^{-t} \cos(t) - e^{-t} \sin(t)) = -2e^{-t} \cos(t)$.
3. Substitute into $y'' + 2y' + 2y$: $(-2e^{-t} \cos(t)) + 2(-e^{-t} \sin(t) + e^{-t} \cos(t)) + 2(e^{-t} \sin(t))$.
4. Simplify: $-2e^{-t} \cos(t) - 2e^{-t} \sin(t) + 2e^{-t} \cos(t) + 2e^{-t} \sin(t) = 0$.

Answer: The expression simplifies to $0$, so the solution is verified.

1. Calculate $\omega = \sqrt{k/m} = \sqrt{8/2} = 2$.
2. General solution: $y(t) = c_1 \cos(2t) + c_2 \sin(2t)$.
3. Apply $y(0)=3$: $c_1(1) + c_2(0) = 3 \implies c_1 = 3$.
4. Apply $y'(0)=0$: $y'(t) = -2c_1 \sin(2t) + 2c_2 \cos(2t) \implies 2c_2 = 0 \implies c_2 = 0$.
5. Calculate $T = 2\pi/\omega = 2\pi/2 = \pi$.

Answer: $y(t) = 3 \cos(2t), T = \pi$

1. Calculate $\omega = \sqrt{25/1} = 5$.
2. General solution: $y(t) = c_1 \cos(5t) + c_2 \sin(5t)$.
3. Apply $y(0)=0$: $c_1 = 0$.
4. Apply $y'(0)=10$: $5c_2 \cos(0) = 10 \implies c_2 = 2$.
5. Calculate $T = 2\pi/\omega = 2\pi/5$.

Answer: $y(t) = 2 \sin(5t), T = 0.4\pi$

1. Calculate $\omega = \sqrt{16/1} = 4$.
2. General solution: $y(t) = c_1 \cos(4t) + c_2 \sin(4t)$.
3. Apply $y(0)=2$: $c_1 = 2$.
4. Apply $y'(0)=4$: $y'(t) = -4c_1 \sin(4t) + 4c_2 \cos(4t) \implies 4c_2 = 4 \implies c_2 = 1$.

Answer: $y(t) = 2 \cos(4t) + \sin(4t)$

1. Divide by 2 to get $y'' + 4y' + 4y = 0$.
2. The characteristic equation is $r^2 + 4r + 4 = 0$.
3. Factor as $(r+2)^2 = 0$, giving a repeated root $r = -2$.
4. Since the discriminant $4^2 - 4(1)(4) = 0$, the system is critically damped.
5. The general solution is $y(t) = (c_1 + c_2 t) e^{-2t}$.

Answer: Critically damped; $y(t) = (c_1 + c_2 t) e^{-2t}$

1. Critical damping occurs when the discriminant $b^2 - 4mk = 0$.
2. Substitute $m=1$ and $k=9$: $b^2 - 4(1)(9) = 0$.
3. Solve for $b$: $b^2 = 36$.
4. Since $b$ must be positive, $b = 6$.

Answer: $b = 6$

1. Characteristic equation: $r^2 + 2r + 5 = 0$.
2. Roots: $r = \frac{-2 \pm \sqrt{4 - 20}}{2} = -1 \pm 2i$.
3. Discriminant is $-16 < 0$, so it is under-damped.
4. General solution: $y(t) = e^{-t}(c_1 \cos(2t) + c_2 \sin(2t))$.
5. Apply $y(0)=1$: $1 = e^0(c_1(1) + c_2(0)) \implies c_1 = 1$.
6. Derivative: $y'(t) = -e^{-t}(c_1 \cos(2t) + c_2 \sin(2t)) + e^{-t}(-2c_1 \sin(2t) + 2c_2 \cos(2t))$.
7. Apply $y'(0)=0$: $0 = -1(c_1) + 1(2c_2) \implies 0 = -1 + 2c_2 \implies c_2 = 1/2$.
8. Final solution: $y(t) = e^{-t}(\cos(2t) + \frac{1}{2}\sin(2t))$.

Answer: Under-damped; $y(t) = e^{-t}(\cos(2t) + \frac{1}{2}\sin(2t))$

1. Solve the homogeneous equation $y'' - 4y' + 3y = 0$. The characteristic equation is $r^2 - 4r + 3 = 0$, which factors to $(r-1)(r-3) = 0$, giving $y_h = c_1 e^t + c_2 e^{3t}$.
2. Guess a particular solution of the form $y_p = At + B$.
3. Compute derivatives: $y_p' = A$ and $y_p'' = 0$.
4. Substitute into the ODE: $0 - 4(A) + 3(At + B) = 6t$.
5. Group terms: $3At + (3B - 4A) = 6t$.
6. Match coefficients: $3A = 6 \rightarrow A = 2$. Then $3B - 4(2) = 0 \rightarrow 3B = 8 \rightarrow B = 8/3$.
7. Combine: $y(t) = c_1 e^t + c_2 e^{3t} + 2t + 8/3$.

Answer: $y(t) = c_1 e^t + c_2 e^{3t} + 2t + \frac{8}{3}$

1. Solve the homogeneous equation $y'' - 5y' + 6y = 0$. The characteristic equation is $r^2 - 5r + 6 = 0$, which factors to $(r-2)(r-3) = 0$, giving $y_h = c_1 e^{2t} + c_2 e^{3t}$.
2. The forcing function is $2e^{4t}$. Since $e^{4t}$ is not in $y_h$, guess $y_p = Ae^{4t}$.
3. Compute derivatives: $y_p' = 4Ae^{4t}$ and $y_p'' = 16Ae^{4t}$.
4. Substitute into the ODE: $16Ae^{4t} - 5(4Ae^{4t}) + 6(Ae^{4t}) = 2e^{4t}$.
5. Simplify: $(16 - 20 + 6)Ae^{4t} = 2e^{4t} \rightarrow 2Ae^{4t} = 2e^{4t}$.
6. Solve for A: $2A = 2 \rightarrow A = 1$.
7. Combine: $y(t) = c_1 e^{2t} + c_2 e^{3t} + e^{4t}$.

Answer: $y(t) = c_1 e^{2t} + c_2 e^{3t} + e^{4t}$

1. Solve the homogeneous equation $y'' - 3y' + 2y = 0$. The characteristic equation is $r^2 - 3r + 2 = 0$, which factors to $(r-1)(r-2) = 0$, giving $y_h = c_1 e^t + c_2 e^{2t}$.
2. The forcing function is $e^{2t}$. Since $e^{2t}$ is already in $y_h$, the guess $Ae^{2t}$ will fail. Multiply by $t$ and guess $y_p = Ate^{2t}$.
3. Compute derivatives: $y_p' = Ae^{2t} + 2Ate^{2t}$ and $y_p'' = 2Ae^{2t} + 2Ae^{2t} + 4Ate^{2t} = 4Ae^{2t} + 4Ate^{2t}$.
4. Substitute into the ODE: $(4Ae^{2t} + 4Ate^{2t}) - 3(Ae^{2t} + 2Ate^{2t}) + 2(Ate^{2t}) = e^{2t}$.
5. Simplify: $(4A - 3A)e^{2t} + (4At - 6At + 2At)e^{2t} = e^{2t} \rightarrow Ae^{2t} = e^{2t}$.
6. Solve for A: $A = 1$.
7. Combine: $y(t) = c_1 e^t + c_2 e^{2t} + te^{2t}$.

Answer: $y(t) = c_1 e^t + c_2 e^{2t} + te^{2t}$

1. The homogeneous equation $y'' + 9y = 0$ has characteristic roots $r = \pm 3i$, so $y_h = c_1 \cos(3t) + c_2 \sin(3t)$.
2. The forcing function $\cos(3t)$ is a solution to the homogeneous equation, indicating resonance.
3. The guess for the particular solution is $y_p = t(A \cos(3t) + B \sin(3t))$.
4. Computing derivatives: $y_p' = (A \cos(3t) + B \sin(3t)) + t(-3A \sin(3t) + 3B \cos(3t))$.
5. Computing $y_p''$ and substituting into $y'' + 9y = \cos(3t)$ leads to $-6A \sin(3t) + 6B \cos(3t) = \cos(3t)$.
6. Matching coefficients: $-6A = 0 \implies A = 0$ and $6B = 1 \implies B = 1/6$.

Answer: y(t) = c_1 \cos(3t) + c_2 \sin(3t) + \frac{1}{6}t \sin(3t)

1. The homogeneous equation $y'' - 4y' + 4y = 0$ has a repeated root $r = 2$, so $y_h = c_1 e^{2t} + c_2 t e^{2t}$.
2. The forcing function $t e^{2t}$ is already part of the homogeneous solution.
3. The standard guess for $t e^{2t}$ would be $(At + B)e^{2t}$.
4. Because both $e^{2t}$ and $t e^{2t}$ are in $y_h$, we must multiply the guess by $t^2$.
5. The correct guess is $y_p = t^2(At + B)e^{2t} = (At^3 + Bt^2)e^{2t}$.
6. Substituting into the ODE and solving for coefficients yields $A = 1/6$ and $B = 0$.

Answer: y_p(t) = \frac{1}{6}t^3 e^{2t}

1. Homogeneous solution: $y_h = c_1 \cos(t) + c_2 \sin(t)$.
2. For $f_1(t) = 2 \cos(t)$, resonance occurs. Guess $y_{p1} = t(A \cos(t) + B \sin(t))$. Substituting gives $y_{p1} = t \sin(t)$.
3. For $f_2(t) = e^t$, no resonance. Guess $y_{p2} = C e^t$. Substituting gives $C + C = 1 \implies C = 1/2$.
4. The total particular solution is $y_p = t \sin(t) + \frac{1}{2}e^t$.

Answer: y(t) = c_1 \cos(t) + c_2 \sin(t) + t \sin(t) + \frac{1}{2}e^t

1. The homogeneous solutions are $y_{1} = \cos(2t)$ and $y_{2} = \sin(2t)$.
2. The Wronskian is $W = \cos(2t)(2\cos(2t)) - \sin(2t)(-2\sin(2t)) = 2(\cos^2(2t) + \sin^2(2t)) = 2$.
3. Compute $u_{1}' = \frac{-\sin(2t)\sec(2t)}{2} = -\frac{1}{2}\tan(2t)$.
4. Integrate $u_{1} = \int -\frac{1}{2}\tan(2t) dt = \frac{1}{4}\ln|\cos(2t)|$.
5. Compute $u_{2}' = \frac{\cos(2t)\sec(2t)}{2} = \frac{1}{2}$.
6. Integrate $u_{2} = \int \frac{1}{2} dt = \frac{1}{2}t$.
7. Assemble $y_{p} = \frac{1}{4}\ln|\cos(2t)|\cos(2t) + \frac{1}{2}t\sin(2t)$.

Answer: y_{p}(t) = \frac{1}{4}\cos(2t)\ln|\cos(2t)| + \frac{1}{2}t\sin(2t)

1. Homogeneous solutions are $y_{1} = e^t$ and $y_{2} = e^{-t}$.
2. Wronskian $W = e^t(-e^{-t}) - e^{-t}(e^t) = -1 - 1 = -2$.
3. Compute $u_{1}' = \frac{-e^{-t}(e^t/t)}{-2} = \frac{1}{2t}$.
4. Integrate $u_{1} = \frac{1}{2}\ln(t)$.
5. Compute $u_{2}' = \frac{e^t(e^t/t)}{-2} = -\frac{e^{2t}}{2t}$.
6. The integral for $u_{2}$ cannot be expressed in elementary functions, so we leave it as $u_{2} = -\int \frac{e^{2t}}{2t} dt$.
7. Assemble $y_{p} = \frac{1}{2}e^t\ln(t) - e^{-t}\int \frac{e^{2t}}{2t} dt$.

Answer: y_{p}(t) = \frac{1}{2}e^t\ln(t) - e^{-t}\int \frac{e^{2t}}{2t} dt

1. From the lesson example, the general solution is $y(t) = c_{1}\cos(t) + c_{2}\sin(t) - \cos(t)\ln|\sec(t) + \tan(t)|$.
2. Apply $y(0) = 0$: $0 = c_{1}(1) + c_{2}(0) - (1)\ln|1 + 0| \implies c_{1} = 0$.
3. Find $y'(t) = c_{2}\cos(t) + \sin(t)\ln|\sec(t) + \tan(t)| - \cos(t)\frac{\sec(t)\tan(t) + \sec^2(t)}{\sec(t) + \tan(t)}$.
4. Simplify the fraction: $\frac{\sec(t)(\tan(t) + \sec(t))}{\sec(t) + \tan(t)} = \sec(t)$.
5. So $y'(t) = c_{2}\cos(t) + \sin(t)\ln|\sec(t) + \tan(t)| - \cos(t)\sec(t) = c_{2}\cos(t) + \sin(t)\ln|\sec(t) + \tan(t)| - 1$.
6. Apply $y'(0) = 1$: $1 = c_{2}(1) + 0 - 1 \implies c_{2} = 2$.
7. The final solution is $y(t) = 2\sin(t) - \cos(t)\ln|\sec(t) + \tan(t)|$.

Answer: y(t) = 2\sin(t) - \cos(t)\ln|\sec(t) + \tan(t)|

1. The homogeneous equation $y'' + y = 0$ has natural frequency $\omega_0 = 1$.
2. The forcing function $\cos(t)$ has frequency $\omega = 1$. Since $\omega = \omega_0$, resonance occurs.
3. Guess $y_p(t) = t(A \cos(t) + B \sin(t))$.
4. Compute $y_p' = A \cos(t) + B \sin(t) + t(-A \sin(t) + B \cos(t))$.
5. Compute $y_p'' = -2A \sin(t) + 2B \cos(t) - t(A \cos(t) + B \sin(t))$.
6. Substitute into $y'' + y = \cos(t)$: $-2A \sin(t) + 2B \cos(t) = \cos(t)$.
7. Match coefficients: $-2A = 0 \implies A = 0$ and $2B = 1 \implies B = 1/2$.

Answer: y_p(t) = \frac{1}{2}t \sin(t)

1. Natural frequency $\omega_0 = \sqrt{16} = 4$.
2. Forcing frequency $\omega = 3$. Since $3 \neq 4$, there is no resonance.
3. Guess $y_p(t) = A \cos(3t) + B \sin(3t)$.
4. Substitute into ODE: $(-9A + 16A) \cos(3t) + (-9B + 16B) \sin(3t) = 5 \sin(3t)$.
5. Simplify: $7A \cos(3t) + 7B \sin(3t) = 5 \sin(3t)$.
6. Solve: $7A = 0 \implies A = 0$ and $7B = 5 \implies B = 5/7$.

Answer: y_p(t) = \frac{5}{7} \sin(3t)

1. Homogeneous solution: $y_h(t) = C_1 \cos(5t) + C_2 \sin(5t)$.
2. Resonance occurs since $\omega = \omega_0 = 5$. Guess $y_p(t) = t(A \cos(5t) + B \sin(5t))$.
3. Substitute into ODE: $y_p'' + 25y_p = -10A \sin(5t) + 10B \cos(5t) = 20 \cos(5t)$.
4. Solve: $-10A = 0 \implies A = 0$ and $10B = 20 \implies B = 2$. So $y_p(t) = 2t \sin(5t)$.
5. General solution: $y(t) = C_1 \cos(5t) + C_2 \sin(5t) + 2t \sin(5t)$.
6. Apply $y(0) = 0$: $C_1(1) + C_2(0) + 0 = 0 \implies C_1 = 0$.
7. Apply $y'(0) = 0$: $y'(t) = 5C_2 \cos(5t) + 2 \sin(5t) + 10t \cos(5t)$. $5C_2(1) + 0 + 0 = 0 \implies C_2 = 0$.

Answer: y(t) = 2t \sin(5t)

1. Identify the frequency difference: $|\omega_{0} - \omega| = |5 - 4.8| = 0.2$ rad/s.
2. Use the beat period formula: $T_{beat} = \frac{2\pi}{|\omega_{0} - \omega|}$.
3. Compute the value: $T_{beat} = \frac{2\pi}{0.2} = 10\pi$.

Answer: 10\pi \approx 31.42 \text{ s}

1. Identify $\omega_{0}^2 = 9$ and $\omega = 2.9$.
2. The particular solution is $y_p(t) = \frac{1}{9 - 2.9^2} \sin(2.9t)$.
3. Calculate the denominator: $9 - 8.41 = 0.59$.
4. The amplitude is $A = |1 / 0.59|$.

Answer: \approx 1.695

1. Set up the amplitude equation: $10 = \frac{0.1}{4 - \omega^2}$.
2. Rearrange to solve for the denominator: $4 - \omega^2 = \frac{0.1}{10} = 0.01$.
3. Solve for $\omega^2$: $\omega^2 = 4 - 0.01 = 3.99$.
4. Take the square root: $\omega = \sqrt{3.99}$.

Answer: \approx 1.997 \text{ rad/s}

1. The equation is $L q'' + R q' + \frac{1}{C} q = 0$.
2. Substitute values: $2 q'' + 10 q' + \frac{1}{0.1} q = 0$, which simplifies to $2 q'' + 10 q' + 10 q = 0$.
3. The characteristic equation is $2r^2 + 10r + 10 = 0$, or $r^2 + 5r + 5 = 0$.
4. The discriminant is $D = 5^2 - 4(1)(5) = 25 - 20 = 5$.
5. Since $D > 0$, the roots are real and distinct, meaning the system is overdamped.

Answer: 2q'' + 10q' + 10q = 0; overdamped

1. Use the formula $\omega_0 = \frac{1}{\sqrt{LC}}$.
2. Calculate $LC = 0.5 \times 2 \times 10^{-6} = 1 \times 10^{-6}$.
3. Take the square root: $\sqrt{10^{-6}} = 10^{-3}$.
4. Compute the reciprocal: $\omega_0 = \frac{1}{10^{-3}} = 1000$.

Answer: 1000 \text{ rad/s}

1. For critical damping, the discriminant of $Lr^2 + Rr + \frac{1}{C} = 0$ must be zero.
2. This requires $R^2 - 4L\frac{1}{C} = 0$.
3. Substitute $L = 1$ and $C = 0.25$: $R^2 - 4(1)(\frac{1}{0.25}) = 0$.
4. Simplify: $R^2 - 4(4) = 0 \implies R^2 = 16$.
5. Since resistance must be positive, $R = 4$.

Answer: 4 \Omega

1. Write the characteristic equation: $r^3 - 5r^2 + 8r - 4 = 0$.
2. Test for roots: $r=1$ gives $1 - 5 + 8 - 4 = 0$, so $(r-1)$ is a factor.
3. Divide to get $(r-1)(r^2 - 4r + 4) = 0$, which is $(r-1)(r-2)^2 = 0$.
4. The roots are $r=1$ (distinct) and $r=2$ (multiplicity 2).
5. Construct the general solution using the repeated root rule.

Answer: $y(t) = C_1 e^t + (C_2 + C_3 t) e^{2t}$

1. The characteristic equation is $r^3 + 3r^2 + 3r + 1 = 0$, which is $(r+1)^3 = 0$.
2. The root $r=-1$ has multiplicity 3, so $y(t) = (C_1 + C_2 t + C_3 t^2) e^{-t}$.
3. Use $y(0)=1$ implies $C_1 = 1$.
4. Differentiate: $y'(t) = (C_2 + 2C_3 t) e^{-t} - (C_1 + C_2 t + C_3 t^2) e^{-t}$.
5. Use $y'(0)=0$ implies $C_2 - C_1 = 0$ so $C_2 = 1$.
6. Differentiate again and use $y''(0)=0$ implies $2C_3 - 2C_2 + C_1 = 0$ so $2C_3 - 2 + 1 = 0$ so $C_3 = 1/2$.

Answer: $y(t) = (1 + t + \frac{1}{2}t^2) e^{-t}$

1. The characteristic equation is $r^4 + 2r^2 + 1 = 0$.
2. This is a perfect square: $(r^2 + 1)^2 = 0$.
3. The roots are $r = i$ and $r = -i$, each with multiplicity 2.
4. For the first pair of complex roots, we have $\cos(t)$ and $\sin(t)$.
5. Because they are repeated, we multiply by $t$ for the second pair: $t \cos(t)$ and $t \sin(t)$.

Answer: $y(t) = C_1 \cos(t) + C_2 \sin(t) + C_3 t \cos(t) + C_4 t \sin(t)$

1. Solve the homogeneous equation $y'' - 3y' + 2y = 0$. The characteristic equation is $r^{2} - 3r + 2 = 0$, which factors to $(r-1)(r-2)=0$. Thus $y_{h} = c_{1}e^{t} + c_{2}e^{2t}$.
2. Guess a particular solution of the form $y_{p} = At^{2} + Bt + C$.
3. Compute derivatives: $y_{p}' = 2At + B$ and $y_{p}'' = 2A$.
4. Substitute into the ODE: $2A - 3(2At + B) + 2(At^{2} + Bt + C) = 4t^{2}$.
5. Group terms: $2At^{2} + (2B - 6A)t + (2A - 3B + 2C) = 4t^{2}$.
6. Solve the system: $2A = 4 \implies A = 2$; $2B - 6(2) = 0 \implies B = 6$; $2(2) - 3(6) + 2C = 0 \implies 2C = 14 \implies C = 7$.
7. Combine: $y = c_{1}e^{t} + c_{2}e^{2t} + 2t^{2} + 6t + 7$.

Answer: $y = c_{1}e^{t} + c_{2}e^{2t} + 2t^{2} + 6t + 7$

1. The characteristic equation is $r^{2} + 4 = 0$, so $r = \pm 2i$.
2. The general solution is $y = c_{1}\cos(2t) + c_{2}\sin(2t)$.
3. Apply $y(0) = 3$: $c_{1}\cos(0) + c_{2}\sin(0) = 3 \implies c_{1} = 3$.
4. Compute $y' = -2c_{1}\sin(2t) + 2c_{2}\cos(2t)$.
5. Apply $y'(0) = 2$: $-2(3)\sin(0) + 2c_{2}\cos(0) = 2 \implies 2c_{2} = 2 \implies c_{2} = 1$.
6. The solution is $y = 3\cos(2t) + \sin(2t)$.

Answer: $y = 3\cos(2t) + \sin(2t)$

1. Homogeneous solutions are $y_{1} = \cos(t)$ and $y_{2} = \sin(t)$.
2. The Wronskian is $W = \cos(t)\cos(t) - \sin(t)(-\sin(t)) = 1$.
3. The particular solution is $y_{p} = -\cos(t) \int \sin(t)\sec(t) dt + \sin(t) \int \cos(t)\sec(t) dt$.
4. Simplify the integrals: $\int \sin(t)\sec(t) dt = \int \tan(t) dt = -\ln|\cos(t)|$.
5. And $\int \cos(t)\sec(t) dt = \int 1 dt = t$.
6. Combine: $y_{p} = -\cos(t)(-\ln|\cos(t)|) + \sin(t)(t) = \cos(t)\ln|\cos(t)| + t\sin(t)$.

Answer: $y_{p} = \cos(t)\ln|\cos(t)| + t\sin(t)$

1. Use linearity: $\mathcal{L}\{3 e^{-4t} + 2t\} = 3 \mathcal{L}\{e^{-4t}\} + 2 \mathcal{L}\{t\}$.
2. Apply the exponential rule with $a = -4$: $\mathcal{L}\{e^{-4t}\} = \frac{1}{s - (-4)} = \frac{1}{s+4}$.
3. Apply the power rule: $\mathcal{L}\{t\} = \frac{1}{s^2}$.
4. Combine the results: $3(\frac{1}{s+4}) + 2(\frac{1}{s^2})$.

Answer: $\frac{3}{s+4} + \frac{2}{s^2}$

1. Recall the derivative formula: $\mathcal{L}\{f'(t)\} = s F(s) - f(0)$.
2. Substitute the given initial value $f(0) = 5$ into the formula.
3. The expression becomes $s F(s) - 5$.

Answer: $s F(s) - 5$

1. Set up the integral: $\int_{0}^{\infty} e^{-st} (t e^{2t}) dt = \int_{0}^{\infty} t e^{-(s-2)t} dt$.
2. Use integration by parts with $u = t$ and $dv = e^{-(s-2)t} dt$.
3. Then $du = dt$ and $v = \frac{-1}{s-2} e^{-(s-2)t}$.
4. The boundary term $[uv]_0^{\infty}$ is $0$ for $s > 2$.
5. The remaining integral is $\int_{0}^{\infty} \frac{1}{s-2} e^{-(s-2)t} dt$.
6. Integrating again gives $\frac{1}{s-2} [\frac{-1}{s-2} e^{-(s-2)t}]_0^{\infty} = \frac{1}{(s-2)^2}$.

Answer: $\frac{1}{(s-2)^2}$

1. Apply linearity: $\mathcal{L}\{3e^{2t} + 4\sin(3t)\} = 3\mathcal{L}\{e^{2t}\} + 4\mathcal{L}\{\sin(3t)\}$.
2. Use the table for $e^{at}$ with $a=2$: $\mathcal{L}\{e^{2t}\} = \frac{1}{s-2}$.
3. Use the table for $\sin(at)$ with $a=3$: $\mathcal{L}\{\sin(3t)\} = \frac{3}{s^2+9}$.
4. Combine the results: $3\frac{1}{s-2} + 4\frac{3}{s^2+9}$.

Answer: \frac{3}{s-2} + \frac{12}{s^2+9}

1. Transform both sides: $\mathcal{L}\{y'\} - 2\mathcal{L}\{y\} = \mathcal{L}\{0\}$.
2. Substitute the derivative rule: $(sY - y(0)) - 2Y = 0$.
3. Plug in the initial condition $y(0)=5$: $sY - 5 - 2Y = 0$.
4. Factor out $Y$: $Y(s-2) = 5$.
5. Solve for $Y$: $Y(s) = \frac{5}{s-2}$.

Answer: Y(s) = \frac{5}{s-2}

1. Transform the equation: $\mathcal{L}\{y''\} + 4\mathcal{L}\{y\} = 0$.
2. Apply the second derivative rule: $(s^2 Y - sy(0) - y'(0)) + 4Y = 0$.
3. Substitute $y(0)=0$ and $y'(0)=2$: $s^2 Y - s(0) - 2 + 4Y = 0$.
4. Simplify: $s^2 Y + 4Y = 2$.
5. Factor $Y$: $Y(s^2 + 4) = 2$.
6. Solve for $Y$: $Y(s) = \frac{2}{s^2+4}$.

Answer: Y(s) = \frac{2}{s^2+4}

1. Factor the denominator: $s^2 - 9 = (s-3)(s+3)$.
2. Set up partial fractions: $\frac{2}{(s-3)(s+3)} = \frac{A}{s-3} + \frac{B}{s+3}$.
3. Solve for constants: $2 = A(s+3) + B(s-3)$. For $s=3$, $2=6A \implies A=1/3$. For $s=-3$, $2=-6B \implies B=-1/3$.
4. Rewrite $Y(s) = \frac{1/3}{s-3} - \frac{1/3}{s+3}$.
5. Apply inverse transforms: $y(t) = \frac{1}{3} e^{3t} - \frac{1}{3} e^{-3t}$.

Answer: y(t) = \frac{1}{3}(e^{3t} - e^{-3t})

1. Set up partial fractions for repeated root: $\frac{s+5}{(s-1)^2} = \frac{A}{s-1} + \frac{B}{(s-1)^2}$.
2. Solve for constants: $s+5 = A(s-1) + B$.
3. For $s=1$, $6 = B$.
4. Comparing coefficients of $s$: $1 = A$.
5. Rewrite $Y(s) = \frac{1}{s-1} + \frac{6}{(s-1)^2}$.
6. Apply inverse transforms: $y(t) = e^t + 6t e^t$.

Answer: y(t) = (1 + 6t)e^t

1. Set up partial fractions: $\frac{1}{(s-2)(s^2+1)} = \frac{A}{s-2} + \frac{Bs+C}{s^2+1}$.
2. Solve for constants: $1 = A(s^2+1) + (Bs+C)(s-2)$.
3. For $s=2$, $1 = 5A \implies A=1/5$.
4. For $s=0$, $1 = A - 2C \implies 1 = 1/5 - 2C \implies 2C = -4/5 \implies C=-2/5$.
5. Comparing $s^2$ coefficients: $0 = A + B \implies B = -1/5$.
6. Rewrite $Y(s) = \frac{1/5}{s-2} + \frac{-1/5 s - 2/5}{s^2+1}$.
7. Split the quadratic term: $Y(s) = \frac{1}{5} \frac{1}{s-2} - \frac{1}{5} \frac{s}{s^2+1} - \frac{2}{5} \frac{1}{s^2+1}$.
8. Apply inverse transforms: $y(t) = \frac{1}{5} e^{2t} - \frac{1}{5} \cos(t) - \frac{2}{5} \sin(t)$.

Answer: y(t) = \frac{1}{5}(e^{2t} - \cos(t) - 2\sin(t))

1. Transform the equation: $s Y(s) - y(0) + 3 Y(s) = 0$.
2. Substitute $y(0) = 5$: $s Y(s) - 5 + 3 Y(s) = 0$.
3. Solve for $Y(s)$: $(s + 3) Y(s) = 5$, so $Y(s) = \frac{5}{s+3}$.
4. Invert using the table: $\mathcal{L}^{-1} \{ \frac{5}{s+3} \} = 5 e^{-3t}$.

Answer: y(t) = 5 e^{-3t}

1. Transform the equation: $s Y(s) - y(0) - 2 Y(s) = \frac{4}{s}$.
2. Substitute $y(0) = 1$: $(s - 2) Y(s) - 1 = \frac{4}{s}$.
3. Isolate $Y(s)$: $(s - 2) Y(s) = \frac{4}{s} + 1 = \frac{s+4}{s}$, so $Y(s) = \frac{s+4}{s(s-2)}$.
4. Partial fractions: $\frac{s+4}{s(s-2)} = \frac{A}{s} + \frac{B}{s-2}$. Solving gives $A = -2$ and $B = 3$.
5. Invert: $y(t) = -2(1) + 3 e^{2t}$.

Answer: y(t) = 3 e^{2t} - 2

1. Transform the equation: $s^2 Y(s) - s y(0) - y'(0) + 4 Y(s) = 0$.
2. Substitute $y(0) = 0$ and $y'(0) = 2$: $s^2 Y(s) - 2 + 4 Y(s) = 0$.
3. Solve for $Y(s)$: $(s^2 + 4) Y(s) = 2$, so $Y(s) = \frac{2}{s^2 + 4}$.
4. Recognize the form $\frac{a}{s^2 + a^2}$ where $a = 2$.
5. Invert: $y(t) = \sin(2t)$.

Answer: y(t) = \sin(2t)

1. Express the function using the Heaviside step function: $f(t) = u(t-3) e^{2t}$.
2. To use the second shifting theorem, we need the form $u(t-a)g(t-a)$. We rewrite $e^{2t}$ as $e^{2(t-3+3)} = e^{2(t-3)} e^6$.
3. Now we have $f(t) = e^6 u(t-3) e^{2(t-3)}$.
4. The transform of $e^{2t}$ is $1/(s-2)$.
5. Applying the theorem: $\mathcal{L}\{f(t)\} = e^6 e^{-3s} \frac{1}{s-2}$.

Answer: \frac{e^{6-3s}}{s-2}

1. Transform the equation: $(s+2)Y(s) = e^{-s}/s$.
2. Solve for $Y(s)$: $Y(s) = e^{-s} \frac{1}{s(s+2)}$.
3. Partial fraction decomposition of $G(s) = \frac{1}{s(s+2)} = \frac{1}{2}(\frac{1}{s} - \frac{1}{s+2})$.
4. Inverse transform of $G(s)$ is $g(t) = \frac{1}{2}(1 - e^{-2t})$.
5. Apply the shift: $y(t) = u(t-1) g(t-1) = \frac{1}{2} u(t-1) (1 - e^{-2(t-1)})$.

Answer: \frac{1}{2} u(t-1) (1 - e^{-2(t-1)})

1. Identify $G(s) = \frac{1}{s^2 + 9}$.
2. The inverse transform of $G(s)$ is $g(t) = \frac{1}{3} \sin(3t)$.
3. The factor $e^{-2s}$ indicates a shift of $a=2$.
4. Apply the second shifting theorem: $h(t) = u(t-2) g(t-2)$.
5. Substitute $t-2$ into $g(t)$: $h(t) = \frac{1}{3} u(t-2) \sin(3(t-2))$.

Answer: \frac{1}{3} u(t-2) \sin(3(t-2))

1. Identify the base function $f_0(t) = \cos(2t)$.
2. Find its transform $F_0(s) = \frac{s}{s^2 + 4}$.
3. Apply the first shifting theorem with $a = -5$, replacing $s$ with $s - (-5) = s + 5$.
4. The result is $F(s) = \frac{s+5}{(s+5)^2 + 4}$.

Answer: \frac{s+5}{(s+5)^2 + 4}

1. Identify the part without the exponential: $H(s) = \frac{1}{s-2}$.
2. Find the inverse transform of $H(s)$, which is $h(t) = e^{2t}$.
3. Apply the second shifting theorem for $a = 3$.
4. The result is $g(t) = u(t-3)h(t-3) = u(t-3)e^{2(t-3)}$.

Answer: u(t-3)e^{2(t-3)}

1. Identify the base function $f_0(t) = e^t$.
2. Find its transform $F_0(s) = \frac{1}{s-1}$.
3. The function is given as $u(t-1)f_0(t-1)$, which fits the second shifting theorem exactly with $a = 1$.
4. The transform is $e^{-1s}F_0(s) = \frac{e^{-s}}{s-1}$.

Answer: \frac{e^{-s}}{s-1}

1. Transform the equation: $(s^2 + 16)Y(s) = e^{-3s}$.
2. Solve for $Y(s)$: $Y(s) = e^{-3s} \frac{1}{s^2 + 16}$.
3. Identify the inverse transform of $\frac{1}{s^2 + 16}$ as $\frac{1}{4}\sin(4t)$.
4. Apply the second shifting theorem to account for $e^{-3s}$.

Answer: y(t) = \frac{1}{4}\sin(4(t-3))u(t-3)

1. Transform: $(s^2 + 2s + 2)Y(s) = e^{-2s}$.
2. Complete the square: $Y(s) = \frac{e^{-2s}}{(s+1)^2 + 1}$.
3. The inverse transform of $\frac{1}{(s+1)^2 + 1}$ is $e^{-t}\sin(t)$.
4. Apply the shift $t \to t-2$ and multiply by $u(t-2)$.

Answer: y(t) = e^{-(t-2)}\sin(t-2)u(t-2)

1. Transform: $(s^2 + 25)Y(s) = 2e^{-s}$.
2. Solve for $Y(s)$: $Y(s) = 2e^{-s} \frac{1}{s^2 + 25}$.
3. The inverse transform of $\frac{1}{s^2 + 25}$ is $\frac{1}{5}\sin(5t)$.
4. Multiply by the constant 2 and apply the shift $t \to t-1$.

Answer: y(t) = \frac{2}{5}\sin(5(t-1))u(t-1)

1. Take the Laplace transform: $s^2 Y + 4s Y + 3Y = 1$.
2. Solve for $Y(s)$: $H(s) = \frac{1}{s^2 + 4s + 3}$.
3. Factor the denominator: $H(s) = \frac{1}{(s+1)(s+3)}$.
4. Use partial fractions: $\frac{1}{(s+1)(s+3)} = \frac{1}{2} \frac{1}{s+1} - \frac{1}{2} \frac{1}{s+3}$.
5. Invert the transform: $h(t) = \frac{1}{2}(e^{-t} - e^{-3t})u(t)$.

Answer: h(t) = \frac{1}{2}(e^{-t} - e^{-3t})u(t)

1. The impulse response is the inverse Laplace transform of the transfer function: $h(t) = \mathcal{L}^{-1}\{\frac{s}{s^2 + 16}\}$.
2. Recognize the form $\frac{s}{s^2 + \omega^2}$ as the transform of $\cos(\omega t)$.
3. Here $\omega = 4$.
4. Therefore, $h(t) = \cos(4t)u(t)$.

Answer: h(t) = \cos(4t)u(t)

1. Find $h(t)$: $sY + 2Y = 1 \implies H(s) = \frac{1}{s+2} \implies h(t) = e^{-2t}u(t)$.
2. For input $f(t) = u(t)$, the transform is $F(s) = \frac{1}{s}$.
3. The output is $Y(s) = H(s)F(s) = \frac{1}{s(s+2)}$.
4. Partial fractions: $\frac{1}{s(s+2)} = \frac{1}{2} \frac{1}{s} - \frac{1}{2} \frac{1}{s+2}$.
5. Invert: $y(t) = \frac{1}{2}(1 - e^{-2t})u(t)$.

Answer: h(t) = e^{-2t}u(t), y(t) = \frac{1}{2}(1 - e^{-2t})u(t)

1. Set up the integral: $y(t) = \int_{0}^{t} e^{-\tau} e^{-3(t-\tau)} d\tau$.
2. Simplify the exponent: $y(t) = \int_{0}^{t} e^{-\tau} e^{-3t} e^{3\tau} d\tau = e^{-3t} \int_{0}^{t} e^{2\tau} d\tau$.
3. Integrate: $y(t) = e^{-3t} [\frac{1}{2} e^{2\tau}]_{0}^{t}$.
4. Evaluate limits: $y(t) = e^{-3t} (\frac{1}{2} e^{2t} - \frac{1}{2})$.
5. Distribute: $y(t) = \frac{1}{2} e^{-t} - \frac{1}{2} e^{-3t}$.

Answer: y(t) = \frac{1}{2}(e^{-t} - e^{-3t})

1. Set up the integral: $y(t) = \int_{0}^{t} 5 \cdot (t - \tau) d\tau$.
2. Split the integral: $y(t) = 5 \int_{0}^{t} t d\tau - 5 \int_{0}^{t} \tau d\tau$.
3. Integrate with respect to $\tau$: $y(t) = 5 [t\tau]_{0}^{t} - 5 [\frac{1}{2}\tau^2]_{0}^{t}$.
4. Evaluate: $y(t) = 5t^2 - \frac{5}{2}t^2$.
5. Simplify: $y(t) = \frac{5}{2}t^2$.

Answer: y(t) = \frac{5}{2}t^2

1. Set up the integral: $y(t) = \int_{0}^{t} \sin(\tau) \cdot 1 d\tau$.
2. Integrate: $y(t) = [-\cos(\tau)]_{0}^{t}$.
3. Evaluate: $y(t) = -\cos(t) - (-\cos(0))$.
4. Simplify: $y(t) = 1 - \cos(t)$.
5. Interpretation: The output is the running integral (accumulated area) of the input signal.

Answer: y(t) = 1 - \cos(t)

1. Identify $F(s) = \frac{1}{s-2}$ and $G(s) = \frac{1}{s-3}$, so $f(t) = e^{2t}$ and $g(t) = e^{3t}$.
2. Set up the convolution integral: $y(t) = \int_{0}^{t} e^{2\tau} e^{3(t-\tau)} d\tau$.
3. Simplify the integrand: $y(t) = e^{3t} \int_{0}^{t} e^{2\tau} e^{-3\tau} d\tau = e^{3t} \int_{0}^{t} e^{-\tau} d\tau$.
4. Integrate: $y(t) = e^{3t} [ -e^{-\tau} ]_{0}^{t} = e^{3t} (-e^{-t} + 1)$.
5. Distribute: $y(t) = -e^{2t} + e^{3t}$.

Answer: y(t) = e^{3t} - e^{2t}

1. Identify $F(s) = \frac{1}{s^2}$ and $G(s) = \frac{1}{s-1}$, so $f(t) = t$ and $g(t) = e^t$.
2. Set up the integral: $y(t) = \int_{0}^{t} \tau e^{t-\tau} d\tau$.
3. Pull out $e^t$: $y(t) = e^t \int_{0}^{t} \tau e^{-\tau} d\tau$.
4. Use integration by parts on $\int \tau e^{-\tau} d\tau$: $u=\tau, dv=e^{-\tau}d\tau \implies du=d\tau, v=-e^{-\tau}$.
5. The integral is $[-\tau e^{-\tau}]_0^t + \int_0^t e^{-\tau} d\tau = -te^{-t} + [-e^{-\tau}]_0^t = -te^{-t} - e^{-t} + 1$.
6. Multiply by $e^t$: $y(t) = e^t(1 - e^{-t} - te^{-t}) = e^t - 1 - t$.

Answer: y(t) = e^t - t - 1

1. The output in s-domain is $Y(s) = H(s)F(s) = \frac{1}{s^2+4} \cdot \frac{2}{s^2+4} = \frac{2}{(s^2+4)^2}$.
2. Identify $h(t) = \frac{1}{2}\sin(2t)$ and $f(t) = \sin(2t)$.
3. Set up convolution: $y(t) = \int_{0}^{t} \frac{1}{2}\sin(2\tau) \sin(2(t-\tau)) d\tau$.
4. Use the product-to-sum identity $\sin(A)\sin(B) = \frac{1}{2}(\cos(A-B) - \cos(A+B))$.
5. Here $A=2\tau$ and $B=2t-2\tau$, so $A-B = 4\tau-2t$ and $A+B = 2t$.
6. The integral becomes $\frac{1}{4} \int_{0}^{t} (\cos(4\tau-2t) - \cos(2t)) d\tau$.
7. Integrate: $\frac{1}{4} [ \frac{1}{4}\sin(4\tau-2t) - \tau\cos(2t) ]_{0}^{t}$.
8. Evaluate: $\frac{1}{4} ( \frac{1}{4}\sin(2t) - t\cos(2t) - (\frac{1}{4}\sin(-2t) - 0) )$.
9. Since $\sin(-2t) = -\sin(2t)$, we get $\frac{1}{4} ( \frac{1}{2}\sin(2t) - t\cos(2t) )$.

Answer: y(t) = \frac{1}{8}\sin(2t) - \frac{1}{4}t\cos(2t)

1. Take the Laplace transform: $s^2 Y(s) + 4s Y(s) + 13 Y(s) = X(s)$.
2. Factor out $Y(s)$: $(s^2 + 4s + 13) Y(s) = X(s)$.
3. Form the ratio: $H(s) = \frac{1}{s^2 + 4s + 13}$.
4. Find poles using the quadratic formula: $s = \frac{-4 \pm \sqrt{16 - 52}}{2} = -2 \pm 3i$.
5. Since the real part is $-2$ (negative), the system is stable and exhibits decaying oscillations.

Answer: H(s) = \frac{1}{s^2 + 4s + 13}, \text{ poles at } s = -2 \pm 3i, \text{ stable decaying oscillation.}

1. Input transform is $X(s) = \frac{1}{s}$.
2. Output transform is $Y(s) = \frac{s+2}{s(s-1)(s+3)}$.
3. Partial fraction decomposition: $\frac{s+2}{s(s-1)(s+3)} = \frac{A}{s} + \frac{B}{s-1} + \frac{C}{s+3}$.
4. Solving for constants: $A = \frac{2}{(-1)(3)} = -\frac{2}{3}$, $B = \frac{3}{(1)(4)} = \frac{3}{4}$, $C = \frac{-1}{(-3)(-4)} = -\frac{1}{12}$.
5. Inverse transform: $y(t) = -\frac{2}{3} + \frac{3}{4}e^t - \frac{1}{12}e^{-3t}$.

Answer: y(t) = -\frac{2}{3} + \frac{3}{4}e^t - \frac{1}{12}e^{-3t}

1. A pole at $s = -5$ gives a term $e^{-5t}$, which decays to zero.
2. A pole at $s = -1$ gives a term $e^{-t}$, which also decays to zero.
3. A pole at $s = 0$ gives a term $e^{0t} = 1$, which is a constant.
4. In the first system, the output does not return to zero but stays at a constant value (the steady-state response).
5. In the second system, the output returns completely to zero. Thus, the first is marginally stable.

Answer: The pole at $s=0$ creates a constant non-decaying term, preventing the system from returning to the zero equilibrium.

1. Transform the equation: $(s+3)Y(s) = 4/s$.
2. Solve for $Y(s)$: $Y(s) = 4 / (s(s+3))$.
3. Partial fraction decomposition: $Y(s) = (4/3)/s - (4/3)/(s+3)$.
4. Inverse transform: $y(t) = 4/3 - (4/3)e^{-3t}$.

Answer: y(t) = \frac{4}{3}(1 - e^{-3t})

1. Transform the equation: $(s^2 + 4)Y(s) = e^{-2s}$.
2. Solve for $Y(s)$: $Y(s) = e^{-2s} / (s^2 + 4)$.
3. Identify the base transform: $\mathcal{L}^{-1} \{ 1/(s^2+4) \} = 0.5 \sin(2t)$.
4. Apply the t-shift: $y(t) = 0.5 \sin(2(t-2)) u(t-2)$.

Answer: y(t) = 0.5 \sin(2(t-2)) u(t-2)

1. Transform the equation: $(s^2 - 4s + 4)Y(s) - sy(0) - y'(0) + 4y(0) = 0$.
2. Substitute ICs: $(s-2)^2 Y(s) - s - 2 + 4 = 0$.
3. Solve for $Y(s)$: $Y(s) = (s-2) / (s-2)^2 = 1/(s-2)$.
4. Inverse transform: $y(t) = e^{2t}$.

Answer: y(t) = e^{2t}

1. Set $x(2 - x - y) = 0$. This implies $x = 0$ or $x + y = 2$.
2. Set $y(3 - 2x - y) = 0$. This implies $y = 0$ or $2x + y = 3$.
3. Case 1: $x = 0$ and $y = 0$. Point $(0, 0)$.
4. Case 2: $x = 0$ and $2x + y = 3 \implies y = 3$. Point $(0, 3)$.
5. Case 3: $x + y = 2$ and $y = 0 \implies x = 2$. Point $(2, 0)$.
6. Case 4: $x + y = 2$ and $2x + y = 3$. Subtracting the first from the second gives $x = 1$, then $y = 1$. Point $(1, 1)$.

Answer: (0, 0), (0, 3), (2, 0), (1, 1)

1. Identify the coefficients of $x$ and $y$ for the first equation: $-1$ and $2$.
2. Identify the coefficients of $x$ and $y$ for the second equation: $3$ and $-2$.
3. Place these into a $2 \times 2$ matrix $A$.

Answer: \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -1 & 2 \\ 3 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}

1. Check equilibrium: $f(0, 0) = 0$ and $g(0, 0) = -0 = 0$. Yes, it is an equilibrium.
2. Look at the velocity vector: $\vec{v} = (y, -x)$.
3. At $(1, 0)$, $\vec{v} = (0, -1)$ (down). At $(0, 1)$, $\vec{v} = (1, 0)$ (right). At $(-1, 0)$, $\vec{v} = (0, 1)$ (up). At $(0, -1)$, $\vec{v} = (-1, 0)$ (left).
4. The vectors point in a clockwise circle around the origin.

Answer: The point $(0, 0)$ is an equilibrium; trajectories are clockwise circles.

1. Find eigenvalues: $\det(A - \lambda I) = (1-\lambda)(2-\lambda) - 6 = \lambda^2 - 3\lambda - 4 = 0$.
2. Solve for $\lambda$: $(\lambda-4)(\lambda+1) = 0$, so $\lambda_1 = 4, \lambda_2 = -1$.
3. For $\lambda=4$: $\begin{pmatrix} -3 & 2 \\ 3 & -2 \end{pmatrix} \mathbf{v} = 0 \implies 3v_1 = 2v_2$, so $\mathbf{v}_1 = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$.
4. For $\lambda=-1$: $\begin{pmatrix} 2 & 2 \\ 3 & 3 \end{pmatrix} \mathbf{v} = 0 \implies v_1 = -v_2$, so $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.

Answer: $\mathbf{x}(t) = c_1 e^{4t} \begin{pmatrix} 2 \\ 3 \end{pmatrix} + c_2 e^{-t} \begin{pmatrix} 1 \\ -1 \end{pmatrix}$

1. Characteristic equation: $(2-\lambda)^2 - 1 = 0$, so $\lambda^2 - 4\lambda + 3 = 0$.
2. Eigenvalues: $(\lambda-3)(\lambda-1) = 0$, so $\lambda_1 = 3, \lambda_2 = 1$.
3. For $\lambda=3$: $\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix} \mathbf{v} = 0 \implies \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$.
4. For $\lambda=1$: $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \mathbf{v} = 0 \implies \mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.
5. Since both eigenvalues are positive, the origin is an unstable source.

Answer: $\lambda_1=3, \mathbf{v}_1=\begin{pmatrix} 1 \\ 1 \end{pmatrix}; \lambda_2=1, \mathbf{v}_2=\begin{pmatrix} 1 \\ -1 \end{pmatrix}; \text{Unstable Source}$

1. From the lesson, eigenvalues are $4$ and $-2$ with eigenvectors $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$.
2. General solution: $\mathbf{x}(t) = c_1 e^{4t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2 e^{-2t} \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.
3. Initial conditions: $\begin{pmatrix} 1 \\ 1 \end{pmatrix} = c_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2 \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.
4. This gives $c_1 + c_2 = 1$ and $c_1 - c_2 = 1$. Adding them gives $2c_1 = 2 \implies c_1 = 1$, so $c_2 = 0$.

Answer: $x(t) = e^{4t}, y(t) = e^{4t}$

1. Find eigenvalues: $\det(A - \lambda I) = \lambda^2 - 3\lambda + 2 = 0$, so $\lambda_1 = 1, \lambda_2 = 2$.
2. Find eigenvector for $\lambda = 1$: $\begin{pmatrix} -1 & 1 \\ -2 & 2 \end{pmatrix} \mathbf{v} = 0 \implies \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$.
3. Find eigenvector for $\lambda = 2$: $\begin{pmatrix} -2 & 1 \\ -2 & 1 \end{pmatrix} \mathbf{v} = 0 \implies \mathbf{v}_2 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$.
4. Assemble general solution: $\mathbf{x}(t) = c_1 e^t \begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2 e^{2t} \begin{pmatrix} 1 \\ 2 \end{pmatrix}$.

Answer: \mathbf{x}(t) = c_1 e^t \begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2 e^{2t} \begin{pmatrix} 1 \\ 2 \end{pmatrix}

1. Eigenvalues: $(1-\lambda)^2 - 4 = 0 \implies \lambda = 3, -1$.
2. Eigenvectors: For $\lambda=3, \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$. For $\lambda=-1, \mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.
3. General solution: $\mathbf{x}(t) = c_1 e^{3t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2 e^{-t} \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.
4. Initial conditions: $c_1 + c_2 = 3$ and $c_1 - c_2 = 1$. Solving gives $c_1 = 2, c_2 = 1$.

Answer: \mathbf{x}(t) = 2 e^{3t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + e^{-t} \begin{pmatrix} 1 \\ -1 \end{pmatrix}

1. General solution: $\mathbf{x}(t) = c_1 e^{5t} \begin{pmatrix} 2 \\ 1 \end{pmatrix} + c_2 e^{-3t} \begin{pmatrix} 0 \\ 1 \end{pmatrix}$.
2. At $t=0$: $2c_1 + 0c_2 = 4 \implies c_1 = 2$.
3. At $t=0$: $1c_1 + 1c_2 = 5 \implies 2 + c_2 = 5 \implies c_2 = 3$.
4. Substitute constants back into the general solution.

Answer: \mathbf{x}(t) = 2 e^{5t} \begin{pmatrix} 2 \\ 1 \end{pmatrix} + 3 e^{-3t} \begin{pmatrix} 0 \\ 1 \end{pmatrix}

1. Find eigenvalues: $\det(A - \lambda I) = (-3-\lambda)^2 - 4 = \lambda^2 + 6\lambda + 5 = 0$.
2. Solve for $\lambda$: $(\lambda + 1)(\lambda + 5) = 0$, so $\lambda_1 = -1, \lambda_2 = -5$.
3. Since both are negative, the origin is a stable node.
4. Find eigenvectors: For $\lambda = -1$, $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$. For $\lambda = -5$, $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.
5. Trajectories enter the origin tangent to $\mathbf{v}_1$ because $-1$ is closer to zero than $-5$.

Answer: Stable Node; trajectories approach along the line $y=x$.

1. Matrix $A = \begin{pmatrix} 2 & 3 \\ 2 & 1 \end{pmatrix}$.
2. Characteristic equation: $(2-\lambda)(1-\lambda) - 6 = \lambda^2 - 3\lambda - 4 = 0$.
3. Solve for $\lambda$: $(\lambda - 4)(\lambda + 1) = 0$, so $\lambda_1 = 4, \lambda_2 = -1$.
4. Since the eigenvalues have opposite signs, the origin is a saddle point.

Answer: Saddle Point

1. Both eigenvalues are positive, so the origin is an unstable node (source).
2. The general solution is $\mathbf{x}(t) = c_1 e^{0.1t} \mathbf{v}_1 + c_2 e^{10t} \mathbf{v}_2$.
3. Close to the origin (t is very negative), the $e^{10t}$ term is negligible, so trajectories emerge tangent to $\mathbf{v}_1$.
4. Far from the origin (large $t$), the $e^{10t}$ term completely dominates the $e^{0.1t}$ term, so all trajectories eventually become parallel to the $\mathbf{v}_2$ axis.

Answer: Unstable node; trajectories emerge tangent to $\mathbf{v}_1$ and eventually align with $\mathbf{v}_2$.

1. Write the matrix $A = \begin{pmatrix} -1 & 1 \\ -1 & -1 \end{pmatrix}$.
2. Find the characteristic equation: $\det(A - \lambda I) = (-1 - \lambda)^2 + 1 = 0$.
3. Solve for $\lambda$: $(-1 - \lambda)^2 = -1 \implies -1 - \lambda = \pm i \implies \lambda = -1 \pm i$.
4. Identify the real part $\alpha = -1$.
5. Since $\alpha < 0$, the origin is a stable spiral.

Answer: Stable Spiral

1. Write the matrix $A = \begin{pmatrix} 3 & -4 \\ 4 & 3 \end{pmatrix}$.
2. Find the characteristic equation: $(3 - \lambda)^2 + 16 = 0$.
3. Solve for $\lambda$: $(3 - \lambda)^2 = -16 \implies 3 - \lambda = \pm 4i \implies \lambda = 3 \pm 4i$.
4. Identify the real part $\alpha = 3$.
5. Since $\alpha > 0$, the origin is an unstable spiral.

Answer: Unstable Spiral

1. Since $\text{Trace}(A) = 0$, the real part $\alpha = 0$.
2. The characteristic equation is $\lambda^2 - \text{Trace}(A)\lambda + \det(A) = 0$, so $\lambda^2 + 9 = 0$.
3. The eigenvalues are $\lambda = \pm 3i$.
4. Since the eigenvalues are purely imaginary, the origin is a centre.
5. The imaginary part is $\beta = 3$. The period is $T = 2\pi / \beta = 2\pi / 3$.

Answer: Centre with period $2\pi/3$

1. Find eigenvalues: $\det(A - \lambda I) = (2-\lambda)^2 = 0$, so $\lambda = 2$ repeated.
2. Find eigenvector $\mathbf{v}$: $(A - 2I)\mathbf{v} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$, which implies $v_2 = 0$. Let $\mathbf{v} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$.
3. Find generalised eigenvector $\mathbf{w}$: $(A - 2I)\mathbf{w} = \mathbf{v} \implies \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} w_1 \\ w_2 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$, which implies $w_2 = 1$. Let $\mathbf{w} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$.
4. Construct general solution: $\mathbf{x}(t) = c_1 e^{2t} \begin{pmatrix} 1 \\ 0 \end{pmatrix} + c_2 e^{2t} (t \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \end{pmatrix})$.

Answer: \mathbf{x}(t) = c_1 e^{2t} \begin{pmatrix} 1 \\ 0 \end{pmatrix} + c_2 e^{2t} \begin{pmatrix} t \\ 1 \end{pmatrix}

1. The first solution is $\mathbf{x}_1(t) = e^{-5t} \begin{pmatrix} 1 \\ -2 \end{pmatrix}$.
2. The second solution is $\mathbf{x}_2(t) = t e^{-5t} \begin{pmatrix} 1 \\ -2 \end{pmatrix} + e^{-5t} \begin{pmatrix} 1 \\ 0 \end{pmatrix}$.
3. The general solution is the linear combination of these two.

Answer: \mathbf{x}(t) = c_1 e^{-5t} \begin{pmatrix} 1 \\ -2 \end{pmatrix} + c_2 e^{-5t} \begin{pmatrix} t+1 \\ -2t \end{pmatrix}

1. Find eigenvalues: $\det(A - \lambda I) = (4-\lambda)^2 = 0$, so $\lambda = 4$ repeated.
2. Find eigenvectors: $(A - 4I)\mathbf{v} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \mathbf{v} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$. Any vector in $\mathbb{R}^2$ is an eigenvector.
3. Since we have two linearly independent eigenvectors (e.g., $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$), the matrix is NOT defective.
4. Because $\lambda = 4 > 0$ and it is a non-defective repeated eigenvalue, the origin is an unstable star node.

Answer: Not defective; unstable star node.

1. The eigenvalues are complex, which indicates rotation.
2. The real part is $-3$, which is negative.
3. A negative real part with complex eigenvalues corresponds to a stable spiral.

Answer: Stable Spiral

1. The eigenvalues are real.
2. The eigenvalues have opposite signs (one positive, one negative).
3. This configuration defines a saddle point.

Answer: Saddle Point

1. The eigenvalues are real.
2. Both eigenvalues are negative.
3. This configuration defines a stable node.

Answer: Stable Node

1. Compute the trace: $\text{tr}(A) = 2 + (-4) = -2$.
2. Compute the determinant: $\det(A) = (2)(-4) - (-1)(3) = -8 + 3 = -5$.
3. Since $\det(A) < 0$, the equilibrium is a saddle point.

Answer: Saddle point

1. Compute the trace: $\text{tr}(A) = -3 + (-3) = -6$.
2. Compute the determinant: $\det(A) = (-3)(-3) - (1)(2) = 9 - 2 = 7$.
3. Since $\det(A) > 0$ and $\text{tr}(A) < 0$, the system is stable.
4. Compute the discriminant: $\Delta = (-6)^2 - 4(7) = 36 - 28 = 8$.
5. Since $\Delta > 0$, the equilibrium is a node.

Answer: Stable node

1. Compute the trace: $\text{tr}(A) = k + k = 2k$.
2. Compute the determinant: $\det(A) = (k)(k) - (1)(-1) = k^2 + 1$.
3. For a spiral, we need $\Delta < 0$: $(2k)^2 - 4(k^2 + 1) = 4k^2 - 4k^2 - 4 = -4$. This is always negative, so it is always a spiral.
4. For stability, we need $\text{tr}(A) < 0$: $2k < 0$, which means $k < 0$.
5. Also need $\det(A) > 0$: $k^2 + 1 > 0$, which is always true.

Answer: k < 0

1. Since $A$ is diagonal, $A^k$ is simply the matrix with entries $2^k$ and $(-3)^k$ on the diagonal.
2. The series definition $e^{At} = \sum \frac{(At)^k}{k!}$ applies to each diagonal entry independently.
3. The top-left entry is $\sum \frac{(2t)^k}{k!} = e^{2t}$.
4. The bottom-right entry is $\sum \frac{(-3t)^k}{k!} = e^{-3t}$.

Answer: e^{At} = \begin{bmatrix} e^{2t} & 0 \\ 0 & e^{-3t} \end{bmatrix}

1. Find eigenvalues: $\det(A-\lambda I) = \lambda^2 - 1 = 0$, so $\lambda = 1, -1$.
2. Find eigenvectors: for $\lambda=1$, $v_1 = [1, 1]^T$; for $\lambda=-1$, $v_2 = [1, -1]^T$.
3. Construct $P = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$ and $P^{-1} = \frac{1}{2} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$.
4. Compute $e^{At} = P \begin{bmatrix} e^t & 0 \\ 0 & e^{-t} \end{bmatrix} P^{-1}$.
5. Multiply: $e^{At} = \frac{1}{2} \begin{bmatrix} e^t+e^{-t} & e^t-e^{-t} \\ e^t-e^{-t} & e^t+e^{-t} \end{bmatrix}$.

Answer: e^{At} = \begin{bmatrix} \cosh(t) & \sinh(t) \\ \sinh(t) & \cosh(t) \end{bmatrix}

1. The matrix $A$ is diagonal, so $e^{At} = \begin{bmatrix} e^t & 0 \\ 0 & e^{2t} \end{bmatrix}$.
2. The solution is $\mathbf{x}(t) = e^{At} \mathbf{x}(0)$.
3. Substitute the values: $\begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = \begin{bmatrix} e^t & 0 \\ 0 & e^{2t} \end{bmatrix} \begin{bmatrix} 3 \\ 5 \end{bmatrix}$.
4. Perform the multiplication: $x(t) = 3e^t$ and $y(t) = 5e^{2t}$.

Answer: \mathbf{x}(t) = \begin{bmatrix} 3e^t \\ 5e^{2t} \end{bmatrix}

1. Set $\mathbf{x}' = \mathbf{0}$ for the particular solution: $\begin{pmatrix} -1 & 2 \\ 0 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -5 \\ -2 \end{pmatrix}$.
2. From the second row: $-3y = -2 \implies y = 2/3$.
3. From the first row: $-x + 2(2/3) = -5 \implies -x + 4/3 = -15/3 \implies -x = -19/3 \implies x = 19/3$.

Answer: \mathbf{x}_p = \begin{pmatrix} 19/3 \\ 2/3 \end{pmatrix}

1. Substitute $x = a \sin(2t) + b \cos(2t)$ and $y = c \sin(2t) + d \cos(2t)$ into the system.
2. From $x' = y$, we have $2a \cos(2t) - 2b \sin(2t) = c \sin(2t) + d \cos(2t)$, so $d = 2a$ and $c = -2b$.
3. From $y' = -x + 2\cos(2t)$, we have $2c \cos(2t) - 2d \sin(2t) = -a \sin(2t) - b \cos(2t) + 2\cos(2t)$.
4. Substitute $c = -2b$ and $d = 2a$: $-4b \cos(2t) - 4a \sin(2t) = -a \sin(2t) - b \cos(2t) + 2\cos(2t)$.
5. Matching $\sin(2t)$: $-4a = -a \implies a = 0$.
6. Matching $\cos(2t)$: $-4b = -b + 2 \implies -3b = 2 \implies b = -2/3$.
7. Then $d = 2(0) = 0$ and $c = -2(-2/3) = 4/3$.

Answer: \mathbf{x}_p(t) = \begin{pmatrix} -\frac{2}{3}\cos(2t) \\ \frac{4}{3}\sin(2t) \end{pmatrix}

1. Find $\mathbf{x}_p$: $-2x + 1 = 0 \implies x = 1/2$ and $-y + 1 = 0 \implies y = 1$. So $\mathbf{x}_p = \begin{pmatrix} 1/2 \\ 1 \end{pmatrix}$.
2. Homogeneous solution: $\mathbf{x}_h = c_1 e^{-2t} \begin{pmatrix} 1 \\ 0 \end{pmatrix} + c_2 e^{-t} \begin{pmatrix} 0 \\ 1 \end{pmatrix}$.
3. General solution: $\mathbf{x}(t) = c_1 e^{-2t} \begin{pmatrix} 1 \\ 0 \end{pmatrix} + c_2 e^{-t} \begin{pmatrix} 0 \\ 1 \end{pmatrix} + \begin{pmatrix} 1/2 \\ 1 \end{pmatrix}$.
4. Apply ICs: $\begin{pmatrix} 0 \\ 0 \end{pmatrix} = \begin{pmatrix} c_1 + 1/2 \\ c_2 + 1 \end{pmatrix} \implies c_1 = -1/2, c_2 = -1$.

Answer: \mathbf{x}(t) = \begin{pmatrix} \frac{1}{2}(1 - e^{-2t}) \\ 1 - e^{-t} \end{pmatrix}

1. Solve $\det(K - \lambda I) = (3-\lambda)(2-\lambda) - 1 = \lambda^2 - 5\lambda + 5 = 0$.
2. Use the quadratic formula: $\lambda = \frac{5 \pm \sqrt{25 - 20}}{2} = \frac{5 \pm \sqrt{5}}{2}$.
3. The frequencies are $\omega_1 = \sqrt{\frac{5 - \sqrt{5}}{2}}$ and $\omega_2 = \sqrt{\frac{5 + \sqrt{5}}{2}}$.
4. Find eigenvectors by solving $(K - \lambda I)\mathbf{v} = 0$ for each $\lambda$.

Answer: \omega_1 = \sqrt{1.38}, \omega_2 = \sqrt{3.62}, \mathbf{v}_1 \approx \begin{pmatrix} 1 \\ 1.62 \end{pmatrix}, \mathbf{v}_2 \approx \begin{pmatrix} 1 \\ -0.62 \end{pmatrix}

1. From the lesson, eigenvalues are $\lambda_1=1, \lambda_2=3$ with $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.
2. Set $\begin{pmatrix} 0 \\ 1 \end{pmatrix} = c_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2 \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.
3. Solving the system: $c_1 + c_2 = 0$ and $c_1 - c_2 = 1$ gives $c_1 = 1/2, c_2 = -1/2$.
4. The general solution is $\mathbf{x}(t) = \frac{1}{2} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \cos(t) - \frac{1}{2} \begin{pmatrix} 1 \\ -1 \end{pmatrix} \cos(\sqrt{3}t)$.

Answer: \mathbf{x}(t) = \frac{1}{2} \begin{pmatrix} \cos(t) - \cos(\sqrt{3}t) \\ \cos(t) + \cos(\sqrt{3}t) \end{pmatrix}

1. Solve $\det(K - \lambda I) = 0$ where $K$ is the matrix above (setting $k/m = 1$ for simplicity).
2. The determinant is $(2-\lambda)((2-\lambda)^2 - 1) - (-1)(-(2-\lambda)) = (2-\lambda)((2-\lambda)^2 - 2) = 0$.
3. The roots are $\lambda_1 = 2$, $\lambda_2 = 2 - \sqrt{2}$, and $\lambda_3 = 2 + \sqrt{2}$.
4. The frequencies are $\omega_1 = \sqrt{2k/m}$, $\omega_2 = \sqrt{(2-\sqrt{2})k/m}$, and $\omega_3 = \sqrt{(2+\sqrt{2})k/m}$.

Answer: \omega = \sqrt{\frac{k}{m}(2-\sqrt{2})}, \sqrt{\frac{2k}{m}}, \sqrt{\frac{k}{m}(2+\sqrt{2})}

1. The matrix is $A = \begin{pmatrix} -4 & 2 \\ 3 & -3 \end{pmatrix}$.
2. The characteristic equation is $(-4-\lambda)(-3-\lambda) - 6 = 0$, which is $\lambda^2 + 7\lambda + 12 - 6 = \lambda^2 + 7\lambda + 6 = 0$.
3. Factoring gives $(\lambda + 1)(\lambda + 6) = 0$, so $\lambda_1 = -1$ and $\lambda_2 = -6$.
4. The slow eigenvalue is $\lambda_{\text{slow}} = -1$.
5. The half-life is $t_{1/2} = \ln 2 / |-1| \approx 0.693$.

Answer: t_{1/2} \approx 0.693

1. Eigenvalues are $\lambda_1 = -1, \lambda_2 = -4$ with eigenvectors $\mathbf{v}_1 = (1, 2)^T, \mathbf{v}_2 = (1, -1)^T$.
2. General solution: $\begin{pmatrix} x \\ y \end{pmatrix} = c_1 e^{-t} \begin{pmatrix} 1 \\ 2 \end{pmatrix} + c_2 e^{-4t} \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.
3. Initial conditions: $c_1 + c_2 = 1$ and $2c_1 - c_2 = 0$.
4. Adding equations: $3c_1 = 1 \implies c_1 = 1/3$. Then $c_2 = 2/3$.
5. The tissue amount is $y(t) = 2(1/3)e^{-t} - (2/3)e^{-4t} = \frac{2}{3}(e^{-t} - e^{-4t})$.

Answer: y(t) = \frac{2}{3}(e^{-t} - e^{-4t})

1. The slow eigenvalue is $\lambda_{\text{slow}} = -0.2$.
2. The slow-mode component decays as $e^{-0.2t}$.
3. We want $e^{-0.2t} = 0.5$.
4. Taking the natural log: $-0.2t = \ln(0.5) = -\ln 2$.
5. Solving for $t$: $t = \ln 2 / 0.2 = 0.693 / 0.2 = 3.465$.

Answer: t \approx 3.465

1. Compute trace and determinant: $\text{tr}(A) = -4$, $\text{det}(A) = (-1)(-3) - (2)(-1) = 3 + 2 = 5$.
2. Solve characteristic equation: $\lambda^2 + 4\lambda + 5 = 0$. Using quadratic formula, $\lambda = \frac{-4 \pm \sqrt{16 - 20}}{2} = -2 \pm i$.
3. Since the real part is $-2$ and the imaginary part is $1$, the equilibrium is a stable spiral.
4. The general solution takes the form $\mathbf{x}(t) = e^{-2t} (c_1 \mathbf{u}(t) + c_2 \mathbf{w}(t))$ where $\mathbf{u}$ and $\mathbf{w}$ are trigonometric vectors.

Answer: Stable spiral with eigenvalues $\lambda = -2 \pm i$.

1. Matrix $A = \begin{pmatrix} 2 & 3 \\ 2 & 1 \end{pmatrix}$.
2. Characteristic equation: $\text{det}\begin{pmatrix} 2-\lambda & 3 \\ 2 & 1-\lambda \end{pmatrix} = (2-\lambda)(1-\lambda) - 6 = \lambda^2 - 3\lambda - 4 = 0$.
3. Factoring gives $(\lambda - 4)(\lambda + 1) = 0$, so $\lambda_1 = 4, \lambda_2 = -1$.
4. For $\lambda = 4$: $-2x + 3y = 0 \implies \mathbf{v}_1 = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$.
5. For $\lambda = -1$: $3x + 3y = 0 \implies \mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.
6. Since eigenvalues are real and opposite, the origin is a saddle point.

Answer: Saddle point with $\lambda = 4, -1$ and eigenvectors $\begin{pmatrix} 3 \\ 2 \end{pmatrix}, \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.

1. Characteristic equation: $\lambda^2 - \text{tr}(A)\lambda + \text{det}(A) = 0 \implies \lambda^2 + 9 = 0$.
2. Eigenvalues are $\lambda = \pm 3i$.
3. This is a center. The trajectories are ellipses centered at the origin.
4. The motion is a periodic orbit with frequency $\omega = 3$ and period $T = 2\pi/3$.

Answer: Center with eigenvalues $\lambda = \pm 3i$; the trajectory is a periodic elliptical orbit.

1. Write the matrix $A = \begin{pmatrix} 2 & 3 \\ 4 & -1 \end{pmatrix}$.
2. Find the characteristic equation: $\det(A - \lambda I) = (2-\lambda)(-1-\lambda) - 12 = \lambda^2 - \lambda - 14 = 0$.
3. Solve for $\lambda$: $\lambda = \frac{1 \pm \sqrt{1 + 56}}{2} = \frac{1 \pm \sqrt{57}}{2}$.
4. Since the eigenvalues are real and have opposite signs ($\sqrt{57} > 1$), the equilibrium is a saddle point.
5. Find eigenvectors $\mathbf{v}_1, \mathbf{v}_2$ and write $\mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2$.

Answer: Saddle point; $\lambda = \frac{1 \pm \sqrt{57}}{2}$

1. Matrix $A = \begin{pmatrix} -1 & 1 \\ -1 & -1 \end{pmatrix}$.
2. Trace $\text{tr}(A) = -2$, Determinant $\det(A) = (-1)(-1) - (1)(-1) = 2$.
3. Characteristic equation: $\lambda^2 + 2\lambda + 2 = 0$.
4. Eigenvalues: $\lambda = \frac{-2 \pm \sqrt{4 - 8}}{2} = -1 \pm i$.
5. Since the real part is negative and the imaginary part is non-zero, it is a stable spiral.

Answer: Stable spiral

1. Set $\mathbf{x}' = 0$ for the steady state: $0 = A \mathbf{x}_p + \mathbf{b}$.
2. Solve the system: $-2x + y = -3$ and $x - 2y = -3$.
3. From the first eq, $y = 2x - 3$. Substitute into second: $x - 2(2x - 3) = -3$.
4. Simplify: $x - 4x + 6 = -3 \implies -3x = -9 \implies x = 3$.
5. Find $y$: $y = 2(3) - 3 = 3$.

Answer: $\mathbf{x}_p = \begin{pmatrix} 3 \\ 3 \end{pmatrix}$

1. Compute the derivative: $(y_{\text{sum}})' = y_1' + y_2' = y_1^3 + y_2^3$.
2. Compute the cube of the sum: $(y_{\text{sum}})^3 = (y_1 + y_2)^3 = y_1^3 + 3y_1^2 y_2 + 3y_1 y_2^2 + y_2^3$.
3. Observe that $(y_{\text{sum}})' - (y_{\text{sum}})^3 = -3y_1^2 y_2 - 3y_1 y_2^2$, which is not zero.

Answer: The sum is not a solution because the cross-terms in the binomial expansion of $(y_1 + y_2)^3$ prevent the equation from being satisfied.

1. Set $x' = 0$: $x(1-x) = 0$ implies $x = 0$ or $x = 1$.
2. Case $x = 0$: Substitute into $y' = 0 \implies y(2-0) = 0 \implies y = 0$. Point: $(0, 0)$.
3. Case $x = 1$: Substitute into $y' = 0 \implies y(2-1) = 0 \implies y = 0$. Point: $(1, 0)$.

Answer: The equilibrium points are $(0, 0)$ and $(1, 0)$.

1. In a linear system, the only possible closed trajectories are centers, where eigenvalues are purely imaginary.
2. Centers are not isolated; every nearby trajectory is also a closed loop of a different size.
3. A limit cycle must be isolated, meaning nearby trajectories must spiral toward or away from it.
4. Linear systems lack the 'saturation' or 'feedback' terms (like $x^3$ or $xy$) needed to push trajectories toward a specific closed loop.

Answer: Linear systems only have non-isolated centers; limit cycles require nonlinearity to create isolated, attracting or repelling closed orbits.

1. Compute partials: $f_x = 1 - 2x$, $f_y = -1$, $g_x = 2x$, $g_y = 1$.
2. Evaluate at $(0,0)$: $A = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$.
3. Find eigenvalues: $\det(A - \lambda I) = (1 - \lambda)^2 = 0$, so $\lambda = 1, 1$.
4. Since both eigenvalues are positive, the origin is an unstable node (source).

Answer: Unstable node (source)

1. Equilibria: $y = 0$ and $-x - x^3 = 0 \implies x(1 + x^2) = 0 \implies x = 0$. Point is $(0,0)$.
2. Jacobian: $J = \begin{pmatrix} 0 & 1 \\ -1 - 3x^2 & 0 \end{pmatrix}$.
3. Evaluate at $(0,0)$: $A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$.
4. Eigenvalues: $\lambda^2 + 1 = 0 \implies \lambda = \pm i$.
5. Since $\text{Re}(\lambda) = 0$, the linearisation is inconclusive.

Answer: Inconclusive

1. Equilibria: $y = 2x$ and $x + 2(2x) - x^2 = 0 \implies 5x - x^2 = 0 \implies x = 0, 5$. First quadrant point is $(5, 10)$.
2. Jacobian: $J = \begin{pmatrix} 2 & -1 \\ 1 - 2x & 2 \end{pmatrix}$.
3. Evaluate at $(5, 10)$: $A = \begin{pmatrix} 2 & -1 \\ -9 & 2 \end{pmatrix}$.
4. Eigenvalues: $(2 - \lambda)^2 - 9 = 0 \implies 2 - \lambda = \pm 3 \implies \lambda = 5, -1$.
5. Opposite signs indicate a saddle point.

Answer: Saddle point

1. Set $2x - xy = 0 \implies x(2 - y) = 0$. Since $x \neq 0$, $y^* = 2$.
2. Set $0.5xy - 3y = 0 \implies y(0.5x - 3) = 0$. Since $y \neq 0$, $0.5x = 3 \implies x^* = 6$.
3. The Jacobian is $J = \begin{pmatrix} 2-y & -x \\ 0.5y & 0.5x-3 \end{pmatrix}$.
4. At $(6, 2)$, $J = \begin{pmatrix} 2-2 & -6 \\ 0.5(2) & 0.5(6)-3 \end{pmatrix} = \begin{pmatrix} 0 & -6 \\ 1 & 0 \end{pmatrix}$.

Answer: (x^*, y^*) = (6, 2), J = \begin{pmatrix} 0 & -6 \\ 1 & 0 \end{pmatrix}

1. The characteristic equation is $\det(J - \lambda I) = \det\begin{pmatrix} -\lambda & -6 \\ 1 & -\lambda \end{pmatrix} = 0$.
2. This gives $\lambda^2 - (-6)(1) = 0 \implies \lambda^2 + 6 = 0$.
3. The eigenvalues are $\lambda = \pm i\sqrt{6}$.
4. Since the eigenvalues are purely imaginary, the linearised system is a center.

Answer: \lambda = \pm i\sqrt{6}, \text{ linear centre}

1. Compute the total derivative $\frac{dV}{dt} = \frac{\partial V}{\partial x}\frac{dx}{dt} + \frac{\partial V}{\partial y}\frac{dy}{dt}$.
2. $\frac{\partial V}{\partial x} = 0.5 - \frac{3}{x}$ and $\frac{\partial V}{\partial y} = 1 - \frac{2}{y}$.
3. $\frac{dV}{dt} = (0.5 - \frac{3}{x})(2x - xy) + (1 - \frac{2}{y})(0.5xy - 3y)$.
4. $\frac{dV}{dt} = (0.5 - \frac{3}{x})x(2-y) + (1 - \frac{2}{y})y(0.5x-3)$.
5. $\frac{dV}{dt} = (0.5x-3)(2-y) + (y-2)(0.5x-3)$.
6. $\frac{dV}{dt} = (0.5x-3)(2-y) - (2-y)(0.5x-3) = 0$.

Answer: \frac{dV}{dt} = 0

1. The x-nullcline is $x + 0.4y = 50$. Intercepts are $(50, 0)$ and $(0, 125)$.
2. The y-nullcline is $y + 0.3x = 100$. Intercepts are $(333.33, 0)$ and $(0, 100)$.
3. The carrying capacities are $K_1 = 50$ and $K_2 = 100$.
4. Comparing intercepts: $50 < 333.33$ and $100 < 125$.
5. Since the carrying capacities are the inner intercepts of the nullcline system, the interior equilibrium is stable.

Answer: Stable coexistence; equilibrium at $(30, 50)$.

1. x-nullcline: $x + 2y = 10$. Intercepts: $(10, 0), (0, 5)$.
2. y-nullcline: $y + 2x = 10$. Intercepts: $(5, 0), (0, 10)$.
3. The carrying capacities $(10, 0)$ and $(0, 10)$ are the outer intercepts.
4. Checking the condition: $K_1 > K_2/a_{21} \implies 10 > 10/2 = 5$ and $K_2 > K_1/a_{12} \implies 10 > 10/2 = 5$.
5. Both conditions for unstable coexistence are met, making the interior point a saddle.

Answer: Unstable coexistence (saddle point); competitive exclusion occurs.

1. x-nullcline: $x + 0.1y = 10$. Intercepts: $(10, 0), (0, 100)$.
2. y-nullcline: $y + 5x = 10$. Intercepts: $(2, 0), (0, 10)$.
3. The x-nullcline is entirely above and to the right of the y-nullcline in the first quadrant.
4. Therefore, $x' > 0$ whenever $y' > 0$ is possible, and $x$ will always drive $y$ to zero.

Answer: Species x will always survive; species y is excluded.

1. Calculate the energy constant $C = \frac{1}{2}v^2 - \frac{g}{L}\cos(y)$.
2. Substitute the values: $C = \frac{1}{2}(2)^2 - 2\cos(0) = 2 - 2(1) = 0$.
3. The critical energy for the separatrix is $C_{crit} = \frac{g}{L} = 2$.
4. Since $0 < 2$, the energy is below the critical threshold.

Answer: Libration

1. Use the formula $C = \frac{1}{2}v^2 - \frac{g}{L}\cos(y)$.
2. Substitute the values: $C = \frac{1}{2}(0.1)^2 - 1\cos(\pi) = 0.005 - (-1) = 1.005$.
3. The critical energy is $C_{crit} = \frac{g}{L} = 1$.
4. Since $1.005 > 1$, the energy is slightly above the separatrix.

Answer: Rotation

1. Calculate the energy constant: $C = \frac{1}{2}(0)^2 - 3\cos(\frac{\pi}{2}) = 0 - 3(0) = 0$.
2. The critical energy required to reach the top is $C_{crit} = \frac{g}{L} = 3$.
3. Since $0 < 3$, the pendulum does not have enough energy to reach the top.
4. The maximum angle $y_{max}$ occurs when $v = 0$, so $0 - 3\cos(y_{max}) = 0$, meaning $\cos(y_{max}) = 0$, so $y_{max} = \frac{\pi}{2}$.

Answer: No, it will not reach the top.

1. Set up the condition for a conserved quantity: $\frac{\partial H}{\partial x}(xy) + \frac{\partial H}{\partial y}(-x^2) = 0$.
2. Divide by $x$ (assuming $x \neq 0$): $y \frac{\partial H}{\partial x} - x \frac{\partial H}{\partial y} = 0$.
3. This implies the ratio of partials is $\frac{\partial H}{\partial x} / \frac{\partial H}{\partial y} = x/y$.
4. Integrating $\frac{\partial H}{\partial x} = x$ gives $H = \frac{1}{2}x^2 + g(y)$.
5. Integrating $\frac{\partial H}{\partial y} = y$ gives $H = \frac{1}{2}y^2 + f(x)$.
6. Combining these, we find $H(x, y) = \frac{1}{2}(x^2 + y^2)$.

Answer: H(x, y) = \frac{1}{2}(x^2 + y^2)

1. Use the Hamiltonian rules: $x' = \frac{\partial H}{\partial y}$ and $y' = -\frac{\partial H}{\partial x}$.
2. Compute the partial derivative with respect to $y$: $\frac{\partial H}{\partial y} = y$.
3. Compute the partial derivative with respect to $x$: $\frac{\partial H}{\partial x} = e^x$.
4. Substitute these into the rules to get the system.

Answer: x' = y, \quad y' = -e^x

1. The condition is $\frac{\partial H}{\partial x}(y) + \frac{\partial H}{\partial y}(-x - x^3) = 0$.
2. Rearrange to $\frac{\partial H}{\partial x} = x + x^3$ and $\frac{\partial H}{\partial y} = y$.
3. Integrate $\frac{\partial H}{\partial x}$ with respect to $x$: $H = \frac{1}{2}x^2 + \frac{1}{4}x^4 + g(y)$.
4. Integrate $\frac{\partial H}{\partial y}$ with respect to $y$: $H = \frac{1}{2}y^2 + f(x)$.
5. The conserved quantity is $H(x, y) = \frac{1}{2}y^2 + \frac{1}{2}x^2 + \frac{1}{4}x^4$.
6. Since $H$ has a strict local minimum at $(0,0)$, the level sets are closed loops around the origin, making it a center.

Answer: H(x, y) = \frac{1}{2}y^2 + \frac{1}{2}x^2 + \frac{1}{4}x^4; \text{ the origin is a center.}

1. Identify the damping term as $(1 - x^2)y$.
2. At the starting point $x = 0.01$, the term $(1 - x^2)$ is approximately $1 - 0.0001 = 0.9999$.
3. Since the coefficient of $y$ is positive in the $y'$ equation (which corresponds to negative damping in the standard $y'' + cy' + ky = 0$ form), energy is added to the system.
4. Therefore, the trajectory will spiral outward away from the origin.

Answer: It will spiral outward because the damping is negative for small $x$.

1. As $\mu \to 0$, the nonlinear damping term $\mu (1 - y^2) y'$ vanishes.
2. The equation approaches the linear harmonic oscillator $y'' + y = 0$.
3. The trajectories of a linear harmonic oscillator are circles in the phase plane (when scaled).
4. Thus, the limit cycle becomes nearly circular and the attraction to it becomes very slow.

Answer: The limit cycle becomes nearly circular, resembling a linear harmonic oscillator.

1. Look at the damping term $y(x^2 - 1)$.
2. Near the origin ($x \approx 0$), the term is approximately $-y$.
3. In the equation $y' = -x - y$, the damping is positive, meaning energy is removed.
4. Since the origin is stable (trajectories spiral inward), the limit cycle (which exists at $x \approx 1$) must repel trajectories from the inside.
5. Therefore, the limit cycle is unstable.

Answer: The limit cycle is unstable because the origin is a stable spiral.

1. Check the inner boundary $r = 0.5$: $\frac{dr}{dt} = 0.5(1 - 0.25) = 0.375 > 0$, so the flow points outward into $R$.
2. Check the outer boundary $r = 2$: $\frac{dr}{dt} = 2(1 - 4) = -6 < 0$, so the flow points inward into $R$.
3. Identify fixed points: $\frac{dr}{dt} = 0$ at $r=0$ and $r=1$. $\frac{d\theta}{dt} = 1$ means there are no fixed points since $\frac{d\theta}{dt}$ is never zero.
4. Since $R$ is a bounded trapping region with no fixed points, a limit cycle must exist by Poincare-Bendixson.

Answer: The region is trapping and contains no fixed points, thus a limit cycle exists at $r=1$.

1. The theorem requires the region to have no fixed points.
2. However, we can create a smaller trapping region by removing a small disk around the unstable spiral.
3. Since the spiral is unstable, trajectories near it move away from it, meaning the boundary of this small disk acts as an 'outward' boundary.
4. The resulting annular region is now a trapping region with no fixed points.

Answer: Yes, by excluding the unstable fixed point, we create a trapping region without fixed points, guaranteeing a limit cycle.

1. The Poincare-Bendixson theorem specifically applies to systems of two first-order ODEs (2D flows).
2. The Lorenz system consists of three first-order ODEs, meaning it is a 3D flow.
3. In 3D, trajectories can avoid crossing without being forced into a point or a loop.

Answer: The theorem only applies to 2D systems; the Lorenz system is 3D.

1. Set $r + 2y - y^2 = 0$ and solve for $y$ using the quadratic formula: $y = \frac{-2 \pm \sqrt{4 - 4(-1)(r)}}{-2} = 1 \pm \sqrt{1+r}$.
2. The bifurcation occurs when the discriminant is zero: $1+r = 0 \implies r_c = -1$.
3. For $r > -1$, there are two equilibria: $y = 1 + \sqrt{1+r}$ and $y = 1 - \sqrt{1+r}$.
4. Check stability using $f'(y) = 2 - 2y$. For the upper branch $y = 1 + \sqrt{1+r}$, $f'(y) = -2\sqrt{1+r} < 0$, so it is stable.
5. For the lower branch $y = 1 - \sqrt{1+r}$, $f'(y) = 2\sqrt{1+r} > 0$, so it is unstable.

Answer: Bifurcation point $r_c = -1$; stable branch $y = 1 + \sqrt{1+r}$, unstable branch $y = 1 - \sqrt{1+r}$.

1. Find equilibria: $y(r - 2y^2) = 0$. This gives $y = 0$ and $y = \pm \sqrt{r/2}$.
2. For $r < 0$, only $y=0$ exists.
3. For $r > 0$, three equilibria exist: $y=0, y=\sqrt{r/2}, y=-\sqrt{r/2}$.
4. A transcritical bifurcation requires two equilibria to exist on both sides of the bifurcation point and exchange stability.
5. Here, the number of equilibria changes from 1 to 3, which is characteristic of a pitchfork bifurcation, not a transcritical one.

Answer: No, it is a pitchfork bifurcation because the number of equilibria changes from 1 to 3.

1. The equation is $y' = ry - \frac{r}{100}y^2 - 10$.
2. Equilibria occur at $-\frac{r}{100}y^2 + ry - 10 = 0$.
3. This is a quadratic in $y$. A saddle-node bifurcation occurs when the discriminant is zero: $D = r^2 - 4(-\frac{r}{100})(-10) = 0$.
4. Simplify: $r^2 - \frac{40r}{100} = 0 \implies r^2 - 0.4r = 0$.
5. Solving for $r$ (excluding $r=0$ as it would make the equation $y' = -10$): $r = 0.4$.

Answer: The saddle-node bifurcation occurs at $r = 0.4$.

1. Set $3y - 2y^3 = 0$, which factors as $y(3 - 2y^2) = 0$.
2. The equilibria are $y = 0$ and $y = \pm\sqrt{3/2}$.
3. The derivative of the right hand side is $f'(y) = 3 - 6y^2$.
4. At $y = 0$, $f'(0) = 3 > 0$, so it is unstable.
5. At $y = \pm\sqrt{3/2}$, $f'(\pm\sqrt{3/2}) = 3 - 6(3/2) = 3 - 9 = -6 < 0$, so they are stable.

Answer: y = 0 \text{ (unstable)}, y = \pm\sqrt{1.5} \text{ (stable)}

1. The radius of the limit cycle is given by $R = \sqrt{\mu}$.
2. Substituting $\mu = 4$ gives $R = \sqrt{4} = 2$.
3. The angular velocity is $\theta' = 1$.
4. The period $T$ is $2\pi / \theta' = 2\pi / 1 = 2\pi$.

Answer: R = 2, T = 2\pi

1. Compute $x' = -\sin(t)$ and $y' = \cos(t)$.
2. Substitute into $x'$ equation: $1 \cos(t) - \sin(t) - \cos(t)( \cos^2(t) + \sin^2(t)) = \cos(t) - \sin(t) - \cos(t) = -\sin(t)$.
3. Substitute into $y'$ equation: $\cos(t) + 1 \sin(t) - \sin(t)( \cos^2(t) + \sin^2(t)) = \cos(t) + \sin(t) - \sin(t) = \cos(t)$.
4. Both equations are satisfied.

Answer: The trajectory satisfies both ODEs exactly.

1. Substitute $x=0, y=0, z=0$ into the first equation: $x' = 10(0 - 0) = 0$.
2. Substitute into the second equation: $y' = 0(15 - 0) - 0 = 0$.
3. Substitute into the third equation: $z' = (0)(0) - (8/3)(0) = 0$.
4. Since all derivatives are zero, the origin is an equilibrium.

Answer: Yes, $(0,0,0)$ is an equilibrium.

1. Set $x' = 0 \implies x = y$.
2. Set $z' = 0 \implies z = x^2/\beta = x^2/2$.
3. Substitute into $y' = 0$: $x(3 - x^2/2) - x = 0$.
4. Factor: $x(2 - x^2/2) = 0$.
5. Solve $x^2/2 = 2 \implies x^2 = 4 \implies x = \pm 2$.
6. For $x=2$, $y=2$ and $z=2^2/2 = 2$. For $x=-2$, $y=-2$ and $z=(-2)^2/2 = 2$.

Answer: $(2, 2, 2)$ and $(-2, -2, 2)$

1. The Poincare-Bendixson theorem states that in a bounded 2D region, any trajectory that does not approach a fixed point must approach a limit cycle.
2. Chaos is characterized by non-periodic, non-convergent trajectories that never intersect.
3. In 2D, a trajectory cannot 'dodge' another trajectory or a limit cycle without crossing it, which is forbidden by uniqueness of solutions.
4. Therefore, the complex folding and weaving required for a strange attractor requires a third dimension to avoid self-intersection.

Answer: Because 2D bounded trajectories are limited to fixed points or limit cycles; a third dimension is needed to allow non-intersecting, non-periodic paths.

1. Use the formula $\delta(t) = \delta(0) e^{\lambda t}$.
2. Substitute the values: $1 = 10^{-6} e^{0.5 t}$.
3. Divide by $10^{-6}$: $10^6 = e^{0.5 t}$.
4. Take the natural logarithm of both sides: $\ln(10^6) = 0.5 t$.
5. Solve for $t$: $t = \frac{6 \ln(10)}{0.5} = 12 \ln(10) \approx 12 \times 2.3025$.
6. Calculate the final value: $t \approx 27.63$.

Answer: 27.63

1. The time to reach a certain error $\epsilon$ is $t = \frac{1}{\lambda} \ln(\epsilon / \delta(0))$.
2. The extension is $\Delta t = t_{\text{new}} - t_{\text{old}} = \frac{1}{\lambda} \ln(\epsilon / (\delta(0)/100)) - \frac{1}{\lambda} \ln(\epsilon / \delta(0))$.
3. Simplify using log laws: $\Delta t = \frac{1}{\lambda} \ln(100)$.
4. Substitute $\lambda = 0.8$: $\Delta t = \frac{\ln(100)}{0.8}$.
5. Calculate: $\Delta t \approx \frac{4.605}{0.8} \approx 5.76$.

Answer: 5.76

1. A negative Lyapunov exponent $\lambda < 0$ implies that $\delta(t) = \delta(0) e^{\lambda t}$ decays toward zero as $t \to \infty$.
2. This means nearby trajectories converge rather than diverge, which is the opposite of sensitive dependence.
3. In terms of attractors, this indicates that the system is attracted to a stable fixed point or a stable limit cycle.
4. Since chaos requires the trajectories to diverge while remaining bounded (stretching and folding), a negative $\lambda$ precludes the possibility of chaos.

Answer: Nearby trajectories converge exponentially, meaning the system is stable and lacks the sensitive dependence required for chaos.

1. Step 1: $x_1 = 2.5(0.2)(1 - 0.2) = 2.5(0.2)(0.8) = 0.4$.
2. Step 2: $x_2 = 2.5(0.4)(1 - 0.4) = 2.5(0.4)(0.6) = 0.6$.
3. Step 3: $x_3 = 2.5(0.6)(1 - 0.6) = 2.5(0.6)(0.4) = 0.6$.
4. Step 4: $x_4 = 2.5(0.6)(1 - 0.6) = 0.6$.
5. The sequence converges to the fixed point $0.6$.

Answer: x_1=0.4, x_2=0.6, x_3=0.6, x_4=0.6; yes, it converges to 0.6.

1. Step 1: Set $x_{n+1} = x_n = x^*$.
2. Step 2: $x^* = r x^* (1 - x^*)$.
3. Step 3: Divide by $x^*$ (assuming $x^* \neq 0$): $1 = r(1 - x^*)$.
4. Step 4: Solve for $x^*$: $1/r = 1 - x^* \implies x^* = 1 - 1/r$.
5. Step 5: Set $0.75 = 1 - 1/r \implies 1/r = 0.25 \implies r = 4$.

Answer: x^* = 1 - 1/r; r = 4.

1. Step 1: The continuous logistic equation is a first-order ODE in one dimension. In 1D continuous systems, trajectories cannot cross and must move monotonically toward an equilibrium or infinity, preventing oscillation or chaos.
2. Step 2: The logistic map is a discrete-time system. This allows the state to 'jump' over the equilibrium point, creating oscillations.
3. Step 3: For continuous systems to be chaotic, the Poincare-Bendixson theorem implies they need at least three dimensions (like the Lorenz system).
4. Step 4: Therefore, the discrete nature of the map effectively adds the complexity needed for chaos in a single variable.

Answer: Continuous 1D systems are monotonic and cannot oscillate; discrete maps can jump over equilibria, allowing for period doubling and chaos.

1. Set $2x - xy = 0$ and $xy - 3y = 0$.
2. From the second equation, $y(x - 3) = 0$, so $y = 0$ or $x = 3$.
3. If $y = 0$, then $2x = 0$, so $(0,0)$ is an equilibrium.
4. If $x = 3$, then $6 - 3y = 0$, so $y = 2$, giving $(3,2)$ as an equilibrium.
5. Compute Jacobian $J = \begin{pmatrix} 2-y & -x \\ y & x-3 \end{pmatrix}$.
6. At $(0,0)$, $J = \begin{pmatrix} 2 & 0 \\ 0 & -3 \end{pmatrix}$, eigenvalues $2, -3$ (Saddle).
7. At $(3,2)$, $J = \begin{pmatrix} 0 & -3 \\ 2 & 0 \end{pmatrix}$, eigenvalues $\pm i\sqrt{6}$ (Center/Spiral).

Answer: Equilibria at $(0,0)$ (Saddle) and $(3,2)$ (Center/Spiral).

1. Set $x = r x (1 - x)$.
2. Divide by $x$ (assuming $x \neq 0$): $1 = r(1 - x)$.
3. Solve for $x$: $1/r = 1 - x \implies x^* = 1 - 1/r$.
4. Stability requires $|f'(x^*)| < 1$.
5. Compute $f'(x) = r - 2rx$.
6. Evaluate at $x^*$: $f'(1 - 1/r) = r - 2r(1 - 1/r) = r - 2r + 2 = 2 - r$.
7. Instability occurs when $|2 - r| > 1$, which means $r > 3$ (since $r$ is positive).

Answer: Fixed point $x^* = 1 - 1/r$; becomes unstable at $r = 3$.

1. Use the formula $\epsilon_{max} = \epsilon_0 e^{\lambda t}$.
2. Substitute values: $0.1 = 10^{-6} e^{0.12 t}$.
3. Divide by $10^{-6}$: $10^5 = e^{0.12 t}$.
4. Take natural log: $\ln(10^5) = 0.12 t$.
5. Calculate $\ln(10^5) \approx 11.513$.
6. Solve for $t$: $t = 11.513 / 0.12 \approx 95.94$.

Answer: $t_{horizon} \approx 95.94$ time units.

1. Assume $y = \sum a_k x^k$, then $y'' = \sum k(k-1)a_k x^{k-2}$.
2. Substitute into $y'' - y = 0$: $\sum (k+2)(k+1)a_{k+2} x^k - \sum a_k x^k = 0$.
3. Recurrence relation: $a_{k+2} = a_k / ((k+2)(k+1))$.
4. Given $a_0 = 1, a_1 = 1$.
5. Compute $a_2 = a_0 / (2 \cdot 1) = 1/2$.
6. Compute $a_3 = a_1 / (3 \cdot 2) = 1/6$.

Answer: $a_0 = 1, a_1 = 1, a_2 = 1/2, a_3 = 1/6$

1. Substitute $y = \sum a_k x^k$ and $y'' = \sum (k+2)(k+1)a_{k+2} x^k$.
2. The equation becomes $\sum [(k+2)(k+1)a_{k+2} + 4a_k] x^k = 0$.
3. Set the coefficient to zero: $(k+2)(k+1)a_{k+2} + 4a_k = 0$.

Answer: $a_{k+2} = \frac{-4a_k}{(k+2)(k+1)}$

1. Initial conditions give $a_0 = 1$ and $a_1 = 0$.
2. Since $a_1 = 0$, all odd coefficients $a_3, a_5, \dots$ are zero.
3. Compute $a_2 = -a_0 / (2 \cdot 1) = -1/2$.
4. Compute $a_4 = -a_2 / (4 \cdot 3) = -(-1/2) / 12 = 1/24$.
5. The terms are $a_0 x^0$, $a_2 x^2$, and $a_4 x^4$.

Answer: $y = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots$

1. Given $y(0)=0$ and $y'(0)=1$, we have $a_0 = 0$ and $a_1 = 1$.
2. From the $k=0$ term of the ODE, $2 a_2 = 0$, so $a_2 = 0$.
3. Using the recurrence $a_{k+2} = a_{k-1} / ((k+2)(k+1))$, the non-zero terms depend on $a_1$.
4. For $k=2$, $a_4 = a_1 / (4 \cdot 3) = 1/12$.
5. For $k=5$, $a_7 = a_4 / (7 \cdot 6) = (1/12) / 42 = 1/504$.
6. For $k=8$, $a_{10} = a_7 / (10 \cdot 9) = (1/504) / 90 = 1/45360$.

Answer: $y(t) = t + \frac{1}{12}t^4 + \frac{1}{504}t^7 + \frac{1}{45360}t^{10} + \dots$

1. Given $a_0 = 1$ and $a_1 = 0$.
2. The recurrence is $a_{k+2} = a_{k-1} / ((k+2)(k+1))$.
3. First non-zero term after $a_0$ is $a_3 = a_0 / (3 \cdot 2) = 1/6$.
4. The next non-zero term is $a_6 = a_3 / (6 \cdot 5) = (1/6) / 30 = 1/180$.

Answer: $a_6 = \frac{1}{180}$

1. Let $u(t) = y(-t)$.
2. By the chain rule, $u'(t) = -y'(-t)$ and $u''(t) = (-1)^2 y''(-t) = y''(-t)$.
3. Since $y$ satisfies $y''(t) = t y(t)$, then $y''(-t) = (-t) y(-t)$.
4. Substituting these into the expression for $u$, we get $u''(t) = -t u(t)$.

Answer: The substitution $u(t) = y(-t)$ transforms the equation $y'' = ty$ into $u'' = -tu$.

1. Divide by $t$ to get standard form: $y'' + \frac{2}{t}y' + y = 0$.
2. The coefficient $p(t) = \frac{2}{t}$ is discontinuous at $t=0$, so $t=0$ is a singular point.
3. Check for regularity: $\lim_{t \to 0} t p(t) = \lim_{t \to 0} t(\frac{2}{t}) = 2$.
4. Check the second coefficient: $\lim_{t \to 0} t^2 q(t) = \lim_{t \to 0} t^2(1) = 0$.
5. Since both limits are finite, $t=0$ is a regular singular point.

Answer: Regular singular point

1. Assume $y = t^r$, leading to the indicial equation $r(r-1) - 2r + 2 = 0$.
2. Simplify: $r^2 - r - 2r + 2 = r^2 - 3r + 2 = 0$.
3. Factor: $(r-1)(r-2) = 0$, so $r_1 = 1$ and $r_2 = 2$.
4. The general solution is $y(t) = C_1 t^1 + C_2 t^2$.

Answer: $y(t) = C_1 t + C_2 t^2$

1. Assume $y = t^r$, leading to the indicial equation $r(r-1) + 3r + 2.25 = 0$.
2. Simplify: $r^2 + 2r + 2.25 = 0$.
3. Use the quadratic formula: $r = \frac{-2 \pm \sqrt{4 - 4(2.25)}}{2} = \frac{-2 \pm \sqrt{4 - 9}}{2} = \frac{-2 \pm \sqrt{-5}}{2}$.
4. The roots are $r = -1 \pm i \frac{\sqrt{5}}{2}$.
5. Using the complex root formula with $\alpha = -1$ and $\beta = \frac{\sqrt{5}}{2}$, the solution is $y(t) = t^{-1} [C_1 \cos(\frac{\sqrt{5}}{2} \ln t) + C_2 \sin(\frac{\sqrt{5}}{2} \ln t)]$.

Answer: $y(t) = t^{-1} [C_1 \cos(\frac{\sqrt{5}}{2} \ln t) + C_2 \sin(\frac{\sqrt{5}}{2} \ln t)]$

1. Divide by $2x^2$ to get $y'' + \frac{3}{2x}y' + \frac{x^2-1}{2x^2}y = 0$.
2. Compute $p_0 = \lim_{x \to 0} x(\frac{3}{2x}) = 1.5$.
3. Compute $q_0 = \lim_{x \to 0} x^2(\frac{x^2-1}{2x^2}) = -0.5$.
4. Set up the indicial equation: $r(r-1) + 1.5r - 0.5 = 0$.
5. Simplify to $r^2 + 0.5r - 0.5 = 0$.
6. Solve using the quadratic formula: $r = \frac{-0.5 \pm \sqrt{0.25 + 2}}{2} = \frac{-0.5 \pm 1.5}{2}$.

Answer: x=0 is a regular singular point; roots are $r_1 = 0.5, r_2 = -1$.

1. Normalize: $y'' + \frac{2}{x}y' + y = 0$, so $p_0 = 2, q_0 = 0$.
2. Indicial equation: $r(r-1) + 2r + 0 = 0 \implies r^2 + r = 0$. Roots are $r=0, r=-1$.
3. Use $r=0$. The equation is $x y'' + 2y' + x y = 0$.
4. Substitute $y = \sum a_k x^k$: $\sum k(k-1)a_k x^{k-1} + \sum 2k a_k x^{k-1} + \sum a_k x^{k+1} = 0$.
5. Combine first two: $\sum k(k+1)a_k x^{k-1}$.
6. Recurrence: $(k+2)(k+3)a_{k+2} = -a_k$.
7. For $a_0=1$: $a_1=0$, $a_2 = -1/(2 \cdot 3) = -1/6$, $a_3=0$, $a_4 = -a_2/(4 \cdot 5) = 1/120$.

Answer: $y(x) = 1 - \frac{1}{6}x^2 + \frac{1}{120}x^4 - \dots$

1. Normalize: $y'' + \frac{1}{x^2}y = 0$.
2. Check $p_0 = \lim x(0) = 0$ (finite).
3. Check $q_0 = \lim x^2(\frac{1}{x^2}) = 1$ (finite).
4. It is a regular singular point, but the solution is not a polynomial because the indicial equation is $r(r-1) + 0r + 1 = 0$.
5. Simplify: $r^2 - r + 1 = 0$.
6. Solve: $r = \frac{1 \pm \sqrt{1-4}}{2} = \frac{1}{2} \pm i\frac{\sqrt{3}}{2}$.

Answer: Standard power series fail because the exponents are complex; the Frobenius exponents are $r = \frac{1}{2} \pm i\frac{\sqrt{3}}{2}$.

1. Compare the equation to the standard form $x^2 y'' + x y' + (x^2 - n^2) y = 0$.
2. Observe that $n^2 = 9$, which means $n = 3$.
3. The general solution is a linear combination of the first and second kind functions of order 3.

Answer: n = 3, y(x) = C_1 J_3(x) + C_2 Y_3(x)

1. The boundary condition requires $J_0(k \cdot 1) = 0$.
2. This means $k$ must be a zero of the Bessel function $J_0$.
3. For the fundamental mode, we use the first zero.

Answer: k \approx 2.4048

1. Multiply the entire equation by $x^2$ to bring it to standard form.
2. The equation becomes $x^2 y'' + x y' + (x^2 - \frac{1}{4}) y = 0$.
3. Compare this to $x^2 y'' + x y' + (x^2 - n^2) y = 0$.
4. We find $n^2 = 1/4$, so $n = 1/2$.

Answer: Yes, it is a Bessel equation of order n = 1/2

1. Compute $P_2'(x) = 3x$ and $P_2''(x) = 3$.
2. Substitute into $(1-x^2)y'' - 2xy' + 6y$.
3. $(1-x^2)(3) - 2x(3x) + 6(\frac{3}{2}x^2 - \frac{1}{2}) = 3 - 3x^2 - 6x^2 + 9x^2 - 3$.
4. The terms cancel to $0$.

Answer: The equation is satisfied as the result is $0$.

1. Set $n=4$ and $k=0$ in the recurrence relation.
2. $a_2 = \frac{0(1) - 4(5)}{(1)(2)} a_0$.
3. $a_2 = \frac{-20}{2} a_0 = -10 a_0$.
4. The ratio $a_2/a_0$ is $-10$.

Answer: -10

1. For $n=1$, the recurrence relation gives $a_3 = \frac{1(2) - 1(2)}{6} a_1 = 0$.
2. Thus $P_1(x) = a_0 + a_1 x$. However, for $n=1$, the even series $a_0$ must be zero for the solution to be a polynomial of degree 1.
3. So $P_1(x) = a_1 x$.
4. Apply normalization: $P_1(1) = a_1(1) = 1$.
5. Therefore $a_1 = 1$.

Answer: 1

1. Set up the inner product integral: $\int_{0}^{1} (1)(x)(x-1) dx$.
2. Expand the integrand: $\int_{0}^{1} (x^2 - x) dx$.
3. Integrate term by term: $[\frac{1}{3}x^3 - \frac{1}{2}x^2]_{0}^{1}$.
4. Evaluate at the boundaries: $(\frac{1}{3} - \frac{1}{2}) - (0 - 0) = -\frac{1}{6}$.
5. Since the result is not zero, the functions are not orthogonal.

Answer: Not orthogonal

1. Set up the inner product integral: $\int_{-1}^{1} (1)(1)(x) dx$.
2. Integrate $x$: $[\frac{1}{2}x^2]_{-1}^{1}$.
3. Evaluate: $\frac{1}{2}(1)^2 - \frac{1}{2}(-1)^2 = \frac{1}{2} - \frac{1}{2} = 0$.
4. Since the result is zero, the functions are orthogonal.

Answer: Orthogonal

1. Set up the weighted inner product: $\int_{0}^{1} c(1)(2x - 1) dx = 0$.
2. Factor out the constant: $c \int_{0}^{1} (2x - 1) dx = 0$.
3. Integrate: $c [x^2 - x]_{0}^{1} = 0$.
4. Evaluate: $c ((1^2 - 1) - (0^2 - 0)) = c(0) = 0$.
5. The integral is zero regardless of the value of $c$. Therefore, any constant $c > 0$ works.

Answer: Any constant $c > 0$

1. Write out the first few terms: $a_1 = \frac{1}{3}$, $a_2 = \frac{1}{3 \cdot 4}$, $a_3 = \frac{1}{3 \cdot 4 \cdot 5}$.
2. Observe the pattern: $a_k = \frac{1}{3 \cdot 4 \cdot \dots \cdot (k+2)}$.
3. Multiply top and bottom by $2!$ to complete the factorial: $a_k = \frac{2}{ (k+2)! }$.
4. Replace the factorial with the Gamma function: $a_k = \frac{2}{\Gamma(k+3)}$.

Answer: $a_k = \frac{2}{\Gamma(k+3)}$

1. For $k=0$, the term is $\frac{(-1)^0}{0! \Gamma(3/2+1)} (x/2)^{3/2} = \frac{1}{\Gamma(5/2)} (x/2)^{3/2}$.
2. Calculate $\Gamma(5/2) = \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{3\sqrt{\pi}}{4}$.
3. The first term is $\frac{4}{3\sqrt{\pi}} (x/2)^{3/2}$.
4. For $k=1$, the term is $\frac{(-1)^1}{1! \Gamma(3/2+2)} (x/2)^{3/2+2} = \frac{-1}{\Gamma(7/2)} (x/2)^{7/2}$.
5. Calculate $\Gamma(7/2) = \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{15\sqrt{\pi}}{8}$.
6. The second term is $-\frac{8}{15\sqrt{\pi}} (x/2)^{7/2}$.

Answer: $J_{3/2}(x) \approx \frac{4}{3\sqrt{\pi}} (x/2)^{3/2} - \frac{8}{15\sqrt{\pi}} (x/2)^{7/2}$

1. Identify the term $\Gamma(n+k+1)$ in the denominator.
2. Since $n$ and $k$ are non-negative integers, $n+k$ is also a non-negative integer.
3. Apply the identity $\Gamma(m+1) = m!$ where $m = n+k$.
4. Substitute $\Gamma(n+k+1) = (n+k)!$ into the summation.
5. The formula becomes $J_n(x) = \sum_{k=0}^{\infty} \frac{(-1)^k}{k! (n+k)!} (x/2)^{n+2k}$.

Answer: $J_n(x) = \sum_{k=0}^{\infty} \frac{(-1)^k}{k! (n+k)!} \left( \frac{x}{2} \right)^{n+2k}$

1. The system has cylindrical symmetry, so the radial part satisfies Bessel's equation of order zero.
2. The solution is $u(r) = A J_0(kr)$.
3. The boundary condition is $u(1) = 0$, which means $J_0(k) = 0$.
4. For the first mode, $k$ must be the first positive zero of $J_0$.

Answer: Bessel function $J_0(kr)$ with $J_0(k) = 0$.

1. Spherical symmetry with angular dependence leads to Legendre's equation.
2. The solutions are Legendre polynomials $P_n(x)$ where $x = \cos(\theta)$.
3. For $n=2$, the solution is $V(\theta) = C P_2(\cos(\theta))$.
4. Using the formula $P_2(x) = \frac{1}{2}(3x^2 - 1)$, we have $V(\theta) = \frac{C}{2}(3 \cos^2(\theta) - 1)$.

Answer: $V(\theta) = C P_2(\cos(\theta))$

1. The power series for $J_n(x)$ contains a factorial term $(k+n)!$ in the denominator.
2. Factorials are only defined for non-negative integers.
3. The Gamma function extends the factorial such that $\Gamma(k+n+1)$ is defined for any real $k+n$.
4. This allows the series to converge and define a valid function for fractional orders.

Answer: It generalizes the factorial in the series denominator to non-integer values.

1. Divide by $2x^2$ to get $y'' + \frac{1}{2x} y' - \frac{x+1}{2x^2} y = 0$.
2. Identify $P(x) = \frac{1}{2x}$ and $Q(x) = -\frac{x+1}{2x^2}$.
3. Compute $p_0 = \lim_{x \to 0} x P(x) = \frac{1}{2}$ and $q_0 = \lim_{x \to 0} x^2 Q(x) = -\frac{1}{2}$.
4. The indicial equation is $r(r-1) + \frac{1}{2}r - \frac{1}{2} = 0$.
5. Simplify to $r^2 - \frac{1}{2}r - \frac{1}{2} = 0$, or $2r^2 - r - 1 = 0$.
6. Factor as $(2r+1)(r-1) = 0$, giving $r_1 = 1$ and $r_2 = -1/2$.

Answer: r_1 = 1, r_2 = -1/2

1. The general Bessel equation is $x^2 y'' + x y' + (x^2 - n^2) y = 0$.
2. Compare the given equation $x^2 y'' + x y' + (x^2 - 1) y = 0$ to the general form.
3. We see that $n^2 = 1$, which implies $n = 1$ or $n = -1$.
4. By definition, $J_n(x)$ is a solution to the Bessel equation of order $n$.
5. Since $n=1$ satisfies the equation, $J_1(x)$ is indeed a solution.

Answer: Yes, it is a solution because it matches the Bessel equation of order n=1.

1. The coefficient formula is $c_n = \frac{\int_{-1}^{1} f(x) P_n(x) dx}{\int_{-1}^{1} [P_n(x)]^2 dx}$.
2. For $n=2$, $c_2 = \frac{\int_{-1}^{1} f(x) P_2(x) dx}{\int_{-1}^{1} [P_2(x)]^2 dx}$.
3. The general formula for the norm of Legendre polynomials is $\int_{-1}^{1} [P_n(x)]^2 dx = \frac{2}{2n+1}$.
4. Substitute $n=2$ into the norm formula: $\frac{2}{2(2)+1} = \frac{2}{5}$.

Answer: c_2 = \frac{5}{2} \int_{-1}^{1} f(x) P_2(x) dx; \text{ norm squared } = 2/5