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))