Consider the initial value problem \[ y^{\prime}-5 y=20 t+4 e^{t}, \quad y(0)=y_{0} \] (a) Solve the initial value problem. (enter y0 for \( y_{0} \) ). \( y(t)= \) (b) Determine the value of \( y_{0} \) that separates solutions that grow positively as \( t \rightarrow \infty \) from those that grow negatively. \( y_{0}= \)

We start with the differential equation \[ y'-5y=20t+4e^t, \quad y(0)=y_0. \] **Step 1. Multiply through by the integrating factor** The integrating factor is \[ \mu(t)=e^{-5t}. \] Multiplying both sides by \(e^{-5t}\), we have \[ e^{-5t}y' - 5e^{-5t}y = 20te^{-5t}+4e^{-5t}e^t. \] Notice that the left‐hand side is the derivative of \(\left(e^{-5t}y\right)\). Also, simplifying the right‐hand side: \[ 4e^{-5t}e^t=4e^{-4t}. \] Thus, the equation becomes \[ \frac{d}{dt}\left(e^{-5t}y\right)=20te^{-5t}+4e^{-4t}. \] **Step 2. Integrate both sides** Integrate with respect to \(t\): \[ e^{-5t}y=\int\left(20te^{-5t}+4e^{-4t}\right)dt+C, \] where \(C\) is the constant of integration. We split the integral into two parts: \[ I_1=\int 20te^{-5t}dt \quad \text{and} \quad I_2=\int 4e^{-4t}dt. \] **Step 3. Compute \(I_1=\int 20te^{-5t}dt\)** Use integration by parts, letting \[ u=t,\quad dv=20e^{-5t}dt. \] Then, \[ du=dt,\quad v=\int 20e^{-5t}dt=20\left(-\frac{1}{5}e^{-5t}\right)=-4e^{-5t}. \] Now apply integration by parts: \[ I_1=uv-\int v\,du=-4te^{-5t}-\int \left(-4e^{-5t}\right)dt=-4te^{-5t}+4\int e^{-5t}dt. \] The remaining integral is \[ \int e^{-5t}dt=-\frac{1}{5}e^{-5t}, \] so \[ I_1=-4te^{-5t}+4\left(-\frac{1}{5}e^{-5t}\right)=-4te^{-5t}-\frac{4}{5}e^{-5t}. \] **Step 4. Compute \(I_2=\int 4e^{-4t}dt\)** \[ I_2=4\int e^{-4t}dt=4\left(-\frac{1}{4}e^{-4t}\right)=-e^{-4t}. \] Thus, together we have \[ e^{-5t}y=-4te^{-5t}-\frac{4}{5}e^{-5t}-e^{-4t}+C. \] **Step 5. Solve for \(y(t)\)** Multiply the entire equation by \(e^{5t}\): \[ y=-4t-\frac{4}{5}-e^{t}+Ce^{5t}. \] That is, the general solution is \[ y(t)=Ce^{5t}-4t-\frac{4}{5}-e^t. \] **Step 6. Apply the initial condition \(y(0)=y_0\)** Substitute \(t=0\) into the solution: \[ y(0)=C\cdot e^{0}-4\cdot 0-\frac{4}{5}-e^0=C-\frac{4}{5}-1. \] Since \(y(0)=y_0\), we have \[ C-\frac{9}{5}=y_0 \quad \Longrightarrow \quad C=y_0+\frac{9}{5}. \] Thus, the solution to the initial value problem is \[ y(t)=\left(y_0+\frac{9}{5}\right)e^{5t}-4t-\frac{4}{5}-e^t. \] **Step 7. Determine the critical value of \(y_0\) for growth** As \(t\rightarrow\infty\), the term \(\left(y_0+\frac{9}{5}\right)e^{5t}\) dominates the behavior of \(y(t)\), because the other terms either grow much more slowly or decay. For the solution not to grow positively or negatively, the coefficient of \(e^{5t}\) must vanish: \[ y_0+\frac{9}{5}=0. \] Thus, \[ y_0=-\frac{9}{5}. \] --- \[ \boxed{\begin{aligned} \text{(a)}\quad y(t)&=\left(y_0+\frac{9}{5}\right)e^{5t}-4t-\frac{4}{5}-e^t,\\[1mm] \text{(b)}\quad y_0&=-\frac{9}{5}. \end{aligned}} \]