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. $$ y_{0}= $$

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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}}
$$
(a) The solution to the initial value problem is: $$ y(t) = \left(y_0 + \frac{9}{5}\right)e^{5t} - 4t - \frac{4}{5} - e^t $$ (b) The value of $$ y_0 $$ that separates solutions that grow positively as $$ t \rightarrow \infty $$ from those that grow negatively is: $$ y_0 = -\frac{9}{5} $$

Consider the initial value problem

(a) Solve the initial value problem. (enter y0 for ).

(b) Determine the value of that separates solutions that grow positively as from those that grow negatively.

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Consider the initial value problem (a) Solve the initial value problem. (enter y0 for ). (b) Determine the value of that separates solutions that grow positively as from those that grow negatively.