(a) Solve the Initial Value Problem

We are given the differential equation:

$y' - 5y = 15t + 3e^t$ and the initial condition $y(0) = y_0$.

This is a first-order linear differential equation of the form $y' + P(t)y = Q(t)$. To solve it, we can use the method of integrating factors.

Step 1: Find the Integrating Factor

First, rewrite the equation in standard form:

$y' + (-5)y = 15t + 3e^t$

The integrating factor $\mu(t)$ is given by:

$\mu(t) = e^{\int -5 \, dt} = e^{-5t}$

Step 2: Multiply the Differential Equation by the Integrating Factor

Multiplying both sides of the equation by $e^{-5t}$:

$e^{-5t}y' - 5e^{-5t}y = (15t + 3e^t)e^{-5t}$

This simplifies to:

$\frac{d}{dt}(e^{-5t}y) = (15t + 3e^t)e^{-5t}$

Step 3: Solve for $y$

Now, integrate both sides with respect to $t$:

$e^{-5t}y = \int (15t + 3e^t)e^{-5t} \, dt$

Break the right-hand side into two parts:

$e^{-5t}y = \int 15t e^{-5t} \, dt + \int 3e^t e^{-5t} \, dt$

1. For the first term, $\int 15t e^{-5t} \, dt$, we can use integration by parts.

   • Let $u = t$ and $dv = 15e^{-5t} \, dt$.
   • Then, $du = dt$ and $v = \frac{-3}{e^{5t}}$.

   After applying the integration by parts formula, we get:

   [ \int 15t e^{-5t} , dt \ %} ...