Consider the initial value problem

y'-5y=15t+3e^t , y(0)=y_0

(a) Solve the initial value problem. (enter y0 for y_0)

(b) Determine the value of

that separates solutions that grow positively as

from those that grow negatively.

### (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
    \ %}
  ...

Consider the initial value problem y'-5y=15t+3e^t , y(0)=y_0 (a) Solve the initial value problem. (b) Determine the value that separates solutions that grow positively as t goes to infinity from those that grow negatively.

Explore how to solve the initial value problem y'-5y=15t+3e^t with y(0)=y_0 using integrating factors. Learn how to determine the critical value that separates solutions with positive growth from those with negative growth as t approaches infinity.

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