Solve the initial value problem: -5y'+3y=e^(-3t/5), y(0) =a. 

To solve the initial value problem \( -5y' + 3y = e^{-3t/5} \) with the initial condition \( y(0) = a \), we'll use the method of solving first-order linear differential equations. 

### Step 1: Rewrite the equation in standard form
The standard form of a first-order linear differential equation is:

\[
y' + p(t)y = q(t)
\]

We can rewrite the given equation as:

\[
y' - \frac{3}{5}y = -\frac{1}{5}e^{-3t/5}
\]

### Step 2: Find the integrating factor
The integrating factor \( \mu(t) \) is given by:

\[
\mu(t) = e^{\int p(t) \, dt}
\]

Here, \( p(t) = -\frac{3}{5} \), so:

\[
\mu(t) = e^{\int -\frac{3}{5} \, dt} = e^{-3t/5}
\]

### Step 3: Multiply the differential equation by the integrating factor
Multiplying both sides of the equation by \( e^{-3t/5} \):

\[
e^{-3t/5}y' - \frac{3}{5}e^{-3t/5}y = -\frac{1}{5}e^{-6t/5}
\]

The left side is now the derivative of the product \( e^{-3t/5}y \):

\[
\frac{d}{dt}\left(e^{-3t/5}y\right) = -\frac{1}{5}e^{-6t/5}
\]

### Step 4: Integrate both sides with respect to \( t \)
Integrating both sides:

\[
e^{-3t/5}y = \int -\frac{1}{5}e^{-6t/5} \, dt
\]

The integral on the right side is:

\[
\int -\frac{1}{5}e^{-6t/5} \, dt = \frac{1}{6}e^{-6t/5} + C
\]

So:

\[
e^{-3t/5}y = \frac{1}{6}e^{-6t/5} + C
\]

### Step 5: Solve for \( y(t) \)
Multiply both sides by \( e^{3t/5} \) to solve for \( y(t) \):

\[
y(t) = \frac{1}{6}e^{-3t/5} + Ce^{3t/5}
\]

### Step 6: Apply the initial condition
Use the initial condition \( y(0) = a \):

\[
a = \frac{1}{6} + C
\]

So,

\[
C = a - \frac{1}{6}
\]

### Final solution
Substitute \( C \) back into the solution:

\[
y(t) = \frac{1}{6}e^{-3t/5} + \left(a - \frac{1}{6}\right)e^{3t/5}
\]

This is the solution to the initial value problem.

---

Would you like more details or have any questions?

Here are some related questions you might consider:
1. How do you find the integrating factor in a first-order linear differential equation?
2. What is the significance of the initial condition in solving differential equations?
3. How does the solution change if the coefficient of \( y' \) is not constant?
4. Can you explain the physical meaning of this differential equation in an application?
5. What happens to the solution if the right-hand side is a different function of \( t \)?

**Tip:** Always double-check your work when solving differential equations by substituting your solution back into the original equation.

Learn how to solve the initial value problem -5y' + 3y = e^(-3t/5) with detailed step-by-step instructions. This problem involves first-order linear differential equations and integrating factors. Suitable for advanced college-level mathematics.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation