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.

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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.