Let's solve the initial value problem $2y' - 5y = e^{3t/4}$ with the initial condition $y(0) = a$.

Step 1: Rewrite the equation in standard form

The standard form of a first-order linear differential equation is:

$y' + p(t)y = q(t)$

To put the given equation into this form, divide the entire equation by 2:

$y' - \frac{5}{2}y = \frac{1}{2}e^{3t/4}$

Here, $p(t) = -\frac{5}{2}$ and $q(t) = \frac{1}{2}e^{3t/4}$.

Step 2: Find the integrating factor

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

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

So, $p(t) = -\frac{5}{2}$, and the integrating factor is:

$\mu(t) = e^{\int -\frac{5}{2} \, dt} = e^{-5t/2}$

Step 3: Multiply the differential equation by the integrating factor

Multiply both sides of the equation by $e^{-5t/2}$:

$e^{-5t/2}y' - \frac{5}{2}e^{-5t/2}y = \frac{1}{2}e^{3t/4} \cdot e^{-5t/2}$

Simplify the right-hand side:

$e^{-5t/2}y' - \frac{5}{2}e^{-5t/2}y = \frac{1}{2}e^{-\frac{5t}{2} + \frac{3t}{4}} = \frac{1}{2}e^{-\frac{10t}{4} + \frac{3t}{4}} = \frac{1}{2}e^{-\frac{7t}{4}}$

Now, the left side is the derivative of $e^{-5t/2}y$:

$\frac{d}{dt}\left(e^{-5t/2}y\right) = \frac{1}{2}e^{-7t/4}$

Step 4: Integrate both sides with respect to $t$

Integrate both sides:

$e^{-5t/2}y = \int \frac{1}{2}e^{-7t/4} \, dt$

The integral on the right side is:

$\int \frac{1}{2}e^{-7t/4} \, dt = -\frac{2}{7}e^{-7t/4} + C$

So:

$e^{-5t/2}y = -\frac{2}{7}e^{-7t/4} + C$

Step 5: Solve for $y(t)$

Multiply both sides by $e^{5t/2}$ to solve for $y(t)$:

$y(t) = -\frac{2}{7}e^{-7t/4} \cdot e^{5t/2} + Ce^{5t/2} = -\frac{2}{7}e^{(5t/2 - 7t/4)} + Ce^{5t/2}$

Simplify the exponent in the first term:

$y(t) = -\frac{2}{7}e^{\frac{10t - 7t}{4}} + Ce^{5t/2} = -\frac{2}{7}e^{3t/4} + Ce^{5t/2}$

Step 6: Apply the initial condition

Use the initial condition $y(0) = a$:

$a = -\frac{2}{7}e^{0} + C \cdot e^{0} = -\frac{2}{7} + C$

So:

$C = a + \frac{2}{7}$

Final solution

Substitute $C$ back into the solution:

$y(t) = -\frac{2}{7}e^{3t/4} + \left(a + \frac{2}{7}\right)e^{5t/2}$

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. What is the role of the integrating factor in solving linear differential equations?
2. How does changing the initial condition affect the solution of a differential equation?
3. What are some applications of first-order linear differential equations?
4. How does the method change if the differential equation is not linear?
5. What if the equation were non-homogeneous with a different type of forcing function?

Tip: Always check the consistency of the units when dealing with real-world applications of differential equations.