To solve the initial value problem 2y′-5y=e^(3t/4) with the initial condition y(0)=a*y*(0)=*a*, we'll use the method of solving first-order linear differential equations.

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.

---

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.

Learn how to solve the initial value problem 2y′-5y=e^(3t/4) with the initial condition y(0)=a. This problem involves first-order linear differential equations and the method of integrating factors. Suitable for advanced undergraduate students and anyone interested in differential equations.

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