To solve this initial value problem using the Laplace transform, we start by applying the Laplace transform to each term in the differential equation:

Given: $y'' - 3y' + 2y = 7e^t$ with initial conditions $y(0) = 5$ and $y'(0) = 4$.

Step 1: Apply the Laplace Transform

Using the Laplace transform properties for derivatives:

1. $\mathcal{L}\{y''(t)\} = s^2 Y(s) - s y(0) - y'(0)$
2. $\mathcal{L}\{y'(t)\} = s Y(s) - y(0)$
3. $\mathcal{L}\{y(t)\} = Y(s)$

Substitute these into the equation:

$s^2 Y(s) - s \cdot 5 - 4 - 3(s Y(s) - 5) + 2Y(s) = \mathcal{L}\{7e^t\}$

Step 2: Simplify the Right Side

Since $\mathcal{L}\{e^{at}\} = \frac{1}{s - a}$, we have:

$\mathcal{L}\{7e^t\} = \frac{7}{s - 1}$

Step 3: Substitute and Simplify

Now, substitute this and simplify:

$s^2 Y(s) - 5s - 4 - 3s Y(s) + 15 + 2Y(s) = \frac{7}{s - 1}$

Combine terms involving $Y(s)$:

$(s^2 - 3s + 2) Y(s) = \frac{7}{s - 1} + 5s + 11$

So, the expression for $\mathcal{L}\{y(t)\} = Y(s)$ in terms of $s$ is:

$Y(s) = \frac{\frac{7}{s - 1} + 5s + 11}{s^2 - 3s + 2}$

Would you like further details on each step?

1. How do you handle partial fractions in Laplace transforms?
2. What is the Laplace transform of higher-order derivatives?
3. How do initial conditions affect the Laplace transform solution?
4. Can you explain the inverse Laplace transform?
5. How does convolution work in Laplace transforms?

Tip: Always substitute initial conditions early to simplify the expression for $Y(s)$.