Let's go through the initial value problem and solve it step-by-step.

The differential equation given is: $y'' - 3y' + 2y = 7e^{-t}$ with initial conditions: $y(0) = 5, \quad y'(0) = 4.$

Step 1: Take the Laplace Transform of Both Sides

We apply the Laplace Transform to each term in the equation. Recall the following Laplace Transform properties:

1. $\mathcal{L}\{y''(t)\} = s^2 Y(s) - sy(0) - y'(0)$,
2. $\mathcal{L}\{y'(t)\} = s Y(s) - y(0)$,
3. $\mathcal{L}\{y(t)\} = Y(s)$,
4. $\mathcal{L}\{e^{-t}\} = \frac{1}{s + 1}$.

Applying the Laplace Transform to each term:

$\mathcal{L}\{y''\} - 3 \mathcal{L}\{y'\} + 2 \mathcal{L}\{y\} = 7 \mathcal{L}\{e^{-t}\}.$

Substitute the transforms of $y$, $y'$, and $y''$, as well as the initial conditions $y(0) = 5$ and $y'(0) = 4$:

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

Simplifying each term:

1. $s^2 Y(s) - 5s - 4$,
2. $-3s Y(s) + 15$,
3. $2 Y(s)$,
4. Right-hand side: $\frac{7}{s + 1}$.

Now combine these terms:

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

Combine $Y(s)$ terms on the left side:

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

This gives the equation for $\mathcal{L}\{y(t)\} = Y(s)$ that you can use for further solution steps. At this stage, partial fraction decomposition is not yet required.

Would you like to proceed with further details?

Additional Questions to Explore

1. How would you apply partial fraction decomposition to solve for $Y(s)$?
2. What is the inverse Laplace Transform of $Y(s)$ once simplified?
3. How do initial conditions affect the Laplace-transformed differential equations?
4. What are the applications of the Laplace Transform in engineering fields?
5. How would this method change for non-homogeneous differential equations with different forcing functions?

Tip:

In Laplace transforms, always keep initial conditions in mind, as they affect the transformed equation significantly.