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 The final answer should be (5s^2+22s+14)/((s+1)^2(s+2))

To solve this initial value problem using the Laplace transform, we will follow these steps.

The differential equation is:
\[
y'' + 3y' + 2y = -3e^{-t}
\]
with initial conditions:
\[
y(0) = 5, \quad y'(0) = 2
\]

### Step 1: Take the Laplace Transform of Both Sides

Let \(\mathcal{L}\{y(t)\} = Y(s)\). The Laplace transform of derivatives are:
- \(\mathcal{L}\{y'\} = sY(s) - y(0)\)
- \(\mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0)\)

Using the initial conditions \(y(0) = 5\) and \(y'(0) = 2\), the Laplace transform of the left side becomes:
\[
\mathcal{L}\{y'' + 3y' + 2y\} = \left(s^2Y(s) - s \cdot 5 - 2\right) + 3\left(sY(s) - 5\right) + 2Y(s)
\]

### Step 2: Expand and Simplify the Left Side

Expanding each term, we get:
\[
s^2Y(s) - 5s - 2 + 3sY(s) - 15 + 2Y(s)
\]

Combining terms involving \(Y(s)\) and constants:
\[
(s^2 + 3s + 2)Y(s) - 5s - 17
\]

### Step 3: Take the Laplace Transform of the Right Side

For the right side, \(-3e^{-t}\), we use the Laplace transform formula:
\[
\mathcal{L}\{-3e^{-t}\} = -\frac{3}{s + 1}
\]

### Step 4: Set Up the Equation and Solve for \(Y(s)\)

Now we have:
\[
(s^2 + 3s + 2)Y(s) - 5s - 17 = -\frac{3}{s + 1}
\]

Rearrange to solve for \(Y(s)\):
\[
Y(s) = \frac{5s + 17 - \frac{3}{s + 1}}{(s + 1)(s + 2)}
\]

### Step 5: Combine Terms

To make the expression simpler, combine terms over a common denominator:
\[
Y(s) = \frac{5s + 17)(s + 1) - 3}{(s + 1)^2(s + 2)}
\]

Expanding the numerator, we get the answer:
\[
Y(s) = \frac{5s^2 + 22s + 14}{(s + 1)^2(s + 2)}
\]

Thus,
\[
\mathcal{L}\{y(t)\} = \frac{5s^2 + 22s + 14}{(s + 1)^2(s + 2)}
\]

Would you like further steps for inverse transformation or have questions on any part of the process?

---

1. What is the purpose of using Laplace transforms in differential equations?
2. How do initial conditions affect the Laplace transform of a derivative?
3. Why is partial fraction decomposition useful in solving for inverse Laplace transforms?
4. How would the solution change if initial conditions were different?
5. What are common Laplace transforms for basic functions?

**Tip:** Always check initial conditions carefully when applying the Laplace transform to differential equations. They significantly impact the transformed equation.

Use the Laplace transform to solve the initial value problem: y'' + 3y' + 2y = -3e^{-t}, y(0) = 5, y'(0) = 2.

Step-by-step solution to the differential equation y'' + 3y' + 2y = -3e^{-t} with initial conditions y(0) = 5 and y'(0) = 2, using Laplace transforms. This solution covers core concepts like transforming derivatives, simplifying algebraic equations, and applying Laplace properties to solve differential equations.

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