Consider y" + 3y' + 2y = e^-t. Find the transfer function Y that results when solving for yp. a) Y(s) = 1/0, no particular solution exists. b) 1/5 c) 1 d)-t

To find the transfer function \( Y(s) \) for the particular solution \( y_p \) of the differential equation:

\[
y'' + 3y' + 2y = e^{-t}
\]

we can proceed by applying the Laplace transform to both sides of the equation.

1. **Take the Laplace Transform of the Differential Equation**:
   \[
   \mathcal{L}(y'') + 3\mathcal{L}(y') + 2\mathcal{L}(y) = \mathcal{L}(e^{-t})
   \]

   Using the Laplace properties for derivatives:
   - \( \mathcal{L}(y'') = s^2 Y(s) - sy(0) - y'(0) \)
   - \( \mathcal{L}(y') = s Y(s) - y(0) \)
   - \( \mathcal{L}(y) = Y(s) \)

   Assuming initial conditions \( y(0) = 0 \) and \( y'(0) = 0 \), the transformed equation simplifies to:

   \[
   s^2 Y(s) + 3s Y(s) + 2 Y(s) = \frac{1}{s + 1}
   \]

2. **Factor the Left Side and Solve for \( Y(s) \)**:
   We can factor the left side as follows:
   \[
   (s^2 + 3s + 2) Y(s) = \frac{1}{s + 1}
   \]

   Factor \( s^2 + 3s + 2 \) to get \( (s + 1)(s + 2) \). Thus, we have:
   \[
   (s + 1)(s + 2) Y(s) = \frac{1}{s + 1}
   \]

   Divide both sides by \( (s + 1) \):
   \[
   Y(s) = \frac{1}{(s + 1)(s + 2)}
   \]

3. **Simplify the Expression**:
   The transfer function \( Y(s) \) for the particular solution is:

   \[
   Y(s) = \frac{1}{(s + 1)(s + 2)}
   \]

None of the answer options provided matches this form exactly, but \( Y(s) = \frac{1}{(s + 1)(s + 2)} \) is the correct result.

Would you like further details on each step, or do you have any specific questions?

### Relative Questions
1. How does the Laplace transform simplify solving differential equations?
2. What are the general steps for finding a particular solution using Laplace transforms?
3. Why are initial conditions important in solving differential equations?
4. How can we interpret the transfer function in physical systems?
5. What are alternative methods for finding particular solutions?

### Tip
Always check if the given options align with the derived solution, especially when none match exactly—this can be due to simplifications or possible errors in provided answer choices.

Consider y'' + 3y' + 2y = e^-t. Find the transfer function Y that results when solving for yp. a) Y(s) = 1/0, no particular solution exists. b) 1/5 c) 1 d)-t

Discover how to find the transfer function Y(s) for the differential equation y'' + 3y' + 2y = e^-t using Laplace transforms. This problem involves key concepts like differential equations, Laplace transforms, and solving for the transfer function. Detailed step-by-step solution provided.

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