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The image you uploaded contains a problem related to Laplace Transformations. The task involves solving a differential equation using the Laplace transform method. Below is the problem statement and its solution, step by step.

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Problem Statement:

Given the following initial value problem (IVP):

$$
y''(t) + 4y'(t) + 5y(t) = e^{-t}, \quad y(0) = 1, \quad y'(0) = 0,
$$

solve for \( y(t) \) using the Laplace transform method.

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Solution:

#### Step 1: Take the Laplace Transform of the Differential Equation
We start by applying the Laplace transform to both sides of the given differential equation. Recall the properties of the Laplace transform:

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

Using these properties, we transform each term in the differential equation:

- \( \mathcal{L}\{y''(t)\} = s^2 Y(s) - sy(0) - y'(0) \)
- \( \mathcal{L}\{4y'(t)\} = 4(sY(s) - y(0)) \)
- \( \mathcal{L}\{5y(t)\} = 5Y(s) \)
- \( \mathcal{L}\{e^{-t}\} = \frac{1}{s+1} \)

Substitute the initial conditions \( y(0) = 1 \) and \( y'(0) = 0 \):

- \( \mathcal{L}\{y''(t)\} = s^2 Y(s) - s(1) - 0 = s^2 Y(s) - s \)
- \( \mathcal{L}\{4y'(t)\} = 4(sY(s) - 1) = 4sY(s) - 4 \)
- \( \mathcal{L}\{5y(t)\} = 5Y(s) \)

The transformed equation becomes:

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

Simplify the left-hand side:

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

Combine like terms:

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

Rearrange to isolate \( Y(s) \):

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

Factor out \( Y(s) \):

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

#### Step 2: Simplify the Right-Hand Side
Combine the terms in the numerator:

$$
\frac{1}{s+1} + s + 4 = \frac{1 + (s+4)(s+1)}{s+1} = \frac{1 + s^2 + 5s + 4}{s+1} = \frac{s^2 + 5s + 5}{s+1}
$$

Thus, the expression for \( Y(s) \) becomes:

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

#### Step 3: Perform Partial Fraction Decomposition
To simplify \( Y(s) \), we decompose it into partial fractions. Assume:

$$
\frac{s^2 + 5s + 5}{(s+1)(s^2 + 4s + 5)} = \frac{A}{s+1} + \frac{Bs + C}{s^2 + 4s + 5}
$$

Multiply through by the denominator \( (s+1)(s^2 + 4s + 5) \):

$$
s^2 + 5s + 5 = A(s^2 + 4s + 5) + (Bs + C)(s+1)
$$

Expand the right-hand side:

$$
s^2 + 5s + 5 = A(s^2 + 4s + 5) + Bs^2 + Bs + Cs + C
$$

$$
s^2 + 5s + 5 = As^2 + 4As + 5A + Bs^2 + Bs + Cs + C
$$

Combine like terms:

$$
s^2 + 5s + 5 = (A + B)s^2 + (4A + B + C)s + (5A + C)
$$

Equate coefficients of \( s^2 \), \( s \), and the constant term:

1. Coefficient of \( s^2 \): \( A + B = 1 \)
2. Coefficient of \( s \): \( 4A + B + C = 5 \)
3. Constant term: \( 5A + C = 5 \)

Solve this system of equations:

From \( A + B = 1 \):

$$
B = 1 - A
$$

Substitute \( B = 1 - A \) into \( 4A + B + C = 5 \):

$$
4A + (1 - A) + C = 5
$$

$$
3A + 1 + C = 5
$$

$$
3A + C = 4 \quad \text{(Equation 1)}
$$

Substitute \( B = 1 - A \) into \( 5A + C = 5 \):

$$
5A + C = 5 \quad \text{(Equation 2)}
$$

Subtract Equation 1 from Equation 2:

$$
(5A + C) - (3A + C) = 5 - 4
$$

$$
2A = 1 \implies A = \frac{1}{2}
$$

Substitute \( A = \frac{1}{2} \) back into \( 3A + C = 4 \):

$$
3\left(\frac{1}{2}\right) + C = 4
$$

$$
\frac{3}{2} + C = 4
$$

$$
C = 4 - \frac{3}{2} = \frac{8}{2} - \frac{3}{2} = \frac{5}{2}
$$

Substitute \( A = \frac{1}{2} \) into \( B = 1 - A \):

$$
B = 1 - \frac{1}{2} = \frac{1}{2}
$$

Thus, the partial fraction decomposition is:

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

#### Step 4: Inverse Laplace Transform
Now, take the inverse Laplace transform of each term:

1. \( \mathcal{L}^{-1}\left\{\frac{\frac{1}{2}}{s+1}\right\} = \frac{1}{2}e^{-t} \)

2. For the second term, rewrite \( \frac{\frac{1}{2}s + \frac{5}{2}}{s^2 + 4s + 5} \):

Complete the square in the denominator:

$$
s^2 + 4s + 5 = (s+2)^2 + 1
$$

Thus:

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

Split into two fractions:

$$
\frac{\frac{1}{2}(s+2)}{(s+2)^2 + 1} + \frac{\frac{3}{2}}{(s+2)^2 + 1}
$$

Take the inverse Laplace transform of each part:

- \( \mathcal{L}^{-1}\left\{\frac{\frac{1}{2}(s+2)}{(s+2)^2 + 1}\right\} = \frac{1}{2}e^{-2t}\cos(t) \)
- \( \mathcal{L}^{-1}\left\{\frac{\frac{3}{2}}{(s+2)^2 + 1}\right\} = \frac{3}{2}e^{-2t}\sin(t) \)

Combine these results:

$$
\mathcal{L}^{-1}\left\{\frac{\frac{1}{2}s + \frac{5}{2}}{s^2 + 4s + 5}\right\} = \frac{1}{2}e^{-2t}\cos(t) + \frac{3}{2}e^{-2t}\sin(t)
$$

#### Step 5: Combine All Terms
The complete solution is:

$$
y(t) = \frac{1}{2}e^{-t} + \frac{1}{2}e^{-2t}\cos(t) + \frac{3}{2}e^{-2t}\sin(t)
$$

#### Final Answer:
$$
\boxed{y(t) = \frac{1}{2}e^{-t} + \frac{1}{2}e^{-2t}\cos(t) + \frac{3}{2}e^{-2t}\sin(t)}
$$