To solve the initial-value problem using the Laplace transform, we first take the Laplace transform of both sides of the differential equation. This will convert the differential equation into an algebraic equation in terms of \( Y(s) \), the Laplace transform of \( y(t) \). We then solve for \( Y(s) \) and take the inverse Laplace transform to find \( y(t) \).
To solve the initial-value problem using the Laplace transform, we will follow these steps:
The given differential equation is:
\[
y^{\prime} + 5y = e^{3t}
\]
Taking the Laplace transform of both sides, we have:
\[
\mathcal{L}\{y^{\prime} + 5y\} = \mathcal{L}\{e^{3t}\}
\]
Using the linearity of the Laplace transform, this becomes:
\[
\mathcal{L}\{y^{\prime}\} + 5\mathcal{L}\{y\} = \mathcal{L}\{e^{3t}\}
\]
The Laplace transform of the derivative \(y^{\prime}\) is:
\[
\mathcal{L}\{y^{\prime}\} = sY(s) - y(0)
\]
where \(Y(s)\) is the Laplace transform of \(y(t)\).
Substituting \(y(0) = 2\), we have:
\[
sY(s) - 2 + 5Y(s) = \frac{1}{s-3}
\]
Combine terms involving \(Y(s)\):
\[
(s + 5)Y(s) - 2 = \frac{1}{s-3}
\]
Rearrange to solve for \(Y(s)\):
\[
(s + 5)Y(s) = \frac{1}{s-3} + 2
\]
\[
Y(s) = \frac{1}{s-3} \cdot \frac{1}{s+5} + \frac{2}{s+5}
\]
For the first term:
\[
\frac{1}{(s-3)(s+5)} = \frac{A}{s-3} + \frac{B}{s+5}
\]
Solving for \(A\) and \(B\), we have:
\[
1 = A(s+5) + B(s-3)
\]
Setting \(s = 3\), we find:
\[
1 = A(3+5) \Rightarrow A = \frac{1}{8}
\]
Setting \(s = -5\), we find:
\[
1 = B(-5-3) \Rightarrow B = -\frac{1}{8}
\]
Thus:
\[
\frac{1}{(s-3)(s+5)} = \frac{1/8}{s-3} - \frac{1/8}{s+5}
\]
Now, we have:
\[
Y(s) = \frac{1/8}{s-3} - \frac{1/8}{s+5} + \frac{2}{s+5}
\]
Taking the inverse Laplace transform:
\[
y(t) = \frac{1}{8}e^{3t} - \frac{1}{8}e^{-5t} + 2e^{-5t}
\]
Combine like terms:
\[
y(t) = \frac{1}{8}e^{3t} + \frac{15}{8}e^{-5t}
\]
The solution to the initial-value problem is:
\[
\boxed{y(t) = \frac{1}{8}e^{3t} + \frac{15}{8}e^{-5t}}
\]
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