1. **State the problem:** We want to find the Laplace transform $Y$ of the solution $y(t)$ to the initial value problem (IVP): $$\frac{d^2y}{dt^2} + k^2 y = e^{5t}, \quad y(0) = 0, \quad y'(0) = 0.$$ 2. **Recall the Laplace transform properties:** - The Laplace transform of $y(t)$ is $Y = \mathcal{L}\{y(t)\}$. - The Laplace transform of the second derivative is: $$\mathcal{L}\left\{\frac{d^2y}{dt^2}\right\} = s^2 Y - s y(0) - y'(0).$$ - The Laplace transform of $e^{at}$ is: $$\mathcal{L}\{e^{at}\} = \frac{1}{s - a}, \quad \text{for } s > a.$$ 3. **Apply the Laplace transform to both sides of the differential equation:** $$\mathcal{L}\left\{\frac{d^2y}{dt^2}\right\} + k^2 \mathcal{L}\{y\} = \mathcal{L}\{e^{5t}\}$$ Substitute the transforms: $$s^2 Y - s y(0) - y'(0) + k^2 Y = \frac{1}{s - 5}.$$ 4. **Use initial conditions $y(0) = 0$ and $y'(0) = 0$:** $$s^2 Y + k^2 Y = \frac{1}{s - 5}.$$ 5. **Factor out $Y$:** $$Y (s^2 + k^2) = \frac{1}{s - 5}.$$ 6. **Solve for $Y$:** $$Y = \frac{1}{(s - 5)(s^2 + k^2)}.$$ 7. **Find $y(t)$ by inverse Laplace transform:** We use partial fraction decomposition: $$\frac{1}{(s - 5)(s^2 + k^2)} = \frac{A}{s - 5} + \frac{B s + C}{s^2 + k^2}.$$ Multiply both sides by $(s - 5)(s^2 + k^2)$: $$1 = A (s^2 + k^2) + (B s + C)(s - 5).$$ 8. **Equate coefficients:** For $s^2$ term: $A + B = 0$. For $s$ term: $-5 B + C = 0$. For constant term: $A k^2 - 5 C = 1$. From $A + B = 0$, we get $B = -A$. From $-5 B + C = 0$, substitute $B = -A$: $$-5(-A) + C = 0 \Rightarrow 5 A + C = 0 \Rightarrow C = -5 A.$$ Substitute $C$ into constant term equation: $$A k^2 - 5(-5 A) = 1 \Rightarrow A k^2 + 25 A = 1 \Rightarrow A (k^2 + 25) = 1 \Rightarrow A = \frac{1}{k^2 + 25}.$$ Then: $$B = -A = -\frac{1}{k^2 + 25}, \quad C = -5 A = -\frac{5}{k^2 + 25}.$$ 9. **Rewrite $Y$ with partial fractions:** $$Y = \frac{1}{k^2 + 25} \cdot \frac{1}{s - 5} - \frac{1}{k^2 + 25} \cdot \frac{s}{s^2 + k^2} - \frac{5}{k^2 + 25} \cdot \frac{1}{s^2 + k^2}.$$ 10. **Use inverse Laplace transforms:** - $\mathcal{L}^{-1}\left\{\frac{1}{s - a}\right\} = e^{a t}$ - $\mathcal{L}^{-1}\left\{\frac{s}{s^2 + b^2}\right\} = \cos(b t)$ - $\mathcal{L}^{-1}\left\{\frac{b}{s^2 + b^2}\right\} = \sin(b t)$ Rewrite the last term: $$\frac{1}{s^2 + k^2} = \frac{k}{k} \cdot \frac{1}{s^2 + k^2} = \frac{1}{k} \cdot \frac{k}{s^2 + k^2}.$$ So: $$\mathcal{L}^{-1}\left\{\frac{1}{s^2 + k^2}\right\} = \frac{1}{k} \sin(k t).$$ 11. **Final solution:** $$y(t) = \frac{1}{k^2 + 25} e^{5 t} - \frac{1}{k^2 + 25} \cos(k t) - \frac{5}{k (k^2 + 25)} \sin(k t).$$