To solve this problem, we need to follow these steps:

Step 1: Set up the Laplace Transform

The differential equation is: $y'' + 4y = f(t),$ where $f(t) = \begin{cases} \sin(\pi t), & 0 \leq t < 1 \\ 0, & t \geq 1 \end{cases}$ and the initial conditions are $y(0) = 0$ and $y'(0) = 0$.

Taking the Laplace transform of both sides, we get: $\mathcal{L}\{y''\} + 4\mathcal{L}\{y\} = \mathcal{L}\{f(t)\}.$ Using the properties of the Laplace transform for derivatives, we have: $s^2 Y(s) - s y(0) - y'(0) + 4 Y(s) = F(s).$ Since $y(0) = 0$ and $y'(0) = 0$, this simplifies to: $(s^2 + 4) Y(s) = F(s).$

Step 2: Laplace Transform of $f(t)$

The function $f(t)$ involves a piecewise definition with a sine term. We can express it in terms of the unit step function $u(t - 1)$: $f(t) = \sin(\pi t) - \sin(\pi t) u(t - 1).$ The Laplace transform of $f(t)$ becomes: $F(s) = \mathcal{L}\{\sin(\pi t)\} - \mathcal{L}\{\sin(\pi t) u(t - 1)\}.$ Using the Laplace transform of $\sin(\pi t)$, which is $\frac{\pi}{s^2 + \pi^2}$, and the second shifting theorem, we get: $F(s) = \frac{\pi}{s^2 + \pi^2} - e^{-s} \cdot \frac{\pi}{s^2 + \pi^2}.$

Step 3: Solve for $Y(s)$

Now, we substitute $F(s)$ into our transformed equation: $(s^2 + 4) Y(s) = \frac{\pi}{s^2 + \pi^2} - e^{-s} \cdot \frac{\pi}{s^2 + \pi^2}.$ Thus, $Y(s) = \frac{\pi}{(s^2 + 4)(s^2 + \pi^2)} - \frac{\pi e^{-s}}{(s^2 + 4)(s^2 + \pi^2)}.$

Step 4: Inverse Laplace Transform to Find $y(t)$

Now, we need to find $y(t)$ by taking the inverse Laplace transform of $Y(s)$.

1. For the first term $\frac{\pi}{(s^2 + 4)(s^2 + \pi^2)}$, use the partial fraction decomposition provided in the note: $\frac{\pi}{(s^2 + \pi^2)(s^2 + 4)} = \frac{\pi}{\pi^2 - 4} \left( \frac{1}{s^2 + 4} - \frac{1}{s^2 + \pi^2} \right).$ This corresponds to inverse transforms involving $\sin(2t)$ and $\sin(\pi t)$.

2. For the second term $-\frac{\pi e^{-s}}{(s^2 + 4)(s^2 + \pi^2)}$, the factor $e^{-s}$ indicates a shift by 1 in $t$. This will involve using the step function $u(t - 1)$ in the inverse transform.

Combining these results will yield the solution for $y(t)$ in terms of sine functions and the step function $u(t - 1)$.

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Related Questions

1. How can we handle piecewise functions in the Laplace domain?
2. What is the significance of the initial conditions in Laplace transforms?
3. How does the second shifting theorem apply to Laplace transforms?
4. What role does partial fraction decomposition play in inverse Laplace transforms?
5. How are unit step functions used in modeling discontinuities in differential equations?

Tip

When working with inverse Laplace transforms, always consider simplifying the expression first using partial fractions, as it makes the inverse process more straightforward.