# Solving an ODE with a piecewise function

How can I solve $$y'' + y = \begin{cases} \sin(t) & 0 \leq t < \pi \\ \cos(t) & \pi \leq t < \infty \end{cases}$$

using Laplace transforms? I also have the initial conditions $$y(0) = 1$$ and $$y'(0) = 0$$.

I'm familiar with doing Laplace transforms when the functions on the RHS are much simpler; however, I'm sort of confused about how to handle the piecewise function.

I tried doing the integral definition of Laplace transform, but it got really messy, so I think there is a better way to do it. The book I'm using had some examples which cleverly used the heaviside function, but I'm not sure about how I would go about doing that since we are dealing with trignometric functions here.

I would really appreciate any help.

• Solve over $[0,\pi]$. Obtain $y (\pi)$ and use it as the initial condition for $t \geq \pi$. – Rodrigo de Azevedo Mar 26 at 5:39

\begin{align} f(t) &= \sin(t)\cdot\big[u(t)-u(t-\pi)\big] + \cos(t)\cdot u(t-\pi) \\ &= \sin(t)\cdot u(t) + \big[-\sin(t) + \cos(t)\big]\cdot u(t-\pi) \\ &= \sin(t)\cdot u(t) + \big[-\sin\big((t-\pi)+\pi\big) + \cos\big((t-\pi)+\pi\big)\big]\cdot u(t-\pi) \\ &= \sin(t)\cdot u(t) + \big[\sin(t-\pi) - \cos(t-\pi)\big]\cdot u(t-\pi) \\ \end{align}
where $$u(t)$$ denotes the Heaviside step function. Using the shift theorem we obtain the Laplace transform
$$F(s) = \frac{1}{s^2+1} + e^{-\pi s}\left[\frac{1}{s^2+1} - \frac{s}{s^2+1}\right]$$