Laplace Transform IVP problem $y'''(t)-y(t) = 2$ with initial conditions $y(0)=0$, $y'(0)=0$, $y''(0)=1$.

Question

I'm trying to solve an IVP using Laplace Transforms and I'm stuck on the inverse. The problem is

$$y'''(t)-y(t) = 2$$ with initial conditions $y(0)=0$, $y'(0)=0$, $y''(0)=1$.

I solved for $\mathcal{L}\left\{y(t)\right\}$ using partial fractions, and I got $-2 / s + 2 / (3(s-1)) + (((4/3)s + (2/3)) / ((s+(1/2))^2 + (3/4))) + 1 /((s+(1/2))^2 + (3/4)))$. I believe there should be a cos and sin function in the inverse, I'm just stuck on finding them. I apologize I am not very familiar with the formatting. I appreciate any help.

Answer 1

With $y'''-y=2$ then \begin{align} {\cal L}(y) &= \dfrac{s+2}{s(s^3-1)} \\ &= -\dfrac{2}{s}+\dfrac{1}{s-1}+\dfrac{s+\frac12-\frac12}{(s+\frac12)^2+\frac34} \\ &= -\dfrac2s+\dfrac{1}{s-1}+\dfrac{s+\frac12}{(s+\frac12)^2+\frac34}-\frac12\dfrac{2}{\sqrt{3}}\dfrac{\frac{\sqrt{3}}{2}}{(s+\frac{1}{2})^2+\frac{3}{4}} \\ y&= -2+e^t+e^{-\frac12t}\cos\dfrac{\sqrt{3}}{2}t-\dfrac{1}{\sqrt{3}}e^{-\frac12t}\sin\dfrac{\sqrt{3}}{2}t \end{align}