# Almost done solving IVP using Laplace transform .. Need advice/guidance

I have the following IVP: $$\begin{cases}y''-2y'+5y = -8e^{-t},\\ \ \\ y(0) = 2\\ \ \\ y'(0) = 12\end{cases}$$

I will show my steps so far.

After taking the transform of each term I get:

$$s^2Y(s) - 2sY(s) + 5Y(s) - 2s - 8 = \frac {-8}{s+1}$$

I simplified the L.H.S and got: $$Y(s)(s^2-2s+5) = \frac {-8}{s+1} + 2s+ 8$$

Then the R.H.S becomes:

$$Y(s)(s^2-2s+5) = \frac {2s^2+10s}{s+1}$$

I then get:

$$Y(s) = \frac {2s^2+10s}{(s+1)(s^2-2s+5)}$$

I then used partial fraction decomposition and got $A = -1, B = 3$, and $C = 5$

I then have the following:

$$Y(s) = \frac {-1}{s+1} + \frac {3s+5}{s^2-2s+5}$$

I am now stuck finding the inverse Laplace transform of the second term.

Please can someone check my work and let me know if I did this right and, if so, what is that inverse Laplace transform I mentioned at the last step?

Help is greatly appreciated. Thank you.

You have $s^2-2s+5=(s-1)^2+4$. So the first term looks like $$\frac {3s+5}{(s-1)^2+4} =\frac {3(s-1)+8}{(s-1)^2+4} =\frac {3(s-1)}{(s-1)^2+4} +\frac {8}{(s-1)^2+4}.$$ Since $\frac{s}{s^2+4}$ is the transform of $\cos 2t$ and $\frac2{s^2+4}$ of $\sin 2t$, with the shifting rule you get $$\mathcal L^{-1}\left[\frac {3s+5}{s^2-2s+5}\right] =\mathcal L^{-1}\left[\frac {3s+5}{(s-1)^2+4}\right] =3e^t\cos2t+4e^t\sin 2t$$

• Thank you. Surprised I didn't think to complete the square Apr 30 '18 at 4:32

$$\frac {3s+5}{s^2-2s+5}=$$

$$\frac {3s+5}{(s-1)^2+4}=$$

$$\frac {3(s-1)}{(s-1)^2+4} +\frac {8}{(s-1)^2+4}$$

The inverse laplace transform is

$$3e^t\cos 2t +4e^t\sin 2t$$