# Solving initial value problem using Laplace Transform

Use Laplace transform to solve the following initial–value problems.

a). $y'' + y = e^{−t}\cos 2t, \\ y(0) = 2, y′(0) = 1$

After using the concept of partial fraction and using Elementary Laplace Transforms I get:

$$\frac{S}{10(S^2+1)} + \frac{1}{5(s^2 + 1)} - \frac{S}{10((S + 1)^2)+4}$$

For the first and second fraction by inverse Laplace transform I get:

$$\frac{1}{10}\;\cos(t) +\frac{1}{5}\; \sin(t)\:$$ The third one is confusing me. I was trying $e^{at} \cos bt$. But there is a plus sign and no value for "a". Any ideas or suggestions to help me out here.

Assuming you have got the fractions correctly, the third fraction could be done as follows: $$\frac{s}{(s+1)^2+4}=\frac{s\color{red}{+1}\color{blue}{-1}}{(s+1)^2+4}=\frac{s\color{red}{+1}}{(s+1)^2+4}+\frac{\color{blue}{-1}}{(s+1)^2+4}\\=\frac{S}{S^2+4}|_{S\to s\color{red}{+1}}\color{blue}{-}\frac{1}2\frac{2}{S^2+4}|_{S\to s+1}$$ $$=e^{-t}\cos 2t\color{blue}{-}\frac{1}2e^{-t}\sin 2t$$
Note that: $$\mathcal{L}(e^{at}f(t))=F(s)|_{s\to s-1}$$
• @Avinesh: Welcome! $(-:$ Sep 13, 2013 at 9:12
• How did you get -1/2 in front of $e^{-t}$ Sep 13, 2013 at 9:13