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.


1 Answer 1


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: $$\mathcal{L}(e^{at}f(t))=F(s)|_{s\to s-1}$$

  • $\begingroup$ Amazing.Thanks alot for your help. $\endgroup$
    – Avinesh
    Sep 13, 2013 at 9:07
  • $\begingroup$ @Avinesh: Welcome! $(-:$ $\endgroup$
    – Mikasa
    Sep 13, 2013 at 9:12
  • $\begingroup$ How did you get -1/2 in front of $e^{-t}$ $\endgroup$
    – Avinesh
    Sep 13, 2013 at 9:13
  • 1
    $\begingroup$ Got it. To balance out 2 in the numerator. $\endgroup$
    – Avinesh
    Sep 13, 2013 at 9:16
  • $\begingroup$ I wish to post my final solution (Scanned image) for verification. Am I welcome to do so. $\endgroup$
    – Avinesh
    Sep 15, 2013 at 1:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .