Find the inverse Laplace transform of the giveb function by using the convolution theorem.

$$F(x) = \frac{s}{(s+1)(s^2+4)}$$

If I use partial fractions I get: $$\frac{s+4}{5(s^2+4)} - \frac{1}{5(x+1)}$$

which gives me Laplace inverses:

$$\frac{1}{5}(\cos2t + \sin2t) -\frac{1}{5} e^{-t}$$

But the answer is: $$f(t) = \int^t_0 e^{-(t -\tau)}\cos(2\tau) d\tau$$

How did they get that?

  • $\begingroup$ Where did you use the convolution theorem in your argument? $\endgroup$
    – tomcuchta
    Commented Dec 8, 2012 at 8:11
  • $\begingroup$ @tomcuchta I have the inverses so to put it in convolution form then all I have to do is plug it in accordingly but even if I do that my inverses are not correct, since the correct answer above doesn't represent the inverse I gave. $\endgroup$
    – Q.matin
    Commented Dec 8, 2012 at 8:13

1 Answer 1


Related techniques (I), (II). Using the fact about the Laplace transform $L$ that

$$ L(f*g)=L(f)L(g)=F(s)G(s)\implies (f*g)(t)=L^{-1}(F(s)G(s)) .$$

In our case, given $ H(s)=\frac{1}{(s+1)}\frac{s}{(s^2+4)}$

$$F(s)=\frac{1}{s+1}\implies f(t)=e^{-t},\quad G(s)=\frac{s}{s^2+4}\implies g(t)=\cos(2t).$$

Now, you use the convolution as

$$ h(t) = \int_{0}^{t} e^{-(t-\tau)}\cos(2\tau) d\tau . $$

  • $\begingroup$ @Q.matine: If you evaluate the last integral in the answer above by a proper substitution first and using integral by parts finally; you will have your own solution: $$h(t)=\frac{1}{5}(\cos 2t + \sin 2t) -\frac{1}{5} e^{-t}$$. $\endgroup$
    – Mikasa
    Commented Dec 8, 2012 at 8:46
  • $\begingroup$ I made it more difficult than it was. I dont know why I took the partial fractions. Thanks a lot Mhenni!! $\endgroup$
    – Q.matin
    Commented Dec 8, 2012 at 8:48
  • $\begingroup$ @BabakSorouh so what is the point of taking the convolution when I got the solution you just wrote by just taking partial fractions and the inverse laplace? $\endgroup$
    – Q.matin
    Commented Dec 8, 2012 at 8:51
  • 2
    $\begingroup$ @Q.matin: You are welcome. Note that,convolution techniques are very useful and they are used in solving many kinds of differential equations . $\endgroup$ Commented Dec 8, 2012 at 8:57
  • 2
    $\begingroup$ @Q.matin: I think there is not a certain reason for that. Maybe, the author wanted to do it by convolution. For example, we know many ode's which are separable and exact. If we are asked to see them as separable, so we will treat them as separable and if they wish to do exact method so do we. Your approach is right. $\endgroup$
    – Mikasa
    Commented Dec 8, 2012 at 9:07

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