$$y''+3y'+2y=\cos(at) $$

where a is a constant. Find solution of this initial value problem in terms of convolution integral.$( y(0)=1$ and $ y'(0)=0 )$

I couldn't solve this problem with convolution integral. If somebody helps me, I'll appreciate.

I found: $$Y(s)=\dfrac 1 {s^2+3s+2} \left(\dfrac s {s^2+a^2}+s+3 \right)$$ with Laplace transformation.After this, I have to use convolution integral. When I use it, the integral becomes too difficult. Actually, I found the result of the integral but I think the result is wrong.

  • $\begingroup$ I hope I didn't change your math $\endgroup$ – Aryadeva Mar 9 '20 at 22:18
  • $\begingroup$ Thanks a lot. This is so helpful. $\endgroup$ – mahmut ali durukan Mar 10 '20 at 16:24
  • $\begingroup$ yx Mahmut ..... $\endgroup$ – Aryadeva Mar 10 '20 at 16:25

Use fraction decomposition $$Y(s)=\dfrac 1 {s^2+3s+2} \left(\dfrac s {s^2+a^2}+s+3 \right)$$ Note that we have: $$s^2+3s+2=(s+2)(s+1)$$ And $$f(s)=\dfrac {s+3} {(s+2)(s+1)}=\dfrac {2} {(s+1)}-\dfrac {1} {(s+2)}$$ Find the inverse Laplace transform of $f(s)$ $$\mathcal{L^{-1}} \{f(s)\}=2e^{-t}-e^{-2t}$$ On the other part $$g(s)=\dfrac s {(s^2+a^2)(s+2)(s+1)}$$ $$g(s)=\dfrac 2 {(s^2+a^2)(s+2)}-\dfrac 1 {(s^2+a^2)(s+1)}$$ Here you have to use the convolution integral formula.For the left fraction you can write: $$h(s)=\dfrac 2 {(s^2+a^2)(s+2)}=\mathcal {L} \left (\dfrac 1a \sin(at)\right ) \mathcal{L}\left (2e^{-2t} \right)$$ $$ \implies h(t)=\frac 2 a\int_0^t \sin(a\tau)e^{-2(t-\tau)}d\tau$$ So that $$g(t)=\frac 2 a\int_0^t \sin(a\tau)e^{-2(t-\tau)}d\tau-\frac 1a\int_0^t \sin(a\tau)e^{-(t-\tau)}d\tau$$ $$g(t)= \dfrac {2e^{-2t}} a\int_0^t \sin(a\tau)e^{2\tau}d\tau- \dfrac {e^{-t}}a\int_0^t \sin(a\tau)e^{\tau}d\tau$$ Evaluate both integrals.

Then you can deduce $y(t)$ since: $$y(t)=g(t)+f(t)$$ $$y(t)=g(t)+2e^{-t}-e^{-2t}$$

  • $\begingroup$ You found g(t) and f(t). But I have trouble with finding y(t). You said Y(s)=f(s)*g(s). You found y(s) and g(s). To find y(t), we need to use convolution again. Am I missing something? $\endgroup$ – mahmut ali durukan Mar 10 '20 at 16:30
  • $\begingroup$ @mahmutalidurukan I added some lines at the end. You only need to evalute the two integrals and deduce the answer. Mahmut $\endgroup$ – Aryadeva Mar 10 '20 at 16:33
  • $\begingroup$ I thought y(t)=f(t)*g(t). That's my mistake. Huge favor. Thanks again $\endgroup$ – mahmut ali durukan Mar 10 '20 at 16:36
  • $\begingroup$ No @mahmutalidurukan it's a sum of functions not a product of functions. So you just need to add both functions to get $y(t)$. Yo're welcome Mahmut $\endgroup$ – Aryadeva Mar 10 '20 at 16:37

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