1
$\begingroup$

guys, I can solve this by using the convolution theorem however when it comes to Laplace I'm stuck somehow. Can someone help me with this, please? $$ \frac{dy}{dt}+2\int_{0}^{t}y(\tau)cosh(t-\tau))d\tau = 4 + \delta (t) \quad ;y(0)=3 $$

Thanks for your help in advance.

$\endgroup$
1
  • $\begingroup$ Welcome to MSE. On this site, saying "I'm stuck" is not enough to guarantee a reply. You should say exactly what you have tried and where you got stuck, so someone can provide the right help for you. $\endgroup$ – Peter Phipps Apr 6 '20 at 18:58
1
$\begingroup$

$$\frac{dy}{dt}+2\int_{0}^{t}y(\tau)\cosh (t-\tau))d\tau = 4 + \delta (t) \\ y(0)=3$$ Note that: $$\mathcal{L}(\delta (t))=1 \text { and } \mathcal{L}(\cosh t)=\dfrac s{s^2-1}$$ Now apply the Laplace Transform: $$sY(s)-y(0)+2Y(s)\dfrac s{s^2-1} = \dfrac 4 s + 1$$ $$sY(s)\left ( 1+\dfrac 2{s^2-1} \right ) = \dfrac 4 s + 4$$ $$Y(s)=\dfrac {4(s+1)(s^2-1)}{s^2(s^2+1)}$$ $$Y(s)=4(s+1) \left (\dfrac 2{s^2+1}-\dfrac 1{s^2} \right)$$ $$Y(s)=4\left (-\dfrac 1{s}-\dfrac 1{s^2}+\dfrac {2s}{s^2+1}+\dfrac 2{s^2+1} \right)$$ Apply inverse Laplace Transform. $$\boxed {y(t)=4\left (-1-t+2 \cos t+2 \sin t \right)}$$

$\endgroup$
3
  • $\begingroup$ I do not get it why there is $s^2+1$ in the third equation's denominator, could you explain it, please? $\endgroup$ – sheldoniscute Apr 7 '20 at 11:18
  • $\begingroup$ Look at the table its the Laplace transform of cosh function. Third equation has $s^2-1$ at the denominator $\endgroup$ – Aryadeva Apr 7 '20 at 14:43
  • $\begingroup$ Is it here ? $sY(s)\left ( 1+\dfrac 2{s^2-1} \right ) = \dfrac 4 s + 4$ Then $sY(s)\left ( \dfrac {s^2+1}{s^2-1} \right ) = \dfrac 4 s + 4$ @sheldoniscute $\endgroup$ – Aryadeva Apr 7 '20 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.