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.
$$\frac{dy}{dt}+2\int_{0}^{t}y(\tau)\cosh (t-\tau))d\tau = 4 + \delta (t) \\ y(0)=3$$ 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. $$\boxed {y(t)=4\left (-1-t+2 \cos t+2 \sin t \right)}$$