Solve $y'(t)=\operatorname{sin}(t)+\int_0^t y(x)\operatorname{cos}(t-x)dx$ by Laplace transform

Question

Question:

Solve $y'(t)=\operatorname{sin}(t)+\int_0^t y(x)\operatorname{cos}(t-x)dx$ such that $y(0)=0$

My try:
I applied Laplace transform on both sides of the equation.

$ sL\{y(t)\} = \frac{1}{s^2+1}+L\{cos(t)*y(t)\} \implies sL\{y(t)\}=\frac{1}{s^2+1}+L\{cos(t)\}\times L\{y(t)\} $
$\implies L\{y(t)\} = \frac{s^2-1}{(s^3-s-1)(s^2+1)} $ (*)

Now, I'm stuck on applying the inverse Laplace transform on (*) to find $y(t)$.

Any idea?

Answer 1

$sL\{y(t)\} = \frac{1}{s^2+1}+L\{cos(t)*y(t)\} $

$sL\{y(t)\} = \frac{1}{s^2+1}+L\{y(t)\}*\frac s {s^2+1} $

$L\{y(t)\}(s- \frac s {s^2+1})= \frac{1}{s^2+1}$

Here you made a mistake I guess

$L\{y(t)\}( \frac {s^3-s+s} {s^2+1})= \frac{1}{s^2+1} $

$L\{y(t)\}= \frac{1}{s^3} $

Answer 2

Hint. You are on the right track. But please check your results, since from your identity $$ sL\{y(t)\}(s)=\frac{1}{s^2+1}+L\{cos(t)\}(s)\times L\{y(t)\} (s) $$ using $$ L\{cos(t)\}(s)=\frac{s}{s^2+1} $$ I rather get $$ L\{y(t)\}(s)=\frac1{s^3} $$ which is now standard to solve.