# Solve y′(t)=sin(t)+int_0^t y(x)cos(t−x)dx such that y(0)=0

Solve ${y}^{\prime }\left(t\right)=\mathrm{sin}\left(t\right)+{\int }_{0}^{t}y\left(x\right)\mathrm{cos}\left(t-x\right)dx$ by Laplace transform
My try:
I applied Laplace transform on both sides of the equation.
$sL\left\{y\left(t\right)\right\}=\frac{1}{{s}^{2}+1}+L\left\{cos\left(t\right)\ast y\left(t\right)\right\}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}sL\left\{y\left(t\right)\right\}=\frac{1}{{s}^{2}+1}+L\left\{cos\left(t\right)\right\}×L\left\{y\left(t\right)\right\}$
Now, I'm stuck on applying the inverse Laplace transform on (*) to find $y\left(t\right)$
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Zackary Hatfield
$sL\left\{y\left(t\right)\right\}\left(s\right)=\frac{1}{{s}^{2}+1}+L\left\{cos\left(t\right)\right\}\left(s\right)×L\left\{y\left(t\right)\right\}\left(s\right)$
using
$L\left\{cos\left(t\right)\right\}\left(s\right)=\frac{s}{{s}^{2}+1}$
I rather get
$L\left\{y\left(t\right)\right\}\left(s\right)=\frac{1}{{s}^{3}}$
which is now standard to solve.
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evitagimm9h
$sL\left\{y\left(t\right)\right\}=\frac{1}{{s}^{2}+1}+L\left\{cos\left(t\right)\ast y\left(t\right)\right\}$
$sL\left\{y\left(t\right)\right\}=\frac{1}{{s}^{2}+1}+L\left\{y\left(t\right)\right\}\ast \frac{s}{{s}^{2}+1}$
$L\left\{y\left(t\right)\right\}\left(s-\frac{s}{{s}^{2}+1}\right)=\frac{1}{{s}^{2}+1}$
Here you made a mistake I guess
$L\left\{y\left(t\right)\right\}\left(\frac{{s}^{3}-s+s}{{s}^{2}+1}\right)=\frac{1}{{s}^{2}+1}$
$L\left\{y\left(t\right)\right\}=\frac{1}{{s}^{3}}$