Question

1. (5 pts.) y'- y - sin (2x)

Answer 1

​Given,

$y'-y=sin(2x)$

​now taking laplace transform on both sides we get

$(s-1)Y(s)=\frac{2}{s^{2}+4}$

$Y(s)=\frac{2}{(s-1)(s^{2}+4)}$

now taking laplace inverse we get

$y(x)=L^{-1}\left\{\frac{2}{\left(s-1\right)\left(s^2+4\right)}\right\}$

this can be written as

$y(x)=L^{-1}\left\{\frac{-2s-2}{5\left(s^2+4\right)}+\frac{2}{5\left(s-1\right)}\right\}$

$y(x)=L^{-1}\left\{-\frac{2s}{5\left(s^2+4\right)}+\frac{2}{5\left(s-1\right)}-\frac{2}{5\left(s^2+4\right)}\right\}$

$y(x)=-\frac{2}{5}\cos \left(2x\right)-\frac{1}{5}\sin \left(2x\right)+\frac{2}{5}e^x$