Hi

Ok i have attempted the following question, but need help on what should be done next:

$\displaystyle y'+8y=sin(x)$

p(x)=8

$\displaystyle q(x)=sin(x)$

$\displaystyle e^{\int p(x)dx}$ $\displaystyle = e^{\int 8xdx}$ $\displaystyle = e^{4x^2}$

$\displaystyle e^{4x^2}y'+8ye^{4x^2}=e^{4x^2}sinx$

$\displaystyle \frac{d(e^{4x^2}y)}{dx}=e^{4x^2}sin(x)$

$\displaystyle e^{4x^2} = \int e^{4x^2} sin(x)$

i used integration by parts

$\displaystyle u=e^{4x^2} du=8xe^{4x^2}$

$\displaystyle dv=sin(x)$ $\displaystyle v=-cos(x)$

$\displaystyle =-e^{4x^2}cos(x) - \int -8xe^{4x^2}cos(x)$

what should i do now??