problem: Linear equation question the question is:
$$(y^2+xy^3)\mathrm dx + (5y^2-xy+y^3\sin(y))\mathrm dy = 0$$
can any body tell me how to solve this linear equation?? when I tried to solve this the expression of integrating factor becomes too much difficult, may be i calculated it wrong... 
Any help will be appreciated. 
Thanks! 

$(y^2+xy^3)~dx+(5y^2-xy+y^3\sin y)~dy=0$
$(xy^3+y^2)~dx=(xy-5y^2-y^3\sin y)~dy$
$\left(x+\dfrac{1}{y}\right)\dfrac{dx}{dy}=\dfrac{x}{y^2}-\dfrac{5}{y}-\sin y$
Let $u=x+\dfrac{1}{y}$ ,
Then $x=u-\dfrac{1}{y}$
$\dfrac{dx}{dy}=\dfrac{du}{dy}+\dfrac{1}{y^2}$
$\therefore u\left(\dfrac{du}{dy}+\dfrac{1}{y^2}\right)=\dfrac{1}{y^2}\left(u-\dfrac{1}{y}\right)-\dfrac{5}{y}-\sin y$
$u\dfrac{du}{dy}+\dfrac{u}{y^2}=\dfrac{u}{y^2}-\dfrac{1}{y^3}-\dfrac{5}{y}-\sin y$
$u\dfrac{du}{dy}=-\dfrac{1}{y^3}-\dfrac{5}{y}-\sin y$
$u~du=\left(-\dfrac{1}{y^3}-\dfrac{5}{y}-\sin y\right)~dy$
$\int u~du=\int\left(-\dfrac{1}{y^3}-\dfrac{5}{y}-\sin y\right)~dy$
$\dfrac{u^2}{2}=\dfrac{1}{2y^2}-5\ln y+\cos y+c$
$\left(x+\dfrac{1}{y}\right)^2=\dfrac{1}{y^2}-10\ln y+2\cos y+C$
