:Problem

Solve the following equation:

$\displaystyle x^4y'=-x^3y-sec(xy)$

:Solution

Rearrange:

$\displaystyle (x^3y+sec(xy))dx+x^4dy=0$

$\displaystyle sec(xy)+x^3ydx+x^4dy=0$

$\displaystyle sec(xy)+x^3(ydx+xdy)=0$

$\displaystyle sec(xy)+x^3d(xy)=0$

Devide by $\displaystyle sec(xy)x^3$

$\displaystyle x^{-3}+cos(xy)d(xy)=0$

Now, I stopped.

There is no dx with 1/x^3 so that i can integrate it!

any help?