Find the general solution of the equation:
1= [3y*cos(y^2)*dy/dx] + [3e^(-x) * ycos(y^2) * dy/dx]
really really stuck on this, any help would be appreciated!
If you factor the RHS you get
$\displaystyle
1 = 3y \cos \left(y^2\right) \left(1 + e^{-x}\right) \frac{dy}{dx}
$
at which point we recognize this is separable, so
$\displaystyle
\frac{dx}{1+e^{-x}} = 3y \cos \left(y^2\right)dy
$
or
$\displaystyle
\frac{e^x\,dx}{e^{x}+1} = 3y \cos \left(y^2\right)dy
$.
Now integrate each side.