I'm trying to solve the following D.E: (2yexp(y/x) + x)dy/dx = 2x + y
Here is my current working:
dy/dx = (2x+y) / (2yexp(y/x) + x)
sub y = ux
then dy/dx = (du/dx)x + u
dy/dx - u = (du/dx) x
(2 + u) / (2uexp(u) + 1) - u = (du/dx) x
2(1-u^2exp(u))/(2uexp(u) + 1) = (du/dx) x
So then I have to integrate 1/x dx (easy) but then also (2uexp(u) + 1) / 2 (1-u^2exp(u)) du . Actually this second integration is possible according to wolfram alpha but I can't see what method to use.
Any help would be much appreciated.