Posted by anonymous on .
solve the following differential equation: (x^2+xy)dy=(xyy^2)dx

university maths 
drwls,
See if this works:
http://www.math.colostate.edu/~gerhard/M345/CHP/ch2_6.pdf
It does not look like it meets the "exact differential" requirements 
university maths 
Steve,
Good reference. Try looking at the last example:
Substitute y = xv ⇒ dy = v dx+x dv
(xyy^2)dx  (x^2+xy)dy = 0
(x^2v  x^2v^2)dx  (x^2 + x^2v)(vdx + xdv) = 0
2/x dx + (1+v)/v^2 dv = 0
2lnx + (lnv  1/v) = C
ln(x^2v)  1.v = C
vx^2 = Cexp(1/v)
y = vx
xy = Ce^{x/y}
Check my math, as always. 
university math 
drwls,
I want to congratulate and thank Steve for that very impressive solution.