# I Find the General solution of H.D.E

1. Feb 13, 2017

### Kbenjamin827

((2(xy)^1/2) - y)y' + x dy = 0

ive been at it for hours, substituting y=ux but im not sure if its definable to have both dy and y'. did the teacher just make a mistake?

2. Mar 2, 2017

### JJacquelin

The mistake is in the equation itself. You wrote ((2(xy)^1/2) - y)y' + x dy = 0 which is non-sens : the first term is finite while the second term is infinitesimal.