Find the general solution for the following differential equation:
y'' + x(y')^2 = 0.
The Attempt at a Solution
Usually for these sort of DE you could use the substitution v(x) = y'(x) and this would simplify such an equation to a first order DE. The (y')^2 part is throwing me off as this would give the equation:
v(x)' + x(v(x))^2 = 0.
This would be grand if it weren't for the v(x)^2 term. Any ideas on how to get around this? Thanks.