Solve:
y''=yy'/x
And show that y=c_{3} is a solution of the problem too.
where y'=dy/dx
How to solve it?
Well I have a stab at it that might lead somewhere. But I can't finish it. Perhaps it will give someone an idea.
Do you have initial conditions on this?
Note that $\displaystyle \frac{1}{2} \frac{d}{dx}(y^2) = y y'$ and that $\displaystyle x y'' = \frac{d}{dx}(xy') - y'$.
Then
$\displaystyle y'' = \frac{y y'}{x}$
$\displaystyle x y'' = y y'$
$\displaystyle \int x y'' dx = \int y y' dx$
$\displaystyle \int \left ( \frac{d}{dx}(x y') - y' \right ) dx = \frac{1}{2}y^2 $
$\displaystyle xy' - \int y' dx = \frac{1}{2}y^2$
$\displaystyle xy' - y = \frac{1}{2}y^2 + a$
where a is the accumulated integration constant.
My hope was for a Bernoulli equation, but we have that extra constant that seems to screw things up. I can't find a way to proceed from here.
-Dan
Edit: I was hoping that, since y = constant was a solution (leading to a =0 in the above equation) would help, but it leads to a Bernoulli equation where y = constant is not a solution to the Bernoulli, so this doesn't help.