Question

y''y + (y')^2 = 0

Answer 1

does anyone know how to find the general solution for this one?

Answer 2

I think you have to use a sort of substitution. Let \(v=y'\) be a function of \(y\) (so treat \(y\) as the independent variable).

From this you have
\[y''=\frac{d}{dx}\left[\frac{dy}{dx}\right]=\frac{d}{dx}\left[v(y)\right]=\frac{dv}{dx}\]
Multiplying by \(\dfrac{dy}{dy}\), you can write it as
\[y''=\frac{dv}{dy}\cdot\frac{dy}{dx}=v~v'\]
So, subbing all the relevant expressions into the original equation gives you
\[(v~v')y+v^2=0\]
From here, you can factor:
\[v\left(v'~y+v\right)=0\]
which gives you two equations to solve, (1) \(v=0\) and (2) \(v'y+v=0\).

Answer 3

This technique works because the DE doesn't have the independent variable \(x\) in it .
So the derivative \(y'\) is a function of \(y\) only \(y'=v(y)\)
I my text book its called an \(x\)-absent DE