Find general solution $y''y^3=1$ I'm having this question in my homework assignment in Linear Algebra and diffrential equation class, and trying to find the general solution for this second ODE.
$$y''y^3 = 1$$
Using substitution I said $p = y'$ and $p' = y''  \rightarrow  \frac{dp}{dx}= \frac{dp}{dy} \times \frac{dy}{dx}$
Then we get $y^3 \frac{dp}{dx}$ now what's the next step should I integrate or differentiate the equation? What's the trick on these type of ODEs?
Thanks in advance! 

Update This is now a complete solution.
Let $v(x)=y(x)^2$. Then $v'=2yy'$ and so (here we assume that $v\neq 0$)
$$
v''=2(y')^2+2yy''=\frac{1}{2}\frac{(v')^2}{y^2}+\frac{2}{y^2}=\frac{1}{2}\frac{(v')^2}{v}+\frac{2}{v},
$$
or
$$
vv''-\frac{1}{2}(v')^2-2=0
$$
This differential equation can be solved as follows. Differentiating, we find that
$$
v'v''+vv'''-v'v''=0,
$$
so $vv'''=0$. Hence $v'''=0$. But then $v$ must be a polynomial of degree $2$.
Since we differentiated we cannot expect any polynomial of degree $2$ to work. We insert a second degee polynomial $v=a+bx+cx^2$ into the second order differential equation and look for conditions on $a$, $b$ and $c$ that gives us a solution. The condition becomes
$$
2ac-\frac{b^2}{2}-2=0.
$$
Solving for $c$, we find that
$$
v(x)=a+bx+\frac{4+b^2}{4a}x^2.
$$
Then, one could go back to $y$ to get
$$
y(x)=\pm\sqrt{a+bx+\frac{4+b^2}{4a}x^2}.
$$
