The differential equation $\dot{y}\ddot{y}=xy$

Question

I was trying to solve

$$\dot{y}\ddot{y}=xy , y(0)=0 , \dot{y}(0)=0$$

I was wondering why I seemed to get the wrong answer. Here is my work.

$$\dot{y}\ddot{y}=xy$$

Let $$u(x,y)=\dot{y}$$

So $$\frac{\partial u}{\partial x}=\ddot{y}$$

So we get

$$u\frac{\partial u}{\partial x}=xy$$

Integrating both sides yields:

$$\frac{1}{2}u^2=\frac{1}{2}x^2y+C_{1}$$

$$y(0)=0\longrightarrow{C_{1}=0}$$

So we get

$$u^2=x^2y$$

$$\dot{y}=x\sqrt{y}$$

$$2\int\frac{1}{2\sqrt{y}}\,dy=\int x\,dx$$

$$2\sqrt{y}=\frac{1}{2}x^2+C_{2}$$

$$\dot{y}(0)=0\longrightarrow{C_{2}=0}$$

$$\sqrt{y}=\frac{1}{4}x^2$$

$$y=\frac{1}{16}x^4$$

The correct solution would be

$$y=\frac{1}{48}x^4$$

What went wrong?

Answer 1

If you set $$ \dot y(x)=u(x,y(x)) $$ where $u$ is specific to a solution, then taking the derivative for $x$ on both sides requires the total derivative on the right, or the application of the chain rule. That is $$ \ddot y(x)=\partial_xu+(\partial_yu)\,\dot y=\partial_xu(x,y(x))+\partial_yu(x,y(x))\,u(x,y(x)) $$

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Inserting that into the DE gives the PDE $$ uu_x+u^2u_y=xy. $$ This has a Lagrange system for characteristic curves $$ \frac{dx}{z}=\frac{dy}{z^2}=\frac{dz}{xy},~~~z=u(x,y) $$ I do not see how this could decouple to get constants along the characteristics.