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

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?

• First of all, if you're using $\dot y$ to represent $dy/dx$, then please write $u=\dot y$. No partial derivatives should appear, anyhow, since these are functions of $x$. Anyhow, your solution is not valid. You cannot keep $y$ in the problem; once you substitute $u=\dot y$, you must write the differential equation in terms of $u$ and $x$ only. When you integrate with respect to $x$, you have no idea how to integrate $xy$, but you pretended $y$ was constant. Commented Jul 4, 2023 at 17:33
• I'm not sure for this step $$u\frac{\partial u}{\partial x}=xy \implies \frac{1}{2}u^2=\frac{1}{2}x^2y+C_{1}$$
– user
Commented Jul 4, 2023 at 17:33
• The step "Integrating both sides" is wrong : you treated $y$ as it was a constant with respect to $x$. Commented Jul 4, 2023 at 18:05
• Does this answer your question? Solving the differential $\frac{y'y'''}{y''} = x$ Commented Jul 4, 2023 at 18:22
• @Gonçalo Yes, not exactly the same, but close enough to suggest an good approach. Commented Jul 4, 2023 at 18:54

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 $$\ddot y(x)=\partial_xu+(\partial_yu)\,\dot y=\partial_xu(x,y(x))+\partial_yu(x,y(x))\,u(x,y(x))$$
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.