# How can i solve this second-order differential equation?

My Problem is, this given second order non-linear differential equation: $$yy^{\prime\prime}=2(y^{\prime})^{2}-2y^{\prime}$$

I am really stuck with second-order differential equations.

My Approach: i was given the advise to solve first $$y^{\prime}=p(y)$$ and to determine $p$ but i didn't managed to get it.

I thought this: $$yy^{\prime\prime}=2(y^{\prime})^{2}-2y^{\prime}$$ could lead to: $$2y^{\prime}=2(y^{\prime})^{2}-yy^{\prime\prime}$$ $$y^{\prime}=(y^{\prime})^{2}-\frac{yy^{\prime\prime}}{2}$$

But i don't get it and i can't solve the equation.

• @TonyPiccolo: I think you mean $\dfrac{\tan\left(\sqrt{C_1} (x+C_2)\right)}{\sqrt{C_1}}$ Jun 24 '13 at 11:33
• $y''=\frac{dy'}{dx} = \frac{dy}{dx}\frac{dy'}{dy}$ Jun 24 '13 at 11:35
• @Robert Israel: Yes, thank you. I removed the comment because I was not sure. Jun 24 '13 at 16:11

Hint: You can write your differential equation in the form $$\dfrac{y''}{y' - 1} = 2 \dfrac{y'}{y}$$ and both sides can be integrated...

• No, $\int 2 \dfrac{y'}{y}\ dx = 2 \ln y + C$. And similarly for $\int \dfrac{y''}{y'-1}\ dx$. Jun 24 '13 at 14:47
• oh okay. But $\int \dfrac{y''}{y'-1}\ dx=ln(y'-1)+C$... and this would result in $$ln(y'-1)+C=2 \ln y+C$$ ... right? Jun 24 '13 at 19:25
• Right (of course, the two $C$'s are not necessarily equal). Now exponentiate to get rid of the logs. Jun 24 '13 at 20:02
• okay, should i get this in the end? $$y=\tan (x+C)$$ Jun 24 '13 at 20:11
• $y' = 1 + A y^2$ is a separable differential equation. Jun 26 '13 at 6:46

ODE has trivial solution $y=0$.

Now we have to solve case $y\neq 0$:

After dividing equation with $y^3$: $$\frac{y y''-2y'^2}{y^3}=\frac{-2y'}{y^3},$$ we get $(\frac{y'}{y^2})'$ on the left side and $(\frac{1}{y^2})'$ on the right side. Therefore, $$\frac{y'}{y^2}=\frac{1}{y^2}+C,$$ $$y'=1+Cy^2,$$ $$\frac{dy}{1+Cy^2}=dx.$$ After integration, we have $$\frac{1}{\sqrt{C}}\tan^{-1}(\sqrt{C}y)=x+D.$$ Finally, $$\frac{1}{\sqrt{C}}\tan{(\sqrt{C}(x+D))}=y.$$