Solving the nonlinear ODE y'y''=-1

[apolski]

Hello, i am trying to solve this nonlinear ODE

y'y''=-1

can someone help me?

p.s maybe 2y'y''=-2 => (y'y')'=-2...

 

[Dick]

That's a good start. Now integrate ((y')^2)'=(-2).

 

[apolski]

οκ. $$(y')^2=-2x+c$$ and

$$y(x)=\int\sqrt{-2x+c}\hspace{3}dx+c_2=$$$$-\frac{2\sqrt{2}}{3}\sqrt{(c_1 - x)^3}+c_2$$

but i see that $$y(x)=\frac{2\sqrt{2}}{3}\sqrt{(c_1 - x)^3}+c_2$$ is sol'n too. How to show that?

 

[Dick]

If (y')^2=c-2x then y' is either +sqrt(c-2x) or -sqrt(c-2x). There are two solutions.

 

[apolski]

Thanx!