Solving Nonlinear ODE: y'y''=-1 - Get Help Now

[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. [tex](y')^2=-2x+c[/tex] and

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

but i see that [tex]y(x)=\frac{2\sqrt{2}}{3}\sqrt{(c_1 - x)^3}+c_2[/tex] 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!