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apolski
Jan5-09, 12:12 PM
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
Jan5-09, 01:31 PM
That's a good start. Now integrate ((y')^2)'=(-2).

apolski
Jan5-09, 01:52 PM
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
Jan5-09, 01:59 PM
If (y')^2=c-2x then y' is either +sqrt(c-2x) or -sqrt(c-2x). There are two solutions.

apolski
Jan5-09, 02:00 PM
Thanx! o:)