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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...
That's a good start. Now integrate ((y')^2)'=(-2).
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?
If (y')^2=c-2x then y' is either +sqrt(c-2x) or -sqrt(c-2x). There are two solutions.
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