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I want to solve the ODE $yy''=(y')^2-y'$ with the initial conditions $y(0)=1, y'(0)=2$.

My attempt:

$$yy''=(y')^2-y'$$ $$(\frac {y'}y)'=(\frac 1y)'$$ $$\frac {y'}y=\frac 1y+c$$

This holds for all $x$. Plugging the initial conditions for $x=0$, we get $c=1$. $$y'=1+y$$ Solving this got me to $y=-1$. It seems like a singular solution, but I used all methods I know and didn't get anything else.

Your ideas, please?


edit: I'm sorry about the silly question. I put my effort on getting through the hard part and lost my ability to perform simple calculations. Thank you for your answers.

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Your solution of $y'=1+y$ is incorrect. Hint: $$ {y'\over 1+y} = {d\over dx}\log(1+y) $$

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  • $\begingroup$ You are right. I can seperate variables. I tried other methods and forgot about this one... Thanks! $\endgroup$ – Whyka Feb 9 '16 at 23:05
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The full solution with the homogeneous part is $$ y=-1+C·e^x $$ and from the initial conditions $C=2$.

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  • $\begingroup$ Oh, of course. I forgot to add the homogenous part. Well, thank you $\endgroup$ – Whyka Feb 9 '16 at 23:04
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You got the hard part right. But the solution to $$y'=1+y$$ is $$\frac{dy}{1+y}=dx \\x + \ln K 1+y\\ y = K e^x -1 $$ And the initial conditions force $K=2$ so the answer is $$y = 2 e^x -1$$

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  • $\begingroup$ I put so much effort into the hard part that I forgot how to solve simple ODEs! Thanks $\endgroup$ – Whyka Feb 9 '16 at 23:08

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