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I have this: $$uu''-u'^2=u'$$ $$u'=\frac{du}{dx}$$

I can't solve it. I've done the recommended substitution $u'=v(u)$, so I have: $$u''=v'=dv/dx=\frac{dv}{du}\frac{du}{dx}=\frac{dv}{du}u'$$ Substituting that in the initial equation I get $$u\frac{dv}{du}=v+1$$ So: $v=uc-1$ I don't know how to go on o get $u(x)$

EDIT: I've done this: $$u'(x)=v(u)$$ $$\frac{dv}{du}=\frac{du'}{du}=0$$ That last step is where I have doubts. $$uu''-u'^2=u'\Longrightarrow uu'\frac{dv}{du}-u'^2=u'\Longrightarrow-u'^2=u'$$ So $u(x)=-x+c,\;\;c\in\mathbb{R}$ I'm missing solutions, right?

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I got $v=Ku-1$. Now you just continue with $du/dx=Ku-1$ and separate again. –  Maesumi Jan 19 '13 at 15:22

1 Answer 1

up vote 2 down vote accepted

(Assuming everything you've done so far is correct...)

By definition, $v(u)=u'$, so all you have to do now is solve $u'(x)=cu(x)-1$. This equation is separable, so can be easily solved.

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God I feel so stupid –  MyUserIsThis Jan 19 '13 at 15:22
@MyUserIsThis Don't be so hard on yourself! We all have these moments. –  000 Jan 19 '13 at 15:25
In fact, $u'=cu-1$, not $cu+1$. –  Did Jan 19 '13 at 15:28
@did Yes, that was a typo, it was fine in my paper. Thanks. –  MyUserIsThis Jan 19 '13 at 15:28
@did: The question was edited since my reply; I just copied what I saw, with $u'$ substituted for $v$. Edited. –  Clive Newstead Jan 19 '13 at 15:28

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