How to find y from differential equation:
(y'')^2+y'=xy'' ?:)

Question

How to find y from differential equation: 
(y'')^2+y'=xy'' ?:)

anonymous · Answer

maby, substitution y'=yz and y''=y(z'+z^2)?

anonymous · Answer

notice there are no y term

u= y'

u' ^2 + u=x u'

u' + u/u'=x

anonymous · Answer

but then..?

anonymous · Answer

...u=u'(x-u').. have any idea? :)

anonymous · Answer

I have no idea, but here is how wolfram solves it http://www4a.wolframalpha.com/Calculate/MSP/MSP20719ibiid3766b33e500000g3d326gh6486h02?MSPStoreType=image/gif&s=38&w=350&h=863

anonymous · Answer

tnx for it ;)

jamesj · Answer

Nice problem.

anonymous · Answer

$$(y'')^2+y'=xy'' $$ is a nonlinear 2nd order differential equation. Some type of substitution is needed to put it in a linear form.

jamesj · Answer

@rolands, you have the answer now, yes?

anonymous · Answer

no..

jamesj · Answer

Ok, so here it is.  Using the substitution suggested where u = y', we have
$$ (u')^2 + u = xu' $$

Differentiate now that equation and  we have
$$ 2 u' u'' + u' = u' + xu'' $$
i.e.,
$$ u'' (2u' - x) = 0 $$
Consider now each case in turn...

jamesj · Answer

If u'' = 0, then u = C1.x + C2
If 2u' = x, then u = x^2/4 + C3

Now integrate u once more to recover y.

anonymous · Answer

tnx ;)