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An actual 2nd order ODE ! Hard? Maybe?!!

Find the general solution of the differential equation
4xy′′+4x(y′)2+2y′−1=0 \displaystyle 4xy'' + 4x(y')^2+2y' -1= 0 .

It could be difficult right?
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Ano123
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Why is no one helping me? 81 views and still no one.
Is it too hard?
Define u(x)u(x) s.t. u′=uy′u'=uy', then note that:

u′′=u′y′+uy′′⇒4xy′′+4xu′2u2⏟4xy′2=4xu′′u u'' = u'y' + uy'' \Rightarrow 4xy''+ \underbrace{4x\frac{u'^2}{u^2}}_{4xy'^2} =\dfrac{4xu''}{u} ,

Which reduces our non-linear ODE to the more reasonable second order ODE:

4xu′′+2u′−u=04xu'' +2u' - u=0.
(edited 8 years ago)
Reply 3
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Ano123
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Original post by Farhan.Hanif93
Define u(x)u(x) s.t. u′=uy′u'=uy', then note that:

u′′=u′y′+uy′′⇒4xy′′+4xu′2u2⏟4xy′2=4xu′′u u'' = u'y' + uy'' \Rightarrow 4xy''+ \underbrace{4x\frac{u'^2}{u^2}}_{4xy'^2} =\dfrac{4xu''}{u} ,

Which reduces our non-linear ODE to the more reasonable second order ODE:

4xu′′+2u′−u=04xu'' +2u' - u=0.


Then what you gonna do from there though?
Original post by Ano123
Then what you gonna do from there though?

You're asking the wrong question - the question is: what are you going to do from here? The ODE is not too difficult at all from this point; if you want more help, post your working.

If you're looking for a hint: Can you spot a substitution that simplifies this?
Reply 5
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Ano123
OP
15
Original post by Farhan.Hanif93
You're asking the wrong question - the question is: what are you going to do from here? The ODE is not too difficult at all from this point; if you want more help, post your working.

If you're looking for a hint: Can you spot a substitution that simplifies this?


I've already got the answer doing it a different way - I don't need help. Seeing how others do it.

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