Struggling with Solving a Diff Eq? Get Expert Help Here!

[noranne]

Never expected to be pleading for help so soon, and especially not on a differential equation, which I usually am good at. But for whatever reason, I cannot solve this problem:

y*d(y,x,2) + (d(y,x))^2 + 1 = 0

Any help would be greatly appreciated!

ETA: I know I'm supposed to substitute u=d(y,x) and u*d(u,y)=d(y,x,2) but I can't get any farther than that.

$$0 = y \frac {d^2y} {dx^2} + (\frac {dy} {dx})^2 + 1$$

 

[Mindscrape]

Could you clarify your notation, is this an ODE, PDE, what? This looks a nonlinear ODE if I follow you right.

There is a crash course in LaTeX here:
https://www.physicsforums.com/showthread.php?t=8997

 

[noranne]

Sorry, yeah it's an ODE. I know the notation is a little weird but it's the easiest way for me to type it.

 

[Dick]

I think what you want to notice is that (y*y')'=y''*y+y'*y'. So you want u=y*y'. In terms of u you have a first order ode. Once you've solved for u, then it's separable.

 

[noranne]

Yeah, but I can't separate it! Or rather, I can separate it, but I get completely unworkable results. It tends to fall apart when I get to

-ln(u^2 + 1)/2 = ln(y) + C

 

[Dick]

With this substitution the ode becomes u'+1=0. Can you solve that? I think you can.

 

[noranne]

Ohh, okay, I see how that's different than the substitution I was using. Let me try this again (again).

 

[noranne]

Success! $$y = \sqrt{-x^2 + x + 1}$$

Thank you SO much! No one in my class has been able to get that, we've been frantically IMing back and forth all night.

 

[Dick]

Sorry to rain on your parade, but you still haven't got it. Aside from the fact it's simply wrong, a second order ode should have two undetermined constants. Where are they? I think you know the general pattern of the solution. Try and do it again, carefully this time.

 

[noranne]

I guess I didn't mention that I was given two BC ( y(1)=1 and y'(1)=0 ) and I was able to solve for them. And my final answer checks out. Phew.

 

[Dick]

Ok. Guess that works.