Help with a diff eq

1. Aug 29, 2007

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.

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

Last edited: Aug 29, 2007
2. Aug 29, 2007

Mindscrape

3. Aug 29, 2007

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.

4. Aug 29, 2007

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.

5. Aug 29, 2007

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

6. Aug 29, 2007

Dick

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

7. Aug 29, 2007

noranne

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

8. Aug 29, 2007

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.

9. Aug 29, 2007

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.

10. Aug 29, 2007

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.

11. Aug 29, 2007

Dick

Ok. Guess that works.

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