How do I solve this differential equation?

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  • #1
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Homework Statement


Solve the following differential equation:


Homework Equations



2 y (y'')^2 + 2 y y''' y' -2 (y')^2 y'' = - (y')^2



The Attempt at a Solution



I don't know if the following is useful, but if you divide both sides by y^2, the LHS of the above becomes:

(2 y (y'')^2 + 2 y y''' y' -2 (y')^2 y'')/y^2 = ( (2 y' y'')/y )'

Which means the equation to solve is;

( (2 y' y'')/y )' = - (y')^2
 

Answers and Replies

  • #2
SteamKing
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If you divide both sides of your original DE by y^2, it's not obvious how that turns into the equations in section 3. above.
 
  • #3
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@SteamKing

Reverse quotient rule. All I know is that what I wrote is definitely correct. I just have no idea how to solve it.
 
  • #4
STEMucator
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@SteamKing

Reverse quotient rule. All I know is that what I wrote is definitely correct. I just have no idea how to solve it.

Steam king is right, you didn't divide through properly at all.

If this is your equation : ##2(y'')^2y + 2 y'''y'y - 2y''(y')^2 = -(y')^2##

Then dividing through by ##y^2## would yield : ##\frac{2(y'')^2}{y} + \frac{2y'''y'}{y} + 2y'' {(\frac{y'}{y})}^2 = -{(\frac{y'}{y})}^2##
 
  • #5
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Why are you guys so fixated on that line??

That's not even the crux of the question I'm asking!

lol


I obtained this differential equation from applying the Euler-Lagrange equation to the following function:

f(y,y') = (y')^2/y


That's how I know my reverse quotient rule is correct.



So how do I solve the original ODE?



Thanks!
 
  • #6
vela
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Because it's pointless to answer the question if you messed up in the first place, and despite your claim that you know it's right, there's an obvious algebra mistake in the third line. It's probably just a typo, but you should recognize why the helpers want to make sure you aren't wasting their time with the wrong question.
 
  • #7
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6 replies and all of them useless.

Fixated on something frivolous.


I gave you guys the correct differential equation in the first line anyway, so there's absolutely no problem imo.
 
  • #8
vela
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Well, half of those six replies are from you, so.... nyuk, nyuk, nyuk...

How did you manage to get that differential equation from ##f(y,y') = \frac{y'^2}{y}##? In particular, how'd you manage to end up with a third derivative?
 
  • #9
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By plugging (y')^2/y into the Euler-Lagrange Equation
 
  • #10
vela
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You must have done it wrong.
 
  • #11
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I've done it wrong because you can't solve the equation.... right....
 
  • #12
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partial derivative of (y')^2/y wrt y = - (y')^2/y^2

partial derivative of (y')^2/y wrt y' = (2 y' y'')/y

derivative of (2 y' y'')/y wrt x = (2 y (y'')^2 + 2 y y''' y' -2 (y')^2 y'')/y^2


finsihed. everything i've done is correct
 
  • #13
vela
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Are you claiming you calculated
$$\frac{d}{dt} \left(\frac{\partial f}{\partial y'}\right) = \frac{\partial f}{\partial y}?$$ Because if you are, you did it wrong.
 
Last edited:
  • #14
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Yes that's the equation, just swop t for x.


How have I done it wrong? I cross-checked with peers...
 
  • #15
vela
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Your mistake is when you took the partial with respect to y'. It should be
$$\frac{\partial f}{\partial y'} = \frac{2y'}{y}.$$ You seem to have applied the chain rule when you shouldn't have. You're not differentiating with respect to t.

You'll end up with a differential equation that you'll easily be able to solve.
 

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