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# D.E. y" = 2y^3 watch

1. .

The Euler lagrange equation of this functional yields the D.E.

How do I solve this?!

I tried differentiating both sides and letting z=y' but that doesn't really get me anywhere. any ideas..
2. z = y' is the right direction. See if you can't use the chain rule somehow
3. Multiply both sides by 2y
4. (Original post by matt2k8)
Multiply both sides by 2y
I actually did that but that ultimately gives me:

and unless C=0 which I can't see any way of showing I cannot solve that .

Edit: btw, I am given y(a) = y(b) = 0

I am guessing actually that the solution is simply y= 0! as that minimises the functional!
5. Going by what Wolfram Alpha says, it seems the solution doesn't have an expression in terms of elementary functions...
6. I assume the initial conditions are y(a) = y'(a) = 0 to make C = 0 which then lets you finish up.

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Updated: September 26, 2010
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