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I think this is a very basic question. I'm just not sure how to use $\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\partial F}{\partial y'}\right)=\frac{\partial F}{\partial y}$ to find the general solution to: $$\int \left(y^2 − y'^{\\2}- 2y \cos( 2x)\right) \:\mathrm{d}x$$

Thanks

Thomas Russell
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Nicky
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If you simply apply the equation $$\frac{\partial F}{\partial y} - \frac{d}{dx}(\frac{\partial F}{\partial y'}) = 0$$ for $F(x , y , y') = y^2 - y'^2 - 2y\cos(2x)$, you shall get $y'' - y = \cos(2x)$ (Please check the calculation).

Hope you can solve the second order ODE.

Supriyo
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  • Ahh so it's just a simple matter of applying and solving. Does this solve for a stationary solution of y? And i got $y''+y=\cos(2x)$? – Nicky Apr 27 '14 at 09:28
  • I do not see anything more than except solving the ODE – Supriyo Apr 27 '14 at 09:30
  • This is just more of a general question, not specifically related to this question. The solution we get from the Euler Lagrange equation has stationary value for the integral? According to Wolfram at least; http://mathworld.wolfram.com/Euler-LagrangeDifferentialEquation.html – Nicky Apr 27 '14 at 09:35
  • Any chance for assistance here: http://math.stackexchange.com/questions/1066495/minimization-of-variational-total-variation-tv-deblurring – Royi Dec 14 '14 at 06:58