I'm trying to figure out which functional $G$ is being minimized in $y''-y = e^x$. The Euler's equation is given by $$ \frac{d}{dx}\frac{\partial G}{\partial y'} - \frac{\partial G}{\partial y} = 0 $$ So I'm trying to work backward. The $y''$ term could be obtained from $a(y')^2$, as $\frac{d}{dx}\frac{\partial G}{\partial y'} = y''$. Consider the second part of Euler's equation, we have the terms $by^2+c\cdot e^xy+d$. Comparing the guess with given equation, we have $$ G = \frac{1}{2}(y')^2+\frac{1}{2}y^2+e^x y+const. $$ My question is do boundary conditions matter here?
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1Euler's equation is for integral functionals – janmarqz Sep 14 '21 at 01:25
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Your Lagrangian $G$ looks correct but the action functional is $\int_a^bG(y(x),y'(x),x),dx$. Regarding boundary conditions please see this post – Kurt G. Sep 14 '21 at 09:08