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Find the variational equivalent of the BVP $$-au''(x)+xu'(x)+u(x)=f(x)$$ subject to $u(0)=u'(L)=0$ .

We start with $$\delta\int_0^Lfudx=\int_0^Lf\delta udx=\int_0^L(-au''+xu'+u)\delta udx$$ By parts and $u'(L)=u(0)=0$ gives $u'\delta u\bigg|_{x=0}^{x=L}=u'(L)\delta u(L)-u'(0)\delta u(0)=0$ . Hence $$\delta\int_0^Lfudx=\delta\int_0^L\frac{a}{2}\bigg(\frac{du}{dx}\bigg)^2dx+\int_0^Lxu'\delta udx+\delta\int_0^L\frac{u^2}{2}dx$$ Now I can't get the $\delta$ factor out of the second integral to complete the variation . By parts isn't helping since chain rule of $x\delta u$ leaves one $u$ factor out of the $\delta$ . Any help is appreciated .

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