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Suppose we want to solve $u + xu' = 0$, which has the general solution $u = \frac{C}{x}$, by minimizing the length squared of $u + xu'$. This should work due to the positive definite condition of inner products.

The inner product $\int_{-1}^{1} p(x)q(x)\ dx$ appears as good as any. This gives us $\int_{-1}^{1} u^2 + 2xuu' + x^2(u')^2\ dx$. According to the Euler-Lagrange Equation, we want to solve $\frac{\partial (u^2 + 2xuu' + x^2(u')^2)}{\partial u} - \frac{d}{dx}(\frac{\partial (u^2 + 2xuu' + x^2(u')^2)}{\partial u'}) = 0$. Simplifying, $2u + 2xu' - \frac{d}{dx}(2xu + 2x^2u') = 0$, thus $x^2u'' + 2xu' = 0$. Wolfram solves this second order ODE as $u = \frac{C_1}{x} + C_2$.

The problem is that unless $C_2 = 0$, this doesn't satisfy the original ODE. It doesn't even have the right number of arbitrary constants. $C_1$ behaves as it should in a general solution, but $C_2$ demands to be fixed at a particular value. What is wrong with my reasoning and/or calculations?

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The problem is that the Euler-Lagrange equation is incomplete information for an extremal with the free end points (no constraints on $u(0)$ and $u(1)$). It must be completed with the natural boundary conditions. In your case, the latter becomes $$ \frac{d}{du'}L(x,u(x),u'(x))\Big|_{x=1}=0\quad\Leftrightarrow\quad u(1)+u'(1)=0\quad\Leftrightarrow\quad C_1+C_2-C_1=C_2=0. $$

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