I have the functional $$J(y)=\int_0^\infty L(y(x))\,dx +\lambda \left ( \int_0^\infty G(y(x))\,dx-C\right) \tag{1}$$
This gives the constrainted Euler lagrange equation: $$(L+\lambda G)_y=L_y+\lambda G_y=0 \tag{2}$$
If we can find $L_y$ and $G_y$, then we can solve for $y$ in terms of $\lambda$.
However, how do we find $\lambda$ itself? What is the general approach for this?
In principle eq. (2) can be solved for $y$ as a function of $x$ and the Lagrange multiplier $\lambda$.
Next plug the solution $y(x;\lambda)$ into the constraint $$ \int_0^\infty G(y(x))\,dx~=~C$$ to obtain an equation for $\lambda$.