This is a concept solution scheme derived from a particular example that I have not been able to generalise sufficiently. The objective is to find a particular solution to a certain second-order equation. Rewrite the equation in the self-adjoint form: $2\left(\left(x+1\right)^4u'\right)'+4(x+1)^2u=0$ $\left(Pu'\right)'-Qu=0 \qquad (1)$ where $P(x)=2\left(x+1\right)^4$ $Q(x)=4\left(x+1\right)^2.$ Now (1) can be interpreted as the Jacobi equation for a certain variational problem $S[y]=\int_a^b F\left(x,y,y'\right)dx \to \min, \qquad y(a)=A, y(b)=B$ so that $P(x) = \left. \frac{\partial^2 F}{\partial y'^2} \right|_{y=y_0(x)}$ $Q(x) = \left. \frac{\partial^2 F}{\partial y^2}\right|_{y=y_0(x)}-\left.\frac{\partial^2 F}{\partial y\partial y'}\right|_{y=y_0(x)}$ where $y_0(x)$ is a solution of the Euler-Lagrange equation for the same functional. Suppose we have figured out that for the given example: $F=\frac{y^2}{y'^4}.$ Consider the following problem $S[y]=\int_0^1\frac{y^2}{y'^4}dx, \qquad y(0)=1,y(1)=\frac{1}{2}$ for which the given equation will be a Jacobi one. Now it is easy to write out the Euler-Lagrange equation. In fact we can write the first integral straightaway, since $F$ does not depend on $x$: $y'F_{y'}-F=C$ $y'\frac{2y'}{y^{4}}-\frac{y'^{2}}{y^{4}}=C^{2}$ $\frac{y'}{y^{2}}=C$ $-\frac{1}{y}=Cx+C_{0}$ Applying boundary condition $y(0)=1$ we obtain a family of solutions: $y(x,C)=\frac{1}{Cx+1}$ Now a solution of the Jacobi equation is really an envelope of the family of solutions of Euler-Lagrange equation and can be found by differentiating with respect to parameter $C$: $u(x)=\frac{\partial y(x,C)}{\partial C}=-\frac{x}{\left(Cx+1\right)^{2}}$ Finally setting $C=1$ by virtue of the boundary conditions: $u(x)=-\frac{x}{\left(x+1\right)^{2}}$ It can be verified by direct substitution that the above function is a solution of the subject equation. Since the equation is of the second order, the process of constructing a general solution from here is straightforward.
Now my concern is: does there exist a way to work backwards from the expressions of $P$ and $Q$ to the function $F$? If this is accomplished then this might result in another indirect method of solving a certain class of second-order ODE. If however, this requires too much pre-knowledge about the solutions of E-L, then this example is bound to remain a case of reverse-engineering.