# $u''+\frac{4}{x+1}u'+\frac{2}{\left(x+1\right)^{2}}u=0$ variational solution

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.

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Maybe you can try the very complicated ones such as $(x^2-1)y''+3xy'+xy=0$ using this method to see how the process will be jammed. – doraemonpaul Jun 4 '12 at 2:27
Ok,I'll give it a go. Thanks! – Valentin Jun 4 '12 at 9:43
btw a simple way to solve the equation in the title is to try to remove the factors of $(1+x)$ by trying a test solution $u(x) = v(x)(x+1)^{a}$. For $a=-2$ the eq. greatly simplifies to $v''(x) = 0$ giving $u(x) = \frac{b+cx}{(x+1)^2}$. – Winther Aug 6 '14 at 13:33