# explicit solution for second order homogeneous linear differential equation

If we consider the equation
$(1-x^2)\dfrac{d^2y}{dx^2} -2x \dfrac{dy}{dx}+2y=0, \quad -1<x<1$
how can we find the explicit solution, what should be the method for solution?

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if I write this way: $y^{''}-\dfrac{2x}{1-x^2}y^{'}+\dfrac{2}{1-x^2}y=0$ – hayrit Aug 18 '14 at 4:02
This is Legendre differential equation for $\ell = 1$. The standard Frobenius method will lead to the Legendre polynomials. – achille hui Aug 18 '14 at 4:08

As mentioned in comment, this ODE is Legendre differential equation for $\ell = 1$:

$$\frac{d}{dx}\left[ (1-x^2) \frac{dy(x)}{dx}\right] + \ell(\ell+1) y(x) = 0$$

The standard Frobenius method will give us the Legendre polynomials.

By inspection, it is sort of obvious $y(x) = x$ is one solution for this ODE. In general, if we already know one solution to a second order homogenous ODE, we don't need Frobenius method to find the other one.

Given any second order homogenous ODE in its Sturm-Liouville form

$$\frac{d}{dx}\left[ p(x) \frac{dy(x)}{dx}\right] + q(x)y(x) = 0 \tag{*1}$$

Let $y_1(x)$ be a solution we already knew, let $y_2(x)$ be another solution we seek. We have $$\frac{d}{dx}\left[ p \left(y_1\frac{dy_2}{dx} - y_2\frac{dy_1}{dx}\right)\right] = y_1 \left[ \frac{d}{dx}\!\!\left[p \frac{dy_2}{dx} \right] + q y_2 \right] - y_2 \left[ \frac{d}{dx}\!\!\left[p \frac{dy_1}{dx} \right] + q y_1 \right ] = 0$$ This implies $$p(x) \left(y_1(x)\frac{dy_2(x)}{dx} - y_2(x)\frac{dy_1(x)}{dx}\right) = \text{const}.$$ Let say's we fix the constant at the left to $1$, we can rewrite above as

$$\frac{d}{dx}\left[\frac{y_2(x)}{y_1(x)}\right] = \frac{1}{p(x)y_1^2(x)} \implies y_2(x) = y_1(x) \left[ \int^x \frac{dt}{p(t)y_1(t)^2} + \text{const}\right]$$ is another solution. Apply this to our ODE, the other solution we seek can be chosen as

\begin{align} x \int^x \frac{1}{(1-t^2)t^2} dt &= x \int^x \left[\frac{1}{t^2} + \frac12\left(\frac{1}{1-t}+\frac{1}{1+t}\right)\right] dt\\ &= x \left[ -\frac{1}{x} + \frac12\log\left(\frac{1+x}{1-x}\right)\right]\\ &= x\tanh^{-1}(x)-1 \end{align}

The one metioned in Claude Leibovici's answer.

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+1, interesting references. – Krokop Aug 18 '14 at 18:20

There is one obvious particular solution which is $y=c_1 x$. The second one is much less obvious to me but, using a CAS, I found as general solution $$y=c_1 x+c_2 \left(x \tanh ^{-1}(x)-1\right)$$ I hope and wish this will give you some ideas. As said by achille hui, beside the solution in terms of Legendre polynomials, you can obviously use Frobenius method which leads to the solution I wrote.

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## HINT

Assume $u = (1-x^2)\dfrac{dy}{dx}$. Try to calculate $\dfrac{du}{dx}$.

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$\dfrac{du}{dx}=(1-x^2)\dfrac{d^2y}{dx^2}-2x\dfrac{dy}{dx}$ – hayrit Aug 18 '14 at 4:17
then, $\dfrac{du}{dx}+2y=0$, how will we rewrite $y$? – hayrit Aug 18 '14 at 4:19