# If we consider the equation (1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+2y=0,\ -1<x<1 how can we find the

If we consider the equation
$$\displaystyle{\left({1}-{x}^{{2}}\right)}{\frac{{{d}^{{2}}{y}}}{{{\left.{d}{x}\right.}^{{2}}}}}-{2}{x}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{2}{y}={0},\ -{1}{ < }{x}{ < }{1}$$
how can we find the explicit solution, what should be the method for solution?

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Jillian Edgerton
As mentioned in comment, this ODE is Legendre differential equation for l=1
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{\left({1}-{x}^{{2}}\right)}{\frac{{{\left.{d}{y}\right.}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}}\right]}+{l}{\left({l}+{1}\right)}{y}{\left({x}\right)}={0}$$
The standard Frobenius method will give us the Legendre polynomials.
By inspection, it is sort of obvious $$\displaystyle{y}{\left({x}\right)}={x}$$ 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
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{p}{\left({x}\right)}{\frac{{{\left.{d}{y}\right.}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}}\right]}+{q}{\left({x}\right)}{y}{\left({x}\right)}={0}$$
Let $$\displaystyle{y}_{{1}}{\left({x}\right)}$$ be a solution we already knew, let $$\displaystyle{y}_{{2}}{\left({x}\right)}$$ be another solution we seek. We have
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{p}{\left({y}_{{1}}{\frac{{{\left.{d}{y}\right.}_{{2}}}}{{{\left.{d}{x}\right.}}}}-{y}_{{2}}{\frac{{{\left.{d}{y}\right.}_{{1}}}}{{{\left.{d}{x}\right.}}}}\right)}\right]}$$
$$\displaystyle={y}_{{1}}{\left[{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{p}{\frac{{{\left.{d}{y}\right.}_{{2}}}}{{{\left.{d}{x}\right.}}}}\right]}+{q}{y}_{{2}}\right]}-{y}_{{2}}{\left[{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{p}{\frac{{{\left.{d}{y}\right.}_{{1}}}}{{{\left.{d}{x}\right.}}}}\right]}+{q}{y}_{{1}}\right]}={0}$$
This implies
$$\displaystyle{p}{\left({x}\right)}{\left({y}_{{1}}{\left({x}\right)}{\frac{{{\left.{d}{y}\right.}_{{2}}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}}-{y}_{{2}}{\left({x}\right)}{\frac{{{\left.{d}{y}\right.}_{{1}}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}}\right)}={c}{o}{n}{s}{t}.$$
Let say's we fix the constant at the left to 1, we can rewrite above as
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left[{\frac{{{y}_{{2}}{\left({x}\right)}}}{{{y}_{{1}}{\left({x}\right)}}}}\right]}={\frac{{{1}}}{{{p}{\left({x}\right)}{{y}_{{1}}^{{2}}}{\left({x}\right)}}}}\Rightarrow{y}_{{2}}{\left({x}\right)}$$
$$\displaystyle={y}_{{1}}{\left({x}\right)}{\left[\int^{{x}}{\frac{{{\left.{d}{t}\right.}}}{{{p}{\left({t}\right)}{y}_{{1}}{\left({t}\right)}^{{2}}}}}\right]}+{c}{o}{n}{s}{t}{]}$$
is another solution. Apply this to our ODE, the other solution we seek can be chosen as
$$\displaystyle{x}\int^{{x}}{\frac{{{1}}}{{{\left({1}-{t}^{{2}}\right)}{t}^{{2}}}}}{\left.{d}{t}\right.}={x}\int^{{x}}{\left[{\frac{{{1}}}{{{t}^{{2}}}}}+{\frac{{{1}}}{{{2}}}}{\left({\frac{{{1}}}{{{1}-{t}}}}+{\frac{{{1}}}{{{1}+{t}}}}\right)}\right]}{\left.{d}{t}\right.}$$
$$\displaystyle={x}{\left[-{\frac{{{1}}}{{{x}}}}+{\frac{{12}}{{\log{{\left({\frac{{{1}+{x}}}{{{1}-{x}}}}\right)}}}}}\right]}$$
$$\displaystyle={x}{{\text{tanh}}^{{-{1}}}{\left({x}\right)}}-{1}$$
The one metioned in Claude Leibovici's answer.
###### Not exactly what you’re looking for?
porschomcl
There is one obvious particular solution which is $$\displaystyle{y}={c}_{{1}}{x}$$. The second one is much less obvious to me but, using a CAS, I found as general solution
$$\displaystyle{y}={c}_{{1}}{x}+{c}_{{2}}{\left({x}{{\text{tanh}}^{{-{1}}}{\left({x}\right)}}-{1}\right)}$$
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.
RizerMix

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