Consider the Chebyshev differential equation. (1-x^2)y''-xy'+\lambda^2y=0. where \lambda is a constant

Consider the Chebyshev differential equation
$$\displaystyle{\left({1}-{x}^{{2}}\right)}{y}{''}-{x}{y}'+\lambda^{{2}}{y}={0}$$
where $$\displaystyle\lambda$$ is a constant. Find the power series solutions of this differential equation about $$\displaystyle{x}_{{0}}={0}$$. For what values of the constant $$\displaystyle\lambda$$ these solutions reduce to polynomials, known as Chebyshev polynomials. Compute first four of the Chebyshev polynomials.

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The objective is to find power series solution to Chebyshev differential equation
$$\displaystyle{\left({1}-{x}^{{2}}\right)}{y}{''}-{x}{y}'+\lambda^{{2}}{y}={0}$$
about the point $$\displaystyle{x}_{{0}}={0}$$ and $$\displaystyle\lambda$$ is a constant.
Also, we have to find the values of $$\displaystyle\lambda$$ such that solution reduces to Chebyshev polynomials and we have to estimate first four Chebyshev polynomials.
Compare the equation $$\displaystyle{\left({1}-{x}^{{2}}\right)}{y}{''}-{x}{y}'+\lambda^{{2}}{y}={0}$$ with $$\displaystyle{P}{\left({x}\right)}{y}{''}+{Q}{\left({x}\right)}{y}'+{y}={0}$$.
Here, $$\displaystyle{P}{\left({x}\right)}={1}-{x}^{{2}}$$ and $$\displaystyle{Q}{\left({x}\right)}=-{x}$$
Both the functions $$\displaystyle{P}{\left({x}\right)}={1}-{x}^{{2}}$$ and $$\displaystyle{Q}{\left({x}\right)}=-{x}$$ are analytic at $$\displaystyle{x}_{{0}}={0}$$.
Now, we look for solution of the form $$\displaystyle{y}{\left({x}\right)}={\sum_{{{n}={0}}}^{\infty}}{a}_{{n}}{x}^{{n}}$$
Substitute the solution in the original equation $$\displaystyle{\left({1}-{x}^{{2}}\right)}{y}{''}-{x}{y}'+\lambda^{{2}}{y}={0}$$
$$\displaystyle{y}'{\left({x}\right)}={\sum_{{{n}={0}}}^{\infty}}{n}{a}_{{n}}{x}^{{{n}-{1}}}$$
$$\displaystyle{y}{''}{\left({x}\right)}={\sum_{{{n}={0}}}^{\infty}}{n}{\left({n}-{1}\right)}{a}_{{n}}{x}^{{{n}-{2}}}$$
Therefore we have,
$$\displaystyle{\sum_{{{n}={2}}}^{\infty}}{n}{\left({n}-{1}\right)}{a}_{{n}}{x}^{{{n}-{2}}}-{\sum_{{{n}={2}}}^{\infty}}{n}{\left({n}-{1}\right)}{a}_{{n}}{x}^{{n}}-{\sum_{{{n}={1}}}^{\infty}}{n}{a}_{{n}}{x}^{{n}}+{\sum_{{{n}={0}}}^{\infty}}\lambda^{{2}}{a}_{{n}}{x}^{{n}}={0}$$
Now replace n by n+2 in first term and simplify
$$\displaystyle{\sum_{{{n}={0}}}^{\infty}}{\left({n}+{2}\right)}{\left({n}+{1}\right)}{a}_{{{n}+{2}}}{x}^{{n}}-{\sum_{{{n}={2}}}^{\infty}}{n}{\left({n}-{1}\right)}{a}_{{n}}{x}^{{n}}-{\sum_{{{n}={1}}}^{\infty}}{n}{a}_{{n}}{x}^{{n}}+{\sum_{{{n}={0}}}^{\infty}}\lambda^{{2}}{a}_{{n}}{x}^{{n}}={0}$$
$$\displaystyle{2}{a}_{{2}}+{6}{a}_{{3}}{x}+{\sum_{{{n}={2}}}^{\infty}}{\left({n}+{2}\right)}{\left({n}+{1}\right)}{a}_{{{n}+{2}}}{x}^{{n}}-{\sum_{{{n}={2}}}^{\infty}}{n}{\left({n}-{1}\right)}{a}_{{n}}{x}^{{n}}-{a}_{{1}}{x}-{\sum_{{{n}={2}}}^{\infty}}{n}{a}_{{n}}{x}^{{n}}+\lambda^{{2}}{a}_{{0}}+{x}\lambda^{{2}}{a}_{{0}}+{\sum_{{{n}={2}}}^{\infty}}\lambda^{{2}}{a}_{{n}}{x}^{{n}}={0}$$
$$\displaystyle{2}{a}_{{2}}+\lambda^{{2}}{a}_{{0}}+{\left({6}{a}_{{3}}-{a}_{{1}}+\lambda^{{1}}{a}_{{0}}\right)}{x}+{\sum_{{{n}={2}}}^{\infty}}{\left[{\left({n}+{2}\right)}{\left({n}+{1}\right)}{a}_{{{n}+{2}}}-{\left[{n}{\left({n}-{1}\right)}+{n}-\lambda^{{2}}\right]}{a}_{{n}}\right]}{x}^{{n}}={0}$$
Now, equate the powers of x from both the sides,
$$\displaystyle{2}{a}_{{2}}+\lambda^{{2}}{a}_{{0}}={0}$$
$$\displaystyle{6}{a}_{{3}}-{a}_{{1}}+\lambda^{{2}}{a}_{{0}}={0}$$
$$\displaystyle{\left[{\left({n}+{2}\right)}{\left({n}+{1}\right)}{a}_{{{n}+{2}}}-{\left[{n}{\left({n}-{1}\right)}+{n}-\lambda^{{2}}\right]}{a}_{{n}}\right]}={0}\quad{n}={2},{3},\ldots$$
Now, we solve the values obtained in above equations,
$$\displaystyle{a}_{{2}}=-{\frac{{\lambda^{{2}}}}{{{2}}}}{a}_{{0}}$$
$$\displaystyle{a}_{{3}}={\frac{{{1}-\lambda^{{2}}}}{{{3}!}}}{a}_{{1}}$$
$$\displaystyle{a}_{{{n}+{2}}}={\frac{{{n}^{{2}}-\lambda^{{2}}}}{{{\left({n}+{2}\right)}{\left({n}+{1}\right)}}}}{a}_{{n}}\quad{n}={2},{3},\ldots$$ With the help of general formula, the few more terms are,
$$\displaystyle{a}_{{4}}={\frac{{{\left({2}^{{2}}-\lambda^{{2}}\right)}{\left(-\lambda^{{2}}\right)}}}{{{4}!}}}{a}_{{0}}$$
$$\displaystyle{a}_{{5}}={\frac{{{\left({3}^{{2}}-\lambda^{{2}}\right)}{\left({1}-\lambda^{{2}}\right)}}}{{{5}!}}}{a}_{{1}}$$
$$\displaystyle{a}_{{6}}={\frac{{{\left({4}^{{2}}-\lambda^{{2}}\right)}{\left({2}^{{2}}-\lambda^{{2}}\right)}{\left(-\lambda^{{2}}\right)}}}{{{6}!}}}{a}_{{0}}$$ Therefore, the pattern can be set up as,
$$\displaystyle{a}_{{{2}{n}}}={\frac{{{\left[{\left({2}{n}-{2}\right)}^{{2}}-\lambda^{{2}}\right]}{\left[{\left({2}{n}-{4}\right)}^{{2}}-\lambda^{{2}}\right]}\ldots{\left({2}^{{2}}-\lambda^{{2}}\right)}{\left(-\lambda^{{2}}\right)}}}{{{\left({2}{n}\right)}!}}}{a}_{{0}},\quad{n}={1},{2},\ldots$$
$$\displaystyle{a}_{{{2}{n}+{1}}}={\frac{{{\left[{\left({2}{n}-{1}\right)}^{{2}}-\lambda^{{2}}\right]}{\left[{\left({2}{n}-{3}\right)}^{{2}}-\lambda^{{2}}\right]}\ldots{\left({3}^{{2}}-\lambda^{{2}}\right)}{\left({1}-\lambda^{{2}}\right)}}}{{{\left({2}{n}+{1}\right)}!}}}{a}_{{1}},\quad{n}={1},{2},\ldots$$
Since, our solution was of the form $$\displaystyle{y}{\left({x}\right)}={\sum_{{{n}={0}}}^{\infty}}{a}_{{n}}{x}$$. Thus, the power series solution is
$$\displaystyle{y}={1}+{\sum_{{{n}={1}}}^{\infty}}{\frac{{{\left[{\left({2}{n}-{2}\right)}^{{2}}-\lambda^{{2}}\right]}{\left[{\left({2}{n}-{4}\right)}^{{2}}-\lambda^{{2}}\right]}\ldots{\left({2}^{{2}}-\lambda^{{2}}\right)}{\left(-\lambda^{{2}}\right)}}}{{{\left({2}{n}\right)}!}}}{x}^{{{2}{n}}}+{x}+$$
$$\displaystyle+{\sum_{{{n}={1}}}^{\infty}}{\frac{{{\left[{\left({2}{n}-{1}\right)}^{{2}}-\lambda^{{2}}\right]}{\left[{\left({2}{n}-{3}\right)}^{{2}}-\lambda^{{2}}\right]}\ldots{\left({3}^{{2}}-\lambda^{{2}}\right)}{\left({1}-\lambda^{{2}}\right)}}}{{{\left({2}{n}+{1}\right)}!}}}{x}^{{{2}{n}+{1}}}$$
$$\displaystyle={u}{\left({x}\right)}+{v}{\left({x}\right)}$$
where,
$$\displaystyle{u}{\left({x}\right)}={1}+{\sum_{{{n}={1}}}^{\infty}}{\frac{{{\left[{\left({2}{n}-{2}\right)}^{{2}}-\lambda^{{2}}\right]}{\left[{\left({2}{n}-{4}\right)}^{{2}}-\lambda^{{2}}\right]}\ldots{\left({2}^{{2}}-\lambda^{{2}}\right)}{\left(-\lambda^{{2}}\right)}}}{{{\left({2}{n}\right)}!}}}{x}^{{{2}{n}}}$$
$$\displaystyle{v}{\left({x}\right)}={x}+{\sum_{{{n}={1}}}^{\infty}}{\frac{{{\left[{\left({2}{n}-{1}\right)}^{{2}}-\lambda^{{2}}\right]}{\left[{\left({2}{n}-{3}\right)}^{{2}}-\lambda^{{2}}\right]}\ldots{\left({3}^{{2}}-\lambda^{{2}}\right)}{\left({1}-\lambda^{{2}}\right)}}}{{{\left({2}{n}+{1}\right)}!}}}{x}^{{{2}{n}+{1}}}$$
Also, we have to find the values of $$\displaystyle\lambda$$ such that solution reduces to Chebyshev polynomials and we have to estimate first four Chebyshev polynomials.
When $$\displaystyle\lambda={0},{1},{2},{3}$$ we get four Chebyshev polynomials.
When $$\displaystyle\lambda={0}$$ and $$\displaystyle{a}_{{{2}{n}}}={0}\ \text{ for }\ \geq{1}$$ so,
$$\displaystyle{u}{\left({x}\right)}={1}$$
When $$\displaystyle\lambda={1}$$ and $$\displaystyle{a}_{{{2}{n}+{1}}}={0}\ \text{ for }\ {n}\geq{1}$$ so,
$$\displaystyle{v}{\left({x}\right)}={x}$$
When $$\displaystyle\lambda={2}$$ and $$\displaystyle{a}_{{{2}{n}}}={0}\ \text{ for }\ {n}\geq{2}$$ so,
$$\displaystyle{u}{\left({x}\right)}={1}-{2}{x}^{{2}}$$
When $$\displaystyle\lambda={3}$$ and $$\displaystyle{a}_{{{2}{n}+{1}}}={0}\ \text{ for }\ {n}\geq{2}$$ so,
$$\displaystyle{v}{\left({x}\right)}={x}-{\frac{{{4}}}{{{5}}}}{x}^{{3}}$$
So, first four Chebyshev polynomials are $$\displaystyle{1},{x},{1}-{2}{x}^{{2}}$$ and $$\displaystyle{x}-{\frac{{{4}}}{{{5}}}}{x}^{{3}}$$