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

Chesley

Chesley

Answered question

2021-08-31

Consider the Chebyshev differential equation
(1x2)yxy+λ2y=0
where λ is a constant. Find the power series solutions of this differential equation about x0=0. For what values of the constant λ these solutions reduce to polynomials, known as Chebyshev polynomials. Compute first four of the Chebyshev polynomials.

Answer & Explanation

bahaistag

bahaistag

Skilled2021-09-01Added 100 answers

The objective is to find power series solution to Chebyshev differential equation
(1x2)yxy+λ2y=0
about the point x0=0 and λ is a constant.
Also, we have to find the values of λ such that solution reduces to Chebyshev polynomials and we have to estimate first four Chebyshev polynomials.
Compare the equation (1x2)yxy+λ2y=0 with P(x)y+Q(x)y+y=0.
Here, P(x)=1x2 and Q(x)=x
Both the functions P(x)=1x2 and Q(x)=x are analytic at x0=0.
Now, we look for solution of the form y(x)=n=0anxn
Substitute the solution in the original equation (1x2)yxy+λ2y=0
y(x)=n=0nanxn1
y(x)=n=0n(n1)anxn2
Therefore we have,
n=2n(n1)anxn2n=2n(n1)anxnn=1nanxn+n=0λ2anxn=0
Now replace n by n+2 in first term and simplify
n=0(n+2)(n+1)an+2xnn=2n(n1)anxnn=1nanxn+n=0λ2anxn=0

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