(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Determine the IVP has bounded solution:

Legendre equation:

[tex](1-x^2)y''-2xy'+6y=0[/tex] ; y(0)=0, y'(0)=1

2. Relevant equations

[tex]P_2 (x)=\frac{1}{2}[3x^2 -1][/tex]

[tex]Q_2 (x)=P_2 (x)\int \frac{dx}{[P_2 (x)]^2 (1-x^2)}[/tex]

[tex]y'(x)=c_1 P'_2 (x) + c_2 {P'_2(x)\int \frac{dx}{[P_2 (x)]^2 (1-x^2)}+c_2 P_2 (x)\frac{1}{[P_2 (x)]^2 (1-x^2)}}[/tex]

3. The attempt at a solution

n(n+1)=6 give n=2

General solution is [tex]y(x)=c_1 P_n (x)+c_2 Q_n (x)[/tex]

[tex]P_2 (x)=\frac{1}{2}[3x^2 -1]\Rightarrow P_2 (0)=1[/tex]

[tex]y(0)=0 \Rightarrow c_1 = 0 \Rightarrow y(x)=c_2 Q_2 (x)[/tex] only.

[tex]P'_2 (x)=6x [/tex] therefore

[tex]y'(x)=c_2 {P'_2(x)\int \frac{dx}{[P_2 (x)]^2 (1-x^2)}+P_2 (x)\frac{1}{[P_2 (x)]^2 (1-x^2)}}[/tex]

[tex]y'(x)=c_2 [6x\int\frac{dx}{(3x^2 -1)^2 (1-x^2)}+(3x^2 -1)\frac{1}{(3x^2 -1)^2 (1-x^2)}][/tex]

x=0, the first part of y'(x)=0 because anything multiply by 6x equal zero. Therefore:

For x=0, [tex]y'(x)=c_2 [\frac{1}{(3x^2 -1)(1-x^2)}][/tex]

[tex] y'(0)=-c_2[/tex]

Therefor there is a bounded solution for the problem.

The answer from the book was [tex]Q_2 [/tex] is unbounded on (-1,1) therefore the Legendre differential equation has no bounded solution from the given boundary condition.

What did I do wrong?

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# Homework Help: Please help with Legendre problem.

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