Please help with Legendre problem.

[yungman]

Homework Statement

Determine the IVP has bounded solution:
Legendre equation:
(1-x^2)y''-2xy'+6y=0 ; y(0)=0, y'(0)=1

Homework Equations

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

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

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)}}

The Attempt at a Solution

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

General solution is y(x)=c_1 P_n (x)+c_2 Q_n (x)

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

y(0)=0 \Rightarrow c_1 = 0 \Rightarrow y(x)=c_2 Q_2 (x) only.

P'_2 (x)=6x therefore

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)}}

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)}]

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

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

y'(0)=-c_2


Therefor there is a bounded solution for the problem.

The answer from the book was Q_2 is unbounded on (-1,1) therefore the Legendre differential equation has no bounded solution from the given boundary condition.
What did I do wrong?

 

[yungman]

Anyone please?

 

[vela]

The book meant bounded over the interval when it asked for a bounded solution.

 

[vela]

What's so special about x=0?

 

[yungman]

Because the boundary condition given y(0)=0 and y'(0)=1

I claim Q_2 (0) is bounded. There is a bounded solution at x=0.

I am still not quite sure about the bound or unbound of Q really means.

 

[vela]

The solution to a differential equation is a function, not the function evaluated at some point.