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.