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## Main Question or Discussion Point

Legendre equation:

[tex](1-x^2)y''-2xy'+n(n+1)y=0[/tex] Where -1< x < 1

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

Where [tex]P_n (x)[/tex] is bounded and [tex]Q_n (x)[/tex] is unbounded on (-1,1).

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

I tried to solve for [tex]Q_n (x)[/tex] by substitude [tex]P_n (x)[/tex] in, using partial fraction to simplify and integrate each terms individually.

[tex]Q_n (x)=P_n (x)[ln(x+A) + ln(x+B)+ \frac{1}{x+C}.......][/tex]

As you can see after multiplying [tex]P_n (x)[/tex] in, [tex] ln(x^+_- A)[/tex] is unbound at x=+/-A.

But the second solution can be represented by an

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

where the power series is bounded on (-1,1).

[tex](1-x^2)y''-2xy'+n(n+1)y=0[/tex] Where -1< x < 1

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

Where [tex]P_n (x)[/tex] is bounded and [tex]Q_n (x)[/tex] is unbounded on (-1,1).

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

**Question:**Why is [tex]Q_n (x)[/tex] unbounded on (-1,1)?I tried to solve for [tex]Q_n (x)[/tex] by substitude [tex]P_n (x)[/tex] in, using partial fraction to simplify and integrate each terms individually.

[tex]Q_n (x)=P_n (x)[ln(x+A) + ln(x+B)+ \frac{1}{x+C}.......][/tex]

As you can see after multiplying [tex]P_n (x)[/tex] in, [tex] ln(x^+_- A)[/tex] is unbound at x=+/-A.

**Is that the reason Q is unbound on (-1,1)?**But the second solution can be represented by an

**infinite series**instead of[tex]Q_n (x)=P_n (x)\int \frac{dx}{[P_n (x)]^2 (1-x^2)}[/tex]

where the power series is bounded on (-1,1).

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