Prove $(1-x^2)y_{n+2} - (2n+1)xy_{n+1}+(m^2-n^2)y_n=0$ for y=sin(m \arcsin x)

[utkarshakash]

Homework Statement

For $y=sin(m \arcsin x)$, prove that
$$(1-x^2)y_{n+2} - (2n+1)xy_{n+1}+(m^2-n^2)y_n=0 $$

Homework Equations

The Attempt at a Solution

My first approach would be to find a general expression for y_n. Starting with
$$y_1=m\dfrac{\cos(m \arcsin x)}{\sqrt{1-x^2}} \\
y_2 = m\dfrac{\cos(m \arcsin x)}{\sqrt{1-x^2}(1-x^2)} - m^2 \dfrac{\sin(m \arcsin x)}{1-x^2} $$

The expression seems to get uglier as n increases and it's too difficult to calculate y_3. Also it's not easy to generalize an expression for y_n just by looking at the results obtained.

 

[PeroK]

Have you thought of using induction?

 

[Ray Vickson]

Homework Statement

For $y=sin(m \arcsin x)$, prove that
$$(1-x^2)y_{n+2} - (2n+1)xy_{n+1}+(m^2-n^2)y_n=0 $$

Homework Equations

The Attempt at a Solution

My first approach would be to find a general expression for y_n. Starting with
$$y_1=m\dfrac{\cos(m \arcsin x)}{\sqrt{1-x^2}} \\
y_2 = m\dfrac{\cos(m \arcsin x)}{\sqrt{1-x^2}(1-x^2)} - m^2 \dfrac{\sin(m \arcsin x)}{1-x^2} $$

The expression seems to get uglier as n increases and it's too difficult to calculate y_3. Also it's not easy to generalize an expression for y_n just by looking at the results obtained.

Please do not use $y_n$ for the $n$th derivative of $y$; the standard notation would be $y^{(n)}$, and your use of other notation (without explanation) is too confusing: I could not figure out what your question was asking.

 

[utkarshakash]

Have you thought of using induction?

Let P(n) be the given expression.
$$P(0)=(1-x^2)y_2-3xy_1+(m^2-1)y=0$$
Let P(n) be true. Therefore,
$(1-x^2)y_{n+2} - (2n+1)xy_{n+1}+(m^2-n^2)y_n=0\\ P(n+1)=(1-x^2)y_{n+3} - (2n+3)xy_{n+2}+(m^2-(n+1)^2)y_{n+1}$

How do I proceed further?

 

[PeroK]

What is the most obvious thing you can do with P(n)?