(1−x2)y″−xy′+m2y=0, m is a constant. I have to show that m∈Z⇒deg(y)=m such that y is a solution of the ode.

Expert Answer to "(1−x2)y″−xy′+m2y=0, m is a constant. I have to..."

Landin Harmon

2022-03-28

(1 - x^{2}) y^{″} - x y^{'} + m^{2} y = 0

, m is a constant. I have to show that

m \in Z \Rightarrow d e g (y) = m

such that y is a solution of the ode.

Cindy Reynolds

Beginner2022-03-29Added 7 answers

Step 1
Note that you have the following, which is valid for

n \geq 0

a_{n + 2} = \frac{(n + m) (n - m)}{(n + 2) (n + 1)} a_{n}

Step 2
Let

m \in Z

be any. If m is even, take

y (0) = 1 and y^{'} (0) = 0

, (so

a_{0} = 1 and a_{1} = 0

). Therefore all the coefficients of the form

a_{2 k + 1}

are zero (because of the formula above). Also, as m is an integer,

a_{m + 2} = 0

, and as it is even for all k,

a_{m + 2 k} = 0

. So,

deg y = m

.
Otherwise, if m is odd, let

y (0) = 0 and y^{'} (0) = 1

. Arguing in a similar way, you conclude that deg

y = m

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