Solve in series x^2x2y''+xy'+(x^2-k^2x2k2)y=0?

1 Answer
Jul 12, 2018

The solutions of the equation are the Bessel functions.

Explanation:

Suppose we can express y(x)y(x) as the sum of a power series:

y(x) = sum_(n=0)^oo c_nx^ny(x)=n=0cnxn

for x in IxI, and that we can differentiate term by term, so that:

y'(x) = sum_(n=1)^oo nc_nx^(n-1)

y''(x) = sum_(n=2)^oo n(n-1)c_nx^(n-2)

Substitute now in the original equation:

x^2sum_(n=2)^oo n(n-1)c_nx^(n-2)+x sum_(n=1)^oo nc_nx^(n-1) +(x^2-k^2) sum_(n=0)^oo c_nx^n =0

sum_(n=1)^oo n(n-1)c_nx^n+ sum_(n=2)^oo nc_nx^n +(x^2-k^2) sum_(n=0)^oo c_nx^n = 0

sum_(n=1)^oo n(n-1)c_nx^n+ sum_(n=2)^oo nc_nx^n + sum_(n=0)^oo c_nx^(n+2) -k^2 sum_(n=0)^oo c_nx^n= 0

Extract the terms with n < 2:

sum_(n=2)^oo n(n-1)c_nx^n+ c_1 x + sum_(n=2)^oo nc_nx^n + sum_(n=0)^oo c_nx^(n+2) -k^2 c_0 -k^2c_1x -k^2sum_(n=2)^oo c_nx^n= 0

and scale the index of the the third sum:

sum_(n=2)^oo n(n-1)c_nx^n + sum_(n=2)^oo nc_nx^n + sum_(n=2)^oo c_(n-2)x^n -k^2sum_(n=2)^oo c_nx^n= -c_1 x +k^2( c_0 +c_1x)

Group now the terms in a single sum:

c_1 x -k^2( c_0 +c_1x)+sum_(n=2)^oo (n^2c_n -nc_n+nc_n + c_(n-2)-k^2c_n)x^n = 0

c_1 x -k^2( c_0 +c_1x)+ sum_(n=2)^oo ((n^2-k^2)c_n+ c_(n-2))x^n = 0

As the sum is null the coefficient of each degree must be null, so we get:

{(k^2c_0 =0),((1-k^2)c_1 = 0), ((k^2-n^2)c_n = c_(n-2)):}

which allows to determine the coefficients recursively.

Note that:

1) All the coefficients are zero for n < k

2) The coefficient for n=k can be assigned arbitrarily, and because all non null coefficient are proportional this is equivalent to scaling the function.

3) For n > k as c_n depend only on c_(n-2) all the coefficients with the same parity as k are non null and all the coefficients with the different parity are null.