Suppose we can express #y(x)# as the sum of a power series:
#y(x) = sum_(n=0)^oo c_nx^n#
for #x in I#, 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.
