Question

xd²y/dx² - (2x-1)dy/dx + (x-1) y =0​

Answer 1

Answer:

Let

y(x)=∑n=0 to ∞ (a_n*x^n)

calculate:

y′(x)=∑n=0 to ∞ [na_nx^(n−1)] and y′′(x)=∑n=0 to ∞ [n(n−1)a_n*x^(n−2)]

Substitute these into y" + 1/x y' +y = 0 giving,

∑n=0 to ∞ { [n(n−1)+n]a_n*x^(n−1)} + ∑n=0 to ∞ [a_n*x^(n+1)] = 0

Change the index in the 2nd series to start at 2

∑n=0 to ∞ {[n(n−1)+n]a_n*x^(n−1)}+ ∑n=2 to ∞ {a_(n−2)*x^n}

Rewrite as

a_1 +∑n=2 to ∞ {(n^2a_n + a_(n−2))*x^(n−1)} = 0

This implies a1=0.

Solving for a_n in terms of a_(n-2) gives:

n^2a_n=−a_(n−2)

Finding the pattern:

a_2=−a_0/(2^2) a_3=−a_1/(3^2)

a_4=a_0/(2^2.4^2) a_5=a_1/(3^2.5^2)

a_6=−a0/(2^2.4^2.6^2) a_7=−a_1/(3^2.5^2.7^2)

But, since a_1 = 0, only the terms with even subscripts show up.

y(x) = a_0 + a_0∑k=1 to ∞ {(−1)^k*x^(2k)/[2^{2k}(k!)^2]}

This is the Bessel function of the first kind of order zero,

J_0(x) = ∑k=0 to ∞ {(−1)^k*x^(2k)/[2^{2k}(k!)^2]}