Find the first four nonzero terms in a power series expansion about x0 for a general solution.

Question

(x^2-10x)y''+6y=0; x0=5

(Type an expression in terms of a0 and a1 that includes all terms up to order 3)

Answer 1

Start with equation y''(x) = f(x) = 6y/(10x-x²) about point x_o = 5

Then,

y(x) = a_o + a₁(x-5) + a₂(x-5)² + a₃(x-5)³ + a₄(x-5)⁴................

by inspection

y(5) = a_o and y'(5) = a₁

then,

y''(5) = 6a_o/(50-25) = 6a_o/25 y'''(x) = 6y'/(10x-x²) - 6y(10-2x)/(10x-x²)²

then,

y'''(5) = 6a₁/25 - 0 = 6a₁/25

y''''(x) = 6y''/(10x-x²) -12y'(10-2x)/(10x-x²)² + 12y/(10x-x²)² + 12y(10-2x)2/(10x-x²)³

then

y''''(5) = (6/25)(6a_o/25) + 0 + 12a_o/625 + 0 = 48a_o/625

So,

y(x) = f(x) = f(5) + f'(5)(x-5) + f''(5)(x-5)²/2! + f'''(5)(x-5)³/3! + f''''(5)(x-5)⁴/4! .........

or

y(x) = f(x) = a_o + a₁(x-5) + (6a_o/25)(x-5)²/2! + (6a₁/25)(x-5)³/3! + (48a_o/625)(x-5)⁴/4!...

or

y(x) = f(x) = a_o + a₁(x-5) + (3a_o/25)(x-5)² + (a₁/25)(x-5)³ + (2a_o/625)(x-5)⁴.....

Very tedious.... hopefully no adding errors!