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

Question

y'-4xy=0; x0=-3

(type an expression in terms of a0 that includes all terms up to order 3.)

Answer 1

I am sorry that I did not understand that the series was about x_o = -3 yesterday! Let's start again.

Start with equation y' = 4xy

if,

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

Then, the coefficients of each term are just the Taylor series coefficients!

y(-3) = a_o

y'(-3) = 4(-3)(a_o) = -12a_o y''(x) = 4y + 4xy

y''(-3) = 4a_o + 4(-3)(-12a_o) = 148a_o y'''(x) = 8y' + 4xy''

y'''(-3) = 8(-12a_o) + 4(-3)(148a_o) = -1872a_o

So, y(x) = f(x) = f(-3) + f'(-3)(x+3) + f''(-3)(x+3)²/2! + f'''(-3)(x+3)³/3!

or y(x) = a_o -12a_o(x+3) + 148a_o(x+3)²/2! -1872a_o(x+3)³/3! .......

or y(x) = a_o{1 -12(x+3) + 74(x+3)² -312(x+3)³ }.......