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

Question

6x^2y''-4y'+2y=0; x0=1

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

Answer 1

The equation 6x²y'' - 4y' + 2y = 0 transforms to y'' = (2/3x²)y' -(1/3x²)y

If the solution is centered at x = 1, the y will have the general solution of

y = f(x) = a_o + a₁(x-1) + a₂(x-1)² + a₃(x-1)³ ........

We find that y(1) = a_o, y'(1) = a₁, y''(1) = (2/3)a₁ - (1/3)a_o, y'''(1) = (4/9)a_o - (11/9)a₁

Since the series is also a Taylor's series of the form

f(x) = f(1) + f'(1)(x-1) + f''(1)(x-1)²/2! + f'''(1)(x-1)³/3!.....

Then, substituting all of the derivatives in the equation, we get

y = f(x) = a_o{ 1 - (1/6)(x-1)² + (2/27)(x-1)³ } + a₁{ (x-1) + (1/3)(x-1)² - (11/54)(x-1)³ }