Find the first four nonzero terms in a power series expansion of the solution to the given initial value problem

Question

3y''-4(cosx)y'-4y=0; y(pi/2)=6, y'(pi/2)=2

(Type an expression that includes all terms up to order 4)

Answer 1

Start with the simplified equation y''(x) = f''(x) = { -(4/3)cosx }y' - (4/3)y

and the solution in terms of (x-pi/2)

So,

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

But this equation is also compatible with the Taylor's Series coefficients! Solving for the derivatives at pi/2

y(pi/2) = 6 (given)

y'(pi/2) = 2 (given)

y''(pi/2) = 8

y'''(pi/2) = 0

y''''(pi/2) = -32/3

y'''''(pi/2) = 8/3

y(x) = f(x) = 6 + 2(x-pi/2) + 8(x-pi/2)²/2! + 0 -32/3(x-pi/2)⁴/4! + 8/3(x-pi/2)⁵/5! .........

y = f(x) = 6 + 2(x-pi/2) + 4(x-pi/2)² - (4/9)(x-pi/2)⁴ + (8/159)(x-pi/2)⁵..............