Differential equations

Question

Use power series method to solve;

y"+9y= 0

Answer 1

y'' + 9y = 0;

y = c₀ + c₁x + c₂x² + ... + c_nxⁿ + ... = ∑₀

^∞c_nx^n, y' = ∑ ₁^∞nc_nx^{n - 1}; y'' = ∑₂^∞n(n - 1)c_nx^{n - 2}_;

∑₂^∞n(n - 1)c_nx^{n - 2} + 9∑₀^∞c_nxⁿ = 0; ∑₀^∞[(n + 1)(n + 2)c_{n + 2} + 9c_n]xⁿ = 0.

So, (n + 1)(n + 2)c_{n + 2} + 9c_n = 0, n = 0, 1, 2, 3, ...

c_{n + 2} = - 9c_n/[(n + 1)(n + 2)]

n = 0, c₂ = - 9/(1·2)c₀; n = 1 c₃ = - 9c₁/(2·3)

n = 2, c₄ = - 9c₂/(3·4) = (- 9)²/(1·2·3·4)c₀; n = 3 c₅ = - 9c₃/(4·5) = (- 9)²/(1·2·3·4·5)c₁

...............................................................

n = 2k, c_2k = (- 1)^k3^2kc₀/(2k)!

n = 2k + 1, c_{2k + 1} = (- 1)^k3^{2k + 1}c₁/(2k + 1)!

y = c₀∑₀^∞(- 1)^k·3^2k/(2k)!x^2k + c₁∑₀^∞(- 1)^k3^{2k + 1}/(2k + 1)!x^{2k + 1}

Answer 2

You did not give any initial conditions. So, if y(0) = A and y'(0) = B then the first 6 terms are

y = f(x) = A + Bx - (9A/2)x² - (3B/2)x³ + (27A/8)x⁴ + (27B/40)x⁵ ......... but in searching for a common

term(s), we find that y = A∑₁^∞ { (-1)^n-1(9)^n-1x^(2n-2)/(2n-2)! } + B∑₁^∞ {(-1)^n-1(9)^n-1x^(2n-1)/(2n-1)! }

which, if you solve analytically, you find that y = Acos(3x) + (B/3)sin(3x)