Question

Solve the following ODE using the power series method:
(x + 2)y′′ + yx ′ − y = 0

Answer 1

Answer on Question #82483 – Math – Differential Equations

Question

Solve the following ODE using the power series method:

$$(x + 2) y'' + x y' - y = 0$$

Solution

$$y = \sum_{n=0}^{\infty} a_n x^n$$ $$y' = \sum_{n=1}^{\infty} a_n n x^{n-1}$$ $$y'' = \sum_{n=2}^{\infty} a_n n(n-1) x^{n-2}$$ $$(x + 2) \sum_{n=2}^{\infty} a_n n(n-1) x^{n-2} + x \sum_{n=1}^{\infty} a_n n x^{n-1} - \sum_{n=0}^{\infty} a_n x^n = 0$$ $$\sum_{n=2}^{\infty} a_n n(n-1) x^{n-1} + 2 \sum_{n=2}^{\infty} a_n n(n-1) x^{n-2} + \sum_{n=1}^{\infty} a_n n x^n - \sum_{n=0}^{\infty} a_n x^n = 0$$ $$n - 2 \rightarrow N, n \rightarrow N + 2$$ $$\sum_{n=2}^{\infty} a_n n(n-1) x^{n-1} + 2 \sum_{N+2=2}^{\infty} a_{N+2} (N+2) (N+2-1) x^{N+2-2} +$$ $$+ \sum_{n=1}^{\infty} a_n n x^n - \sum_{n=0}^{\infty} a_n x^n = 0$$ $$n - 1 \rightarrow N, n \rightarrow N + 1$$ $$\sum_{N+1=2}^{\infty} a_{N+1} (N+1) (N+1-1) x^{N+1-1} + 2 \sum_{N=0}^{\infty} a_{N+2} (N+2) (N+1) x^N +$$ $$+ \sum_{n=1}^{\infty} a_n n x^n - \sum_{n=0}^{\infty} a_n x^n = 0$$ $$\sum_{N=1}^{\infty} a_{N+1} (N+1) N x^N + 2 \sum_{N=0}^{\infty} a_{N+2} (N+2) (N+1) x^N +$$ $$+ \sum_{N=1}^{\infty} a_N N x^N - \sum_{N=0}^{\infty} a_N x^N = 0$$ $$\sum_{N=1}^{\infty} (a_{N+1}(N+1)N + 2a_{N+2}(N+2)(N+1) + a_NN - a_N)x^N + 2a_{0+2}(0+2)(0+1) - a_0 = 0$$ $$4a_2 - a_0 = 0$$ $$a_{N+1}(N+1)N + 2a_{N+2}(N+2)(N+1) + a_N(N-1) = 0$$ $$y = a_0 + \sum_{n=1}^{\infty} a_n x^n$$ $$4a_2 - a_0 = 0$$ $$a_{n+2} = -\frac{a_{n+1}(n+1)n + a_n(n-1)}{2(n+2)(n+1)}, \quad n \geq 1$$

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