Question

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

Answer 1

Answer on Question #75304 – Math – Differential Equations Question

Solve the following ODE using power series method.

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

Solution

Let assume the solution in form $y = \sum_{n=0}^{\infty} a_n x^n$, exists

Then

$$y' = \sum_{n=1}^{\infty} n a_n x^{n-1}$$ $$y'' = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}$$

Substitute

$$(x + 2) \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} + x \sum_{n=1}^{\infty} n a_n x^{n-1} - \sum_{n=0}^{\infty} a_n x^n = 0$$ $$\sum_{n=2}^{\infty} 2n(n-1) a_n x^{n-2} + \sum_{n=2}^{\infty} n(n-1) a_n x^{n-1} + \sum_{n=1}^{\infty} n a_n x^n - \sum_{n=0}^{\infty} a_n x^n = 0$$ $$\sum_{N=0}^{\infty} 2(N+2)(N+1) a_{N+2} x^N + \sum_{N=1}^{\infty} N(N+1) a_{N+1} x^N + \sum_{N=1}^{\infty} N a_N x^N - \sum_{N=0}^{\infty} a_N x^N = 0$$ $$\begin{array}{l} 2(0+2)(0+1) a_{0+2} x^0 + \sum_{N=1}^{\infty} 2(N+2)(N+1) a_{N+2} x^N + \\ + \sum_{N=1}^{\infty} N(N+1) a_{N+1} x^N + \sum_{N=1}^{\infty} N a_N x^N - c_0 x^0 - \sum_{N=1}^{\infty} a_N x^N = 0 \end{array}$$ $$4 a_2 - a_0 = 0$$ $$\sum_{N=1}^{\infty} x^N [2(N+2)(N+1) a_{N+2} + N(N+1) a_{N+1} + (N-1) a_N] = 0$$ $$2(N+2)(N+1) a_{N+2} + N(N+1) a_{N+1} + (N-1) a_N = 0, N \geq 1$$ $$a_{N+2} = - \frac{N(N+1) a_{N+1} + (N-1) a_N}{2(N+2)(N+1)}$$

Plug in $a_0 = 1, a_1 = 0$ to find the first solution $y_1(x)$

$$\begin{array}{l} 4 a _ {2} - 1 = 0 \Rightarrow a _ {2} = \frac {1}{4} \\ a _ {3} = - \frac {1 (1 + 1) \left(\frac {1}{4}\right) + (1 - 1) (0)}{2 (1 + 2) (1 + 1)} = - \frac {1}{24} \\ a _ {4} = - \frac {2 (2 + 1) \left(- \frac {1}{24}\right) + (2 - 1) \left(\frac {1}{4}\right)}{2 (2 + 2) (2 + 1)} = 0 \\ a _ {5} = - \frac {3 (3 + 1) (0) + (3 - 1) \left(- \frac {1}{24}\right)}{2 (3 + 2) (3 + 1)} = \frac {1}{480} \\ a _ {6} = - \frac {4 (4 + 1) \left(\frac {1}{480}\right) + (4 - 1) (0)}{4 (4 + 2) (4 + 1)} = - \frac {1}{2880} \\ \end{array}$$ $$y _ {1} (x) = 1 + \frac {1}{4} x ^ {2} - \frac {1}{24} x ^ {3} + \frac {1}{480} x ^ {5} - \frac {1}{2880} x ^ {6} + \dots$$

Plug in $a_0 = 0, a_1 = 1$ to find the second solution $y_2(x)$ $4a_{2} - a_{0} = 0 \Rightarrow a_{2} = 0$

$$\begin{array}{l} a _ {3} = - \frac {1 (1 + 1) (0) + (1 - 1) (1)}{2 (1 + 2) (1 + 1)} = 0 \\ a _ {4} = - \frac {2 (2 + 1) (0) + (2 - 1) (0)}{2 (2 + 2) (2 + 1)} = 0 \\ a _ {5} = - \frac {3 (3 + 1) (0) + (3 - 1) (0)}{2 (3 + 2) (3 + 1)} = 0 \\ a _ {6} = - \frac {4 (4 + 1) (0) + (4 - 1) (0)}{4 (4 + 2) (4 + 1)} = 0 \\ y _ {2} (x) = x \\ \end{array}$$

The general solution is

$$y = c _ {1} y _ {1} + c _ {2} x$$

Answer: $y = c_{1}y_{1} + c_{2}x$ ,

$$\begin{array}{l} y _ {1} (x) = 1 + \frac {1}{4} x ^ {2} - \frac {1}{24} x ^ {3} + \frac {1}{480} x ^ {5} - \frac {1}{2880} x ^ {6} + \dots \\ y _ {2} (x) = x \\ \end{array}$$

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