Question

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

Answer 1

Answer on Question #81521- Math - Differential Equations

Solve the following ODE using the power series method:

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

Solution

We assume there is a solution of the form

$$y = \sum_{n=0}^{\infty} c_n x^n.$$

Then

$$y' = \sum_{n=1}^{\infty} n \cdot c_n x^{n-1},$$

and

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

Substituting in the differential equation, we get

$$(x + 2) \sum_{n=2}^{\infty} (n - 1) n \cdot c_n x^{n-2} + x \sum_{n=1}^{\infty} n \cdot c_n x^{n-1} - \sum_{n=0}^{\infty} c_n x^n = 0$$ $$x \sum_{n=2}^{\infty} (n - 1) n \cdot c_n x^{n-2} + 2 \sum_{n=2}^{\infty} (n - 1) n \cdot c_n x^{n-2} + \sum_{n=1}^{\infty} n \cdot c_n x^n - \sum_{n=0}^{\infty} c_n x^n = 0$$ $$\sum_{n=2}^{\infty} (n - 1) n \cdot c_n x^{n-1} + 2 \sum_{n=2}^{\infty} (n - 1) n \cdot c_n x^{n-2} + \sum_{n=1}^{\infty} n \cdot c_n x^n - \sum_{n=0}^{\infty} c_n x^n = 0$$ $$\sum_{n=1}^{\infty} (n + 1) n \cdot c_{n+1} x^n + 2 \sum_{n=0}^{\infty} (n + 2) (n + 1) \cdot c_{n+2} x^n + \sum_{n=0}^{\infty} (n - 1) \cdot c_n x^n = 0$$ $$\sum_{n=1}^{\infty} \left[ (n + 1) n \cdot c_{n+1} + 2 (n + 1) (n + 2) \cdot c_{n+2} + (n - 1) \cdot c_n \right] x^n = 0$$

This equation is true if the coefficient of $x^n$ is 0:

$$\sum_{n=1}^{\infty} \left[ (n + 1) n \cdot c_{n+1} + 2 (n + 1) (n + 2) \cdot c_{n+2} + (n - 1) \cdot c_n \right] x^n = 0$$ $$n (n + 1) \cdot c_{n+1} + 2 (n + 1) (n + 2) \cdot c_{n+2} + (n - 1) \cdot c_n = 0$$ $$c_{n+2} = \frac{(1 - n) \cdot c_n - n(n + 1) \cdot c_{n+1}}{2(n + 1)(n + 2)}, \quad n = 0, 1, 2, 3 \dots$$

Put $n = 0,1,2,3\ldots$

$n = 0$:

$$c_2 = \frac{c_0}{2 \cdot 1 \cdot 2} = \frac{c_0}{4}$$

$n = 1$:

$$c_3 = \frac{-1 \cdot 2 \cdot c_2}{2 \cdot 2 \cdot 3} = -\frac{c_2}{2 \cdot 3} = -\frac{c_0}{2 \cdot 3 \cdot 4} = -\frac{c_0}{4!}$$

$n = 2$:

$$c_4 = \frac{-1 \cdot c_2 - 2 \cdot 3c_3}{2 \cdot 3 \cdot 4} = \frac{-\frac{c_0}{4} - 6 \cdot \left(-\frac{c_0}{2 \cdot 3 \cdot 4}\right)}{2 \cdot 3 \cdot 4} = 0$$

$n = 3$:

$$c_5 = \frac{-2 \cdot c_3 - 3 \cdot 4c_4}{2 \cdot 4 \cdot 5} = \frac{-2 \cdot \left(-\frac{c_0}{4!}\right)}{2 \cdot 4 \cdot 5} = \frac{c_0}{4 \cdot 5!}$$

$n = 4$:

$$c_6 = \frac{-3 \cdot c_4 - 4 \cdot 5c_5}{2 \cdot 5 \cdot 6} = \frac{-4 \cdot 5 \cdot \frac{c_0}{4 \cdot 5!}}{2 \cdot 5 \cdot 6} = -\frac{c_0}{2 \cdot 6!}$$

The solution is

$$\begin{aligned} y &= c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \cdots = \\ &= c_0 \left(1 + \frac{1}{4} x^2 - \frac{1}{4!} x^3 + \frac{1}{4 \cdot 5!} x^5 - \frac{1}{2 \cdot 6!} x^6 + \cdots\right) \end{aligned}$$

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