Question

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

Answer 1

Answer on Question #81460 – Math – Differential Equations

Question

1. Solve the following ODE using the power series method:

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

Solution

Let

$$y = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + \dots$$

Then

$$y' = a_1 + 2 a_2 x + 3 a_3 x^2 + 4 a_4 x^3 + \dots$$ $$y'' = 2 a_2 + 6 a_3 x + 12 a_4 x^2 + \dots$$ $$x y' = a_1 x + 2 a_2 x^2 + 3 a_3 x^3 + 4 a_4 x^4 + \dots$$ $$(x + 2) y'' = 4 a_2 + (2 a_2 + 12 a_3) x + (6 a_3 + 24 a_4) x^2 + \dots$$

And finally:

$$(x + 2) y'' + x y' - y = (4 a_2 - a_0) + (2 a_2 + 12 a_3) x + (a_2 + 6 a_3 + 24 a_4) x^2 + \dots$$

But since

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

we come to the set of equations:

$$\left\{ \begin{array}{c} (4 a_2 - a_0) = 0 \\ (2 a_2 + 12 a_3) = 0 \\ (a_2 + 6 a_3 + 24 a_4) = 0 \\ \vdots \end{array} \right.$$

Solving it we obtain $a_i$:

$$\left\{ \begin{array}{c} a_2 = a_0 / 4 \\ a_3 = -a_0 / 24 \\ a_4 = 0 \\ \vdots \end{array} \right.$$

So, $a_i$ ($i \neq 1$) depends on $a_0$ only, and if we define

$$h(x) = 1 + \frac{x^2}{4} - \frac{x^3}{24} + \dots$$

then all solutions could be found in the form of

$$y(x) = a_0 h(x) + a_1 x$$

**Answer**

$$y(x) = a_0 h(x) + a_1 x,$$

where

$$h(x) = 1 + \frac{x^2}{4} - \frac{x^3}{24} + \dots$$

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