Question

Solve the following ODE using the power series method: (1 - X^2) y" - 2xy' + 2 y = 0

Answer 1

Answer on Question #50743 – Math – Differential Calculus | Equations

Solve the following ODE using the power series method:

$$(1 - X ^ {\wedge} 2) y'' - 2 x y' + 2 y = 0$$

Solution

Ordinary differential equation

$$(1 - x^2) y'' - 2 x y' + 2 y = 0$$

is equivalent to

$$y'' - \frac{2x}{1 - x^2} y' + \frac{2}{1 - x^2} y = 0.$$

Let

$$y = \sum_{n=0}^{\infty} a_n x^n = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots, \quad y' = \sum_{n=1}^{\infty} n a_n x^{n-1}, \quad y'' = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}$$

Substitute $x = 0$ into $y = \sum_{n=0}^{\infty} a_n x^n$, which results in $y(0) = a_0 = 0$.

Substitute $x = 0$ into $y = \sum_{n=1}^{\infty} n a_n x^{n-1}$, which results in $y'(0) = a_1 = 1$.

We are searching for two linearly independent solutions of differential equation, so let the first one be such that $y(0) = 0, y'(0) = 1$, hence $a_0 = 0$, $a_1 = 1$.

Substitute $y = \sum_{n=0}^{\infty} a_n x^n$, $y' = \sum_{n=1}^{\infty} n a_n x^{n-1}$, $y'' = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}$, into initial differential equation $(1 - x^2) y'' - 2 x y' + 2 y = 0$, which results in

$$\sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - \sum_{n=2}^{\infty} n(n-1) a_n x^n - 2 \sum_{n=1}^{\infty} n a_n x^n + 2 \sum_{n=0}^{\infty} a_n x^n = 0$$ $$\sum_{n=0}^{\infty} (n + 1)(n + 2)a_{n+2}x^n - \sum_{n=2}^{\infty} n(n - 1)a_nx^n - 2\sum_{n=1}^{\infty} na_nx^n + 2\sum_{n=0}^{\infty} a_nx^n = 0$$ $$\sum_{n=2}^{\infty} [(n + 1)(n + 2)a_{n+2} - n(n - 1)a_n - 2na_n + 2a_n]x^n + 2a_2 + 6a_3x - 2a_1x + 2a_0 + 2a_1x = 0$$ $$\sum_{n=2}^{\infty} [(n + 1)(n + 2)a_{n+2} - n(n - 1)a_n - 2na_n + 2a_n]x^n + 2a_2 + 6a_3x + 2a_0 = 0$$

Plug $x = 0$ and $a_0 = 0$ into the last equation, obtain $a_2 = 0$. Rewrite the series as

$$\sum_{n=2}^{\infty} [(n + 1)(n + 2)a_{n+2} - n(n - 1)a_n - 2na_n + 2a_n]x^n + 6a_3x = 0$$

Divide both sides by $x \neq 0$ and obtain

$$\sum_{n=2}^{\infty} [(n + 1)(n + 2)a_{n+2} - n(n - 1)a_n - 2na_n + 2a_n]x^{n-1} + 6a_3 = 0$$

Plug $x = 0$ into the last equation, obtain $6a_3 = 0$, hence $a_3 = 0$.

Thus we have $a_1 = 1$, $a_0 = a_2 = a_3 = 0$.

Rewrite the series as

$$\sum_{n=2}^{\infty} [(n + 1)(n + 2)a_{n+2} - n(n - 1)a_n - 2na_n + 2a_n]x^{n-1} = 0.$$

A power series is identically equal to zero if and only if all of its coefficients are equal to zero.

Hence

$$\begin{array}{l} (n + 1)(n + 2) a_{n+2} - n(n - 1) a_n - 2 n a_n + 2 a_n = 0 \rightarrow \\ \rightarrow \quad a_{n+2} = \frac{n^2 + n - 2}{(n + 1)(n + 2)} a_n = \frac{n - 1}{n + 1} a_n \end{array}$$

Read the recurrence relation for the case $n = 0$: $a_2 = -a_0$. Reading off the recurrence relation for $n = 1$ yields $a_3 = 0$.

Notice that because $a_2 = a_3 = 0$, all rest coefficients must be zero. Therefore the first linear independent solution is $y = x$.

Points $x = 1$ and $x = -1$ are singular points of equation

$$y'' - \frac{2x}{1 - x^2} y' + \frac{2}{1 - x^2} y = 0.$$

The first linearly independent solution is found.

To obtain the second linearly independent solution, the next formula is used:

$$F = \int \frac{f_1}{f_2} dx = \int \frac{-2x}{1 - x^2} dx = \int \frac{d(1 - x^2)}{1 - x^2} dx = \ln |1 - x^2|,$$

general solution can be represented as

$$y = y_1 \left(C_1 + C_2 \int \frac{e^{-F}}{y_1^2} dx\right) = x \left(C_1 + C_2 \int \frac{e^{-ln|1 - x^2|}}{x^2} dx\right),$$

where $C_1$ and $C_2$ are arbitrary real constants.

Let $-1 < x < 1$, then

$$y = y_1 \left(C_1 + C_2 \int \frac{e^{-F}}{y_1^2} dx\right) = x \left(C_1 + C_2 \int \frac{e^{-ln(1 - x^2)}}{x^2} dx\right) =$$ $$\begin{array}{l} = x \left(C _ {1} + C _ {2} \int \frac {d x}{(1 - x ^ {2}) x ^ {2}}\right) = x \left(C _ {1} - C _ {2} \int \frac {d x}{(x ^ {2} - 1) x ^ {2}}\right) \\ = x \left(C _ {1} - C _ {2} \int \left(\frac {1}{x ^ {2} - 1} - \frac {1}{x ^ {2}}\right) d x\right) \\ = x \left(C _ {1} - C _ {2} \left(\frac {1}{2} \ln \left| \frac {x - 1}{x + 1} \right| + \frac {1}{x}\right)\right) \\ = x \left(C _ {1} - C _ {2} \left(\ln \sqrt {\left| \frac {x - 1}{x + 1} \right|} + \frac {1}{x}\right)\right) \\ \end{array}$$

Other method is to search for the second linearly independent solution in the form

$$y = \sum_ {n = 0} ^ {\infty} a _ {n} x ^ {n + r}.$$

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