Question

Solve the following ODE using the power series method: (x^2-1)y"+3xy'+xy=0

Answer 1

Answer on Question #86336 – Math – Differential Equations

Question

Solve the following ODE using the power series method:

$$(x^2 - 1) y'' + 3x y' + x y = 0$$

Solution

We seek the solution of the equation in the form of a power series $y(x) = \sum_{n=0}^{\infty} a_n x^n$. Then $y' = \sum_{n=1}^{\infty} a_n n x^{n-1}$ and $y'' = \sum_{n=2}^{\infty} a_n n (n-1) x^{n-2}$:

$$(x^2 - 1) \sum_{n=2}^{\infty} a_n n (n-1) x^{n-2} + 3x \sum_{n=1}^{\infty} a_n n x^{n-1} + x \sum_{n=0}^{\infty} a_n x^n = 0$$ $$\sum_{n=2}^{\infty} a_n n (n-1) x^n - \sum_{n=2}^{\infty} a_n n (n-1) x^{n-2} + 3 \sum_{n=1}^{\infty} a_n n x^n + \sum_{n=0}^{\infty} a_n x^{n+1} = 0$$ $$\sum_{n=2}^{\infty} a_n n (n-1) x^n - \sum_{n=0}^{\infty} a_{n+2} (n+2) (n+1) x^n + 3 \sum_{n=1}^{\infty} a_n n x^n + \sum_{n=1}^{\infty} a_{n-1} x^n = 0$$ $$\sum_{n=2}^{\infty} a_n n (n-1) x^n - 2a_2 - 6a_3 x - \sum_{n=2}^{\infty} a_{n+2} (n+2) (n+1) x^n + 3a_1 x + 3 \sum_{n=2}^{\infty} a_n n x^n + a_0 x$$ $$+ \sum_{n=2}^{\infty} a_{n-1} x^n = 0$$ $$\left\{ \begin{array}{c} -2a_2 = 0 \\ -6a_3 + 3a_1 + a_0 = 0 \\ a_n n (n-1) - a_{n+2} (n+2) (n+1) + 3a_n n + a_{n-1} = 0, \qquad n = 2,3,4 \dots \end{array} \right.$$ $$\left\{ \begin{array}{c} a_2 = 0 \\ a_3 = \dfrac{3a_1 + a_0}{6} \\ a_n n (n+2) - a_{n+2} (n+2) (n+1) + a_{n-1} = 0, \qquad n = 2,3,4 \dots \end{array} \right.$$ $$\left\{ \begin{array}{c} a_2 = 0 \\ a_3 = \dfrac{3a_1 + a_0}{6} \\ a_{n+2} = \dfrac{a_n n (n+2) + a_{n-1}}{(n+2) (n+1)}, \qquad n = 2,3,4 \dots \end{array} \right.$$ $$\begin{array}{l} y(x) = \sum_{n=0}^{\infty} a_n x^n = a_0 + a_1 x + \dfrac{3a_1 + a_0}{6} x^3 + \sum_{n=2}^{\infty} a_{n+2} x^{n+2} \\ = a_0 + a_1 x + \dfrac{3a_1 + a_0}{6} x^3 + \sum_{n=2}^{\infty} \dfrac{a_n n (n+2) + a_{n-1}}{(n+2) (n+1)} x^{n+2} \end{array}$$

Answer: $y(x) = a_{0} + a_{1}x + \frac{3a_{1} + a_{0}}{6} x^{3} + \sum_{n = 2}^{\infty}\frac{a_{n}n(n + 2) + a_{n - 1}}{(n + 2)(n + 1)} x^{n + 2}$.

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