Question

Use the power series method to obtain one solution of the following ODE:x^2y"+3y'-xy=0

Answer 1

Answer on Question #70017 – Math – Differential Equations

Question

Use the power series method to obtain one solution of the following ODE:

$$x^{2}y'' + 3y' - xy = 0$$

Solution

Assume that $y = \sum_{n=0}^{\infty} a_n x^n$ is a solution. Then we have

$$\begin{array}{l} y' = \sum_{n=1}^{\infty} n a_n x^{n-1} \\ y'' = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}, x^2 y'' = \sum_{n=2}^{\infty} n(n-1) a_n x^n \\ x y = \sum_{n=0}^{\infty} a_n x^{n+1} \\ \end{array}$$

Substituting for $x^2 y'', y'$ and $xy$ in the given differential equation, we obtain the following series.

$$\begin{array}{l} \sum_{n=2}^{\infty} n(n-1) a_n x^n + 3 \sum_{n=1}^{\infty} n a_n x^{n-1} - \sum_{n=0}^{\infty} a_n x^{n+1} = 0 \\ \sum_{n=2}^{\infty} n(n-1) a_n x^n + 3 \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n - \sum_{n=1}^{\infty} a_{n-1} x^n = 0 \\ \end{array}$$ $$\begin{array}{l} \sum_{n=2}^{\infty} n(n-1) a_n x^n + 3(1) a_1 x^0 + 3(2) a_2 x^1 + \sum_{n=2}^{\infty} 3(n+1) a_{n+1} x^n - a_0 x^1 - \\ - \sum_{n=2}^{\infty} a_{n-1} x^n = 0 \\ \end{array}$$ $$\begin{array}{l} 3 a_1 = 0 \Rightarrow a_1 = 0 \\ 6 a_2 - a_0 = 0 \Rightarrow a_2 = \frac{a_0}{6} \\ \end{array}$$ $$\begin{array}{l} \sum_{n=2}^{\infty} n(n-1) a_n x^n + \sum_{n=2}^{\infty} 3(n+1) a_{n+1} x^n + \sum_{n=2}^{\infty} (-1) a_{n-1} x^n = 0 \\ \sum_{n=2}^{\infty} x^n [n(n-1) a_n + 3(n+1) a_{n+1} - a_{n-1}] = 0 \\ \end{array}$$ $$n(n - 1)a_n + 3(n + 1)a_{n+1} - a_{n-1} = 0, n = 2, 3, 4, \dots$$ $$n = 2$$ $$2a_2 + 9a_3 - a_1 = 0 \Rightarrow 9a_3 = -2a_2 \Rightarrow a_3 = -\frac{2}{9}a_2 = -\frac{2}{9}\left(\frac{a_0}{6}\right) = -\frac{a_0}{27}$$ $$n = 3$$ $$6a_3 + 12a_4 - a_2 = 0 \Rightarrow 12a_4 = -6a_3 + a_2 \Rightarrow a_4 = \frac{-6\left(-\frac{a_0}{27}\right) + \frac{a_0}{6}}{12} = \frac{7a_0}{216}$$ $$n = 4$$ $$\begin{array}{l} 12a_4 + 15a_5 - a_3 = 0 \Rightarrow 15a_5 = -12a_4 + a_3 \Rightarrow a_5 = \frac{-12\left(\frac{7a_0}{216}\right) - \frac{a_0}{27}}{15} = \\ = -\frac{23a_0}{810} \end{array}$$ $$n = 5$$ $$\begin{array}{l} 20a_5 + 18a_6 - a_4 = 0 \Rightarrow 18a_6 = -20a_5 + a_4 \Rightarrow \\ \Rightarrow a_6 = \frac{-20\left(-\frac{23a_0}{810}\right) + \frac{7a_0}{216}}{18} = \frac{389a_0}{11664} \end{array}$$ $$\begin{array}{l} y(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + a_5x^5 + a_6x^6 + \dots \\ y(x) = a_0 + 0 + \frac{a_0}{6}x^2 - \frac{a_0}{27}x^3 + \frac{7a_0}{216}x^4 - \frac{23a_0}{810}x^5 + \frac{389a_0}{11664}x^6 + \dots = \\ = a_0\left(1 + \frac{x^2}{6} - \frac{x^3}{27} + \frac{7x^4}{216} - \frac{23x^5}{810} + \frac{389x^6}{11664} + \dots\right) \end{array}$$ $$\begin{array}{l} y_1(x) = 1 + \frac{x^2}{6} - \frac{x^3}{27} + \frac{7x^4}{216} - \frac{23x^5}{810} + \frac{389x^6}{11664} + \dots \\ y_2(x) = 0 \end{array}$$

Answer: $y_1(x) = 1 + \frac{x^2}{6} - \frac{x^3}{27} + \frac{7x^4}{216} - \frac{23x^5}{810} + \frac{389x^6}{11664} + \dots$

Answer provided by https://www.AssignmentExpert.com