# (a) Seek power series solutions of the given differential equation about the giv

(a) Seek power series solutions of the given differential equation about the given point $$x_0$$, find the recurrence relation.

(b) Find the first tour terms in each of two solutions $$\displaystyle{y}_{{{1}}}{\quad\text{and}\quad}{y}_{{{2}}}$$ (unless the series terminates sooner).

(c) By evaluating the Wronskian $$\displaystyle{W}{\left({y}_{{{1}}}{y}_{{{2}}}\right)}{\left({x}_{{{0}}}\right)}$$, show that $$\displaystyle{y}_{{{1}}}{\quad\text{and}\quad}{y}_{{{2}}}$$ form a fundamental set of solutions.

(d) It possible, find the general term in each solution $$\displaystyle{\left({1}+{x}^{{{2}}}\right)}{y}\text{}{4}{x}{y}'+{6}{y}={0},{x}_{{{0}}}={0}$$

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Laith Petty

a) For this equation $$\displaystyle{x}_{{{0}}}$$ is an ordinary point so we look for a soultion in the form of a power series about $$\displaystyle{x}_{{{0}}}$$
$$\displaystyle{y}={\sum_{{{n}={0}}}^{{+\infty}}}{a}_{{{n}}}{x}^{{{n}}}$$
which converges in some interval $$|x|{<}{p}$$ The series for y and y" are given by
$$\displaystyle{y}'={\sum_{{{n}={1}}}^{{+\infty}}}{n}{a}_{{{n}}}{x}^{{{n}-{1}}}$$, $$\displaystyle{y}\text{}{\sum_{{{n}={2}}}^{{+\infty}}}{n}{\left({n}-{1}\right)}{a}_{{{n}}}{x}^{{{n}-{2}}}$$
thus, alter plugging these equalities into the initial differential equation, we obtain
$$\displaystyle{\left({1}+{x}^{{{2}}}\right)}={\sum_{{{n}={2}}}^{{+\infty}}}{n}{\left({n}-{1}\right)}{a}_{{{n}}}{x}^{{{n}-{2}}}-{4}{x}{\sum_{{{n}={1}}}^{{+\infty}}}{n}{a}_{{{n}}}{x}^{{{n}-{1}}}+{6}{\sum_{{{n}={0}}}^{{+\infty}}}{a}_{{{n}}}{x}^{{{n}}}={0}$$
$$\displaystyle\Leftrightarrow{\sum_{{{n}={2}}}^{{+\infty}}}{n}{\left({n}-{1}\right)}{a}_{{{n}}}{x}^{{{n}-{2}}}+{\sum_{{{n}={2}}}^{{+\infty}}}{n}{\left({n}-{1}\right)}{a}_{{{n}}}{x}^{{{n}}}-{4}{\sum_{{{n}={1}}}^{{+\infty}}}{n}{a}_{{{n}}}{x}^{{{n}}}+{6}{\sum_{{{n}={0}}}^{{+\infty}}}{a}_{{{n}}}{x}^{{{n}}}={0}$$
$$\displaystyle\Leftrightarrow{\sum_{{{n}={0}}}^{{+\infty}}}{\left({n}+{2}\right)}{\left({n}+{1}\right)}{a}_{{{n}+{2}}}{x}^{{{n}}}+{\sum_{{{n}={2}}}^{{+\infty}}}{n}{\left({n}-{1}\right)}{a}_{{{n}}}{x}^{{{n}}}-{4}{a}_{{{1}}}{x}-{4}{\sum_{{{n}={2}}}^{{+\infty}}}{n}{a}_{{{n}}}{x}^{{{n}}}+{6}{a}_{{{0}}}+{6}{a}_{{{1}}}{x}+{6}{\sum_{{{n}={2}}}^{{+\infty}}}{a}_{{{n}}}{x}^{{{n}}}={0}$$
$$\displaystyle\Leftrightarrow{2}{a}_{{{2}}}+{6}{a}_{{{3}}}{x}{\sum_{{{n}={2}}}^{{+\infty}}}{\left({n}+{2}\right)}{\left({n}+{1}\right)}{a}_{{{n}+{2}}}{x}^{{{n}}}+{\sum_{{{n}={2}}}^{{+\infty}}}{n}{\left({n}-{1}\right)}{a}_{{{n}}}{x}^{{{n}}}+{2}{a}_{{{1}}}{x}-{4}{\sum_{{{n}={2}}}^{{+\infty}}}{n}{a}_{{{n}}}{x}^{{{n}}}+{6}{a}_{{{0}}}+{6}{\sum_{{{n}={2}}}^{{+\infty}}}{a}_{{{n}}}{x}^{{{n}}}={0}$$
$$\displaystyle\Leftrightarrow{2}{a}_{{{2}}}+{6}{a}_{{{0}}}+{\left({6}{a}_{{{3}}}+{2}{a}_{{{1}}}\right)}{x}+{\sum_{{{n}={2}}}^{{+\infty}}}{\left[{\left({n}+{2}\right)}{\left({n}+{1}\right)}{a}_{{{n}+{2}}}+{n}{\left({n}-{1}\right)}{a}_{{{n}}}-{4}{n}{a}_{{{n}}}+{6}{a}_{{{n}}}\right]}{x}^{{{n}}}={0}$$
Step 2
Equalizing the coefficients on both sides of the equation, we obtain
$$\begin{cases}2a_{2}+6a_{0} = 0\\ 6a_{3}+2a_{1}=0\\\left(n+2\right)\left(n+1\right)a_{n+2}+\left(n^{2}-5n+6\right)a_{n}=0, n\geq2, n\in N \end{cases}$$
$$\Leftrightarrow\begin{cases}a_{2}=-3a_{0}\\ a_{3}=-\frac{1}{3}a_{1}\\a_{n+2}=-\frac{\left(n^{2}-5n+6\right)a_{n}}{\left(n+1\right)\left(n+2\right)}=-\frac{\left(n-3\right)\left(n-2\right)a_{n}}{\left(n+1\right)\left(n+2\right)}, n\geq2, n\in N \end{cases}$$
The last equation is true also for $$\displaystyle{n}={0}$$ and $$\displaystyle{n}={1}$$, thus the recurrence relation is
$$\displaystyle{\left({n}+{2}\right)}{\left({n}+{1}\right)}{a}_{{{n}+{2}}}+{\left({n}-{3}\right)}{\left({n}-{2}\right)}{a}_{{{n}}}={0},{n}\in{N}_{{{0}}}$$