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

defazajx 2021-10-01 Answered

(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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Expert Answer

Laith Petty
Answered 2021-10-02 Author has 5786 answers

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
\(\displaystyle{\left({n}+{2}\right)}{\left({n}+{1}\right)}{a}_{{{n}+{2}}}+{\left({n}-{3}\right)}{\left({n}-{2}\right)}{a}_{{{n}}}={0},{n}\in{N}_{{{0}}}\)

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