# Section 5.2 – Series Solutions Near an Ordinary Point I

Directions. The following are review problems for the section. It is recommended you work the problems yourself, and then click "Answer" to check your answer. If you do not understand a problem, you can click "Video" to learn how to solve it.
1. For the equation $$(x^2+1)y''+xy'-y=0$$
1. Determine a lower bound for the radius of convergence of the series solutions of the differential equation about $$x_0=0$$.
2. Seek its power series solution about $$x_0=0$$; find the recurrence relation.
3. Find the general term of each solution $$y_1(x)$$ and $$y_2(x)$$.
4. Find the first four terms in each of two solutions $$y_1$$ and $$y_2$$. Show that $$W[y_1, y_2](0)\ne 0$$.​

1. The lower bound is 1.
2. $$\displaystyle \sum_{n=0}^{\infty}n(n-1)a_nx^n+\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n+\sum_{n=0}^{\infty}na_nx^n-\sum_{n=0}^{\infty}a_nx^n=0$$ . The recurrence relation is $$(n+2)a_{n+2}+(n-1)a_n=0$$ for all $$n=0,1,2,..$$
3. $$y_1=1+\displaystyle\sum_{k=1}^{\infty}(-1)^{k-1}\dfrac{(2k-3)!!}{(2k)!!}x^{2k}$$ where $$(-1)!!=1, \quad$$ $$y_2=x$$
4. $$y_1=1+\dfrac{1}{2}x^2-\dfrac{1}{(2)(4)}x^4+\dfrac{3(1)}{(2)(4)(6)}x^6-...\quad$$ and $$\quad\quad y_2=x$$
The Wronskian is $W[y_1,y_2](0)= \begin{vmatrix} 1 & 0 \\ 0 & 1 \\ \end{vmatrix} =1$

To see the full video page and find related videos, click the following link.

2. For the equation $$(x^2+1)y''-6y=0$$
1. Determine a lower bound for the radius of convergence of the series solutions of the differential equation about $$x_0=0$$.
2. Seek its power series solution about $$x_0=0$$; find the recurrence relation.
3. Find the general term of each solution $$y_1(x)$$ and $$y_2(x)$$.
4. Find the first four terms in each of two solutions $$y_1$$ and $$y_2$$. Show that $$W[y_1, y_2](0)\ne 0$$.​

1. The lower bound is 1.
2. $$\displaystyle \sum_{n=0}^{\infty}n(n-1)a_nx^n+\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n-\sum_{n=0}^{\infty}6a_nx^n=0$$ . The recurrence relation is $$a_{n+2}=-\dfrac{n-3}{n+1}a_n$$ for all $$n=0,1,2,...$$
3. $$y_1=1+3x^2+x^4+\displaystyle\sum_{k=3}^{\infty}\dfrac{3(-1)^k}{(2k-1)(2k-3)}x^{2k}, \quad \quad$$ $$y_2=x+x^3$$
4. $$y_1=1+3x^2+x^4-\dfrac{1}{5}x^6+... \quad \quad$$ $$y_2=x+x^3$$
The Wronskian is $W[y_1,y_2](0)= \begin{vmatrix} 1 & 0 \\ 0 & 1 \\ \end{vmatrix} =1$

To see the full video page and find related videos, click the following link.

3. Seek a power series solution for the initial value problem $y''-(1+x)y=0,\quad y(0)=1,y'(0)=-2$ up to the terms of degree 5. Then do the same for finding the general solution of the equation.​

$y=1-2x+\dfrac{1}{2}x^2-\dfrac{1}{6}x^3-\dfrac{1}{8}x^4+\dfrac{1}{60}x^5+...\quad$
For the general solution of the equation, we have that $$y=a_0y_1+a_1y_2$$, where $y_1=1+\dfrac{1}{2}x^2+\dfrac{1}{6}x^3+\dfrac{1}{24}x^4+\dfrac{1}{30}x^5+... \quad$ and $$\quad$$ $y_2=x+\dfrac{1}{6}x^3+\dfrac{1}{12}x^4+\dfrac{1}{120}x^5+...$

To see the full video page and find related videos, click the following link.