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## Series Solutions about an Ordinary Point

Consider the following linear homogeneous second order ODE: $$$y''+p(x)y'+q(x)y=0. \;\;\;\;(60)$$$ If $$p$$ and $$q$$ are constant, then we learned how to find the solution. Now we find a series solution when $$p$$ and $$q$$ are analytic at some point $$x_0$$. If $$p$$ and $$q$$ are analytic at $$x_0$$, then $$x_0$$ is called an ordinary point of (60). If one of$$p$$ and $$q$$ is not analytic at $$x_0$$, then $$x_0$$ is called a singular point of (60).

Theorem. If $$x_0$$ is an ordinary point of (60), then the general solution is $y=\sum_{n=0}^{\infty}a_n(x-x_0)^n=a_0y_1(x)+a_1y_2(x),$ where $$y_1$$ and $$y_2$$ are two fundamental power series solutions of (60) about $$x_0$$.

Example. $$y''+y=0$$

1. Find two power series solutions $$y_1$$ and $$y_2$$ about $$0$$.

2. Show that power series solutions $$y_1$$ and $$y_2$$ are fundamental solutions.

3. Write the general solution in terms of power series solutions.

Solution. Suppose we have the following power series solution: $$$y=a_0+a_1 x+a_2 x^2+a_3 x^3+\cdots=\sum_{n=0}^{\infty}a_n x^n \;\;\;\;(61)$$$ Differentiating we get $\begin{eqnarray*} y'=&a_1 +2a_2 x+3a_3 x^2+4a_4x^3+\cdots &=\sum_{n=1}^{\infty}na_nx^{n-1}\\ y''=&2a_2+3\cdot2 a_3x+4\cdot 3 a_4 x^2+\cdots &=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2} \end{eqnarray*}$ Shifting index we get $$$y''=2a_2+3\cdot2 a_3x+4\cdot 3 a_4 x^2+\cdots =\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2} x^{n} \;\;\;\;(62)$$$ Using (61) and (62) in $$y''+y=0$$, we get \begin{align*} \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2} x^{n} +\sum_{n=0}^{\infty}a_n x^n &=0\\ \sum_{n=0}^{\infty}[(n+2)(n+1)a_{n+2}+a_n] x^{n} &=0 \end{align*} Since the series is zero for all $$x$$, each coefficient is zero, i.e., $(n+2)(n+1)a_{n+2}+a_n=0 \text{ for } n=0,1,2,\ldots$ So we have $a_{n+2}=-\frac{a_n}{(n+2)(n+1)} \text{ for } n=0,1,2,\ldots$ Note the pattern \begin{align*} n=0 &\implies a_2 =-\frac{a_0}{2\cdot 1} =-\frac{a_0}{2!} \\ n=2 &\implies a_4 =-\frac{a_2}{4\cdot 3} =\frac{a_0}{4\cdot 3\cdot 2!} =\frac{a_0}{4!}\\ n=4 &\implies a_6 =-\frac{a_4}{6\cdot 5} =-\frac{a_0}{6\cdot 5\cdot 4!} =-\frac{a_0}{6!}\\ &\cdots\\ n=2k-2 &\implies a_{2k} =(-1)^k \frac{a_0}{(2k)!}= \frac{(-1)^k}{(2k)!}a_0,\; k=0,1,2,3,\ldots \end{align*} Similarly \begin{align*} n=1 &\implies a_3 =-\frac{a_1}{3\cdot 2} =-\frac{a_1}{3!} \\ n=3 &\implies a_5 =-\frac{a_3}{5\cdot 4} =\frac{a_1}{5\cdot 4\cdot 3!} =\frac{a_1}{5!}\\ n=5 &\implies a_7 =-\frac{a_5}{7\cdot 6} =-\frac{a_1}{7\cdot 6\cdot 5!} =-\frac{a_1}{7!}\\ &\cdots\\ n=2k-1 &\implies a_{2k+1} =(-1)^k \frac{a_1}{(2k+1)!}= \frac{(-1)^k}{(2k+1)!}a_1,\; k=0,1,2,3,\ldots \end{align*} Thus \begin{align*} y&=a_0+a_1 x+a_2 x^2+a_3 x^3+ a_4 x^4+a_5 x^5+\cdots\\ &= a_0+a_1 x-\frac{a_0}{2!}x^2 -\frac{a_1}{3!}x^3+ \frac{a_0}{4!}x^4 +\frac{a_1}{5!}x^5+\cdots\\ &=a_0\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots \right)+ a_1 \left(\frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots \right)\\ &=a_0 \sum_{n=0}^{\infty}(-1)^n \frac{x^{2n}}{(2n)!} + a_1 \sum_{n=0}^{\infty}(-1)^n \frac{x^{2n+1}}{(2n+1)!}.\\ \end{align*}
(a) We can verify that $$y_1=\displaystyle\sum_{n=0}^{\infty}(-1)^n \frac{x^{2n}}{(2n)!}$$ and $$y_2=\displaystyle\sum_{n=0}^{\infty}(-1)^n \frac{x^{2n+1}}{(2n+1)!}$$ are two power series solutions of $$y''+y=0$$ which are analytic at $$0$$ (i.e., positive radius of convergence). Here the radii of convergence of $$y_1$$ and $$y_2$$ are $$\infty$$ which means they converge for all $$x$$. So our series solutions are valid for all $$x$$.

(b) Note that $$y_1(0)=1$$ and $$y_2(0)=0$$. Also $$y_1'(0)=0$$ and $$y_2'(0)=1$$. Then $W(y_1,y_2)(0)=\,\begin{array}{|cc|}1& 0\\ 0& 1\end{array}=1\neq 0.$ Thus $$y_1$$ and $$y_2$$ are linearly independent and hence fundamental solutions.

(c) The general solution is $y=a_0 \sum_{n=0}^{\infty}(-1)^n \frac{x^{2n}}{(2n)!} + a_1 \sum_{n=0}^{\infty}(-1)^n \frac{x^{2n+1}}{(2n+1)!}.$ (But we already knew the general solution to be $$y=c_1\cos x+ c_2\sin x$$)

Example. $$(1-x)y''+xy'-y=0$$

1. Find the first four nonzero terms in each of two power series solutions $$y_1$$ and $$y_2$$ about $$0$$.

2. Show that power series solutions $$y_1$$ and $$y_2$$ are fundamental solutions.

3. Write the general solution in terms of power series solutions.

Solution. Suppose we have the following power series solution: $$$y=a_0+a_1 x+a_2 x^2+a_3 x^3+\cdots=\sum_{n=0}^{\infty}a_n x^n \;\;\;\;(63)$$$ Differentiating we get $\begin{eqnarray*} y'=&a_1 +2a_2 x+3a_3 x^2+4a_4x^3+\cdots &=\sum_{n=1}^{\infty}na_nx^{n-1}\\ y''=&2a_2+3\cdot2 a_3x+4\cdot 3 a_4 x^2+\cdots &=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2} \end{eqnarray*}$ Shifting index we get $\begin{eqnarray} y'=a_1 +2a_2 x+3a_3 x^2+4a_4x^3+\cdots =\sum_{n=0}^{\infty} (n+1)a_{n+1}x^{n} \;\;\;\;(64)\\ y''=2a_2+3\cdot2 a_3x+4\cdot 3 a_4 x^2+\cdots =\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2} x^{n} \;\;\;\;(65) \end{eqnarray}$ Using (63), (64), and (65) in $$(1-x)y''+xy'-y=0$$, we get \begin{align*} (1-x)\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2} x^{n} +x\sum_{n=0}^{\infty} (n+1)a_{n+1}x^{n}-\sum_{n=0}^{\infty}a_n x^n &=0\\ \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2} x^{n} -\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2} x^{n+1} +\sum_{n=0}^{\infty} (n+1)a_{n+1}x^{n+1}-\sum_{n=0}^{\infty}a_n x^n &=0\\ \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2} x^{n} +\sum_{n=0}^{\infty} [-(n+2)(n+1)a_{n+2}+(n+1)a_{n+1}] x^{n+1} -\sum_{n=0}^{\infty}a_n x^n &=0\\ \end{align*} Shifting index in the middle series we get \begin{align} \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2} x^{n} +\sum_{n=1}^{\infty} [-(n+1)na_{n+1}+na_{n}] x^{n} -\sum_{n=0}^{\infty}a_n x^n &=0 \nonumber\\ \left(2a_2+\sum_{n=1}^{\infty}(n+2)(n+1)a_{n+2} x^{n} \right) +\sum_{n=1}^{\infty} [-(n+1)na_{n+1}+na_{n}] x^{n} +\left( -a_0-\sum_{n=1}^{\infty}a_n x^n \right) &=0\nonumber\\ 2a_2-a_0+ \sum_{n=1}^{\infty} [(n+2)(n+1)a_{n+2}-(n+1)na_{n+1}+na_{n}-a_n] x^{n} &=0\nonumber\\ 2a_2-a_0+ \sum_{n=1}^{\infty} [(n+2)(n+1)a_{n+2}-(n+1)na_{n+1}+(n-1)a_{n}] x^{n} &=0\;\;\;\;(66) \end{align} Since the series is zero for all $$x$$, each coefficient is zero, i.e., \begin{align*} 2a_2-a_0 &= 0\\ (n+2)(n+1)a_{n+2}-(n+1)na_{n+1}+(n-1)a_{n} &= 0 \text{ for } n=1,2,3,\ldots\\ \end{align*} So we have \begin{align*} & \hspace{32pt} && a_2 =\frac{a_0}{2} \\ n=1 &\implies 6a_3-2a_2 =0 &\implies &a_3=\frac{a_2}{3} =\frac{a_0}{6}\\ n=2 &\implies 12a_4-6a_3+a_2 =0 &\implies &a_4=\frac{a_3}{2}-\frac{a_2}{12}=\frac{a_0}{12}-\frac{a_0}{24}=\frac{a_0}{24} \end{align*} Thus \begin{align*} y&=a_0+a_1 x+a_2 x^2+a_3 x^3+ a_4 x^4+a_5 x^5+\cdots\\ &= a_0+a_1 x+\frac{a_0}{2}x^2 +\frac{a_0}{6}x^3+ \frac{a_0}{24}x^4 +\cdots\\ &=a_0\left(1+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\cdots \right)+ a_1 x.\\ \end{align*} (a) We can verify that $$y_1=1+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\cdots$$ and $$y_2=x$$ are two power series solutions about $$0$$.

(b) Note that $$y_1(0)=1$$ and $$y_2(0)=0$$. Also $$y_1'(0)=0$$ and $$y_2'(0)=1$$. Then $W(y_1,y_2)(0)=\,\begin{array}{|cc|}1& 0\\ 0& 1\end{array}=1\neq 0.$ Thus $$y_1$$ and $$y_2$$ are linearly independent and hence fundamental solutions.

(c) The general solution is $y=a_0\left(1+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\cdots \right)+ a_1 x.$ (We can verify that the general solution is $$y=c_1(e^x-x)+ c_2 x$$)

Suppose we have the following power series solution: $$$y=a_0+a_1 x+a_2 x^2+a_3 x^3+\cdots \;\;\;\;(67)$$$ Differentiating we get $$$y'=a_1 +2a_2 x+3a_3 x^2+4a_4x^3+\cdots \;\;\;\;(68)$$$ $$$y''=2a_2+6 a_3x+12 a_4 x^2+\cdots \;\;\;\;(69)$$$ Using (67), (68), and (69) in $$(1-x)y''+xy'-y=0$$, we get \begin{align*} (1-x)(2a_2+6 a_3x+12 a_4 x^2+\cdots) +x(a_1 +2a_2 x+3a_3 x^2+4a_4x^3+\cdots)&\\ -(a_0+a_1 x+a_2 x^2+a_3 x^3+\cdots) &=0\\ (2a_2-a_0)+(6a_3-2a_2)x+(12a_4-6a_3+a_2)x^2+\cdots &=0. \end{align*} Since the series is zero for all $$x$$, each coefficient is zero. Then we have all the coefficients in terms of $$a_0$$ and $$a_1$$: \begin{align*} 2a_2-a_0=0 &\implies a_2 =\frac{a_0}{2} \\ 6a_3-2a_2 =0 &\implies a_3=\frac{a_2}{3} =\frac{a_0}{6}\\ 12a_4-6a_3+a_2 =0 &\implies a_4=\frac{a_3}{2}-\frac{a_2}{12}=\frac{a_0}{12}-\frac{a_0}{24}=\frac{a_0}{24} \end{align*} Thus \begin{align*} y&=a_0+a_1 x+a_2 x^2+a_3 x^3+ a_4 x^4+a_5 x^5+\cdots\\ &= a_0+a_1 x+\frac{a_0}{2}x^2 +\frac{a_0}{6}x^3+ \frac{a_0}{24}x^4 +\cdots\\ &=a_0\left(1+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\cdots \right)+ a_1 x.\\ \end{align*}

Example. $$(1-x)y''+xy'-y=6\sin x$$
Find the general solution in terms of power series with first four nonzero terms.

Solution. By the preceding problem we have the homogeneous solution: $y_h=a_0\left(1+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\cdots \right)+ a_1 x.$ Now to find a particular solution $$y_p=\sum_{n=0}^{\infty}a_n x^n$$ we plug $$\sin x=\displaystyle \frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$$ in (66): $2a_2-a_0+ \sum_{n=1}^{\infty} [(n+2)(n+1)a_{n+2}-(n+1)na_{n+1}+(n-1)a_{n}] x^{n} = \frac{6x}{1!}-\frac{6x^3}{3!}+\frac{6x^5}{5!}-\frac{6x^7}{7!}+\cdots$ Comparing the coefficients of like power terms of $$x$$, we get \begin{align*} 2a_2-a_0=0 &\implies & a_2 =\frac{a_0}{2} \\ 6a_3-2a_2 =6 &\implies &a_3=\frac{a_2}{3}+1 =\frac{a_0}{6}+1\\ 12a_4-6a_3+a_2 =0 &\implies &a_4=\frac{a_3}{2}-\frac{a_2}{12}=\frac{a_0}{24}+\frac{1}{2} \end{align*} Plugging $$a_0=6$$ and $$a_1=1$$, we get $y_p=6+x+3x^2+2x^3+\frac{3}{4}x^4+\cdots.$ Thus the general solution is $y=y_h+y_p=a_0\left(1+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\cdots \right)+ a_1 x+\left(6+x+3x^2+2x^3+\frac{3}{4}x^4+\cdots\right).$

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