Consider the following linear homogeneous second order ODE: \[\begin{equation} y''+p(x)y'+q(x)y=0. \;\;\;\;(60) \end{equation}\] 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: \[\begin{equation} y=a_0+a_1 x+a_2 x^2+a_3 x^3+\cdots=\sum_{n=0}^{\infty}a_n x^n \;\;\;\;(61) \end{equation}\] 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 \[\begin{equation} y''=2a_2+3\cdot2 a_3x+4\cdot 3 a_4 x^2+\cdots =\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2} x^{n} \;\;\;\;(62) \end{equation}\] 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: \[\begin{equation} y=a_0+a_1 x+a_2 x^2+a_3 x^3+\cdots=\sum_{n=0}^{\infty}a_n x^n \;\;\;\;(63) \end{equation}\] 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 \(y=c_1(e^x-x)+ c_2 x\))

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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