7.4 Series Solutions of Differential Equations

1. Assume $y(x)={\displaystyle \sum }_{n=0}^{\infty }{a}_n{x}^n$ (step 1). Then, $y^{\prime }(x)={\displaystyle \sum _{n=1}^{\infty }n}{a}_n{x}^{n-1}$ and $y\text{″}(x)={\displaystyle \sum _{n=2}^{\infty }n}(n-1)a_n{x}^{n-2}$ (step 2). We want to find values for the coefficients $a_n$ such that
   
   $$\begin{array}{ccc}\hfill y\text{″}-y& =\hfill & 0\hfill \\ \hfill {\displaystyle \sum _{n=2}^{\infty }n}(n-1)a_n{x}^{n-2}-{\displaystyle \sum }_{n=0}^{\infty }{a}_n{x}^n& =\hfill & 0\text{(step 3).}\hfill \end{array}$$
   
   We want the indices on our sums to match so that we can express them using a single summation. That is, we want to rewrite the first summation so that it starts with $n=0.$
   To re-index the first term, replace n with $n+2$ inside the sum, and change the lower summation limit to $n=0.$ We get
   
   $${\displaystyle \sum _{n=2}^{\infty }n}(n-1)a_n{x}^{n-2}={\displaystyle \sum _{n=0}^{\infty }(n+2)}(n+1)a_{n+2}{x}^n.$$
   
   This gives
   
   $$\begin{array}{ccc}\hfill {\displaystyle \sum _{n=0}^{\infty }(n+2)}(n+1)a_{n+2}{x}^n-{\displaystyle \sum }_{n=0}^{\infty }{a}_n{x}^n& =\hfill & 0\hfill \\ \hfill {\displaystyle \sum }_{n=0}^{\infty }\left[(n+2)(n+1)a_{n+2}-a_n\right]x^n& =\hfill & 0\text{(step 4).}\hfill \end{array}$$
   
   Because power series expansions of functions are unique, this equation can be true only if the coefficients of each power of x are zero. So we have
   
   $$(n+2)(n+1)a_{n+2}-a_n=0\text{for}n=0,1,2\text{,….}$$
   
   This recurrence relationship allows us to express each coefficient $a_n$ in terms of the coefficient two terms earlier. This yields one expression for even values of n and another expression for odd values of n. Looking first at the equations involving even values of n, we see that
   
   $$\begin{array}{ccc}\hfill a_2& =\hfill & \frac{a_0}{2}\hfill \\ \hfill a_4& =\hfill & \frac{a_2}{4\cdot 3}=\frac{a_0}{4!}\hfill \\ \hfill a_6& =\hfill & \frac{a_4}{6\cdot 5}=\frac{a_0}{6!}\hfill \\ & \text{⋮.}\hfill & \end{array}$$
   
   Thus, in general, when n is even, $a_n=\frac{a_0}{n!}$ (step 5).
   For the equations involving odd values of n, we see that
   
   $$\begin{array}{ccc}\hfill a_3& =\hfill & \frac{a_1}{3\cdot 2}=\frac{a_1}{3!}\hfill \\ \hfill a_5& =\hfill & \frac{a_3}{5\cdot 4}=\frac{a_1}{5!}\hfill \\ \hfill a_7& =\hfill & \frac{a_5}{7\cdot 6}=\frac{a_1}{7!}\hfill \\ & \text{⋮.}\hfill & \end{array}$$
   
   Therefore, in general, when n is odd, $a_n=\frac{a_1}{n!}$ (step 5 continued).
   Putting this together, we have
   
   $$\begin{array}{cc}\hfill y(x)& ={\displaystyle \sum _{n=0}^{\infty }{a}_n}{x}^n\hfill \\ & =a_0+a_{1}x+\frac{a_0}{2}{x}^2+\frac{a_1}{3!}{x}^3+\frac{a_0}{4!}{x}^4+\frac{a_1}{5!}{x}^5+\text{⋯.}\hfill \end{array}$$
   
   Re-indexing the sums to account for the even and odd values of n separately, we obtain
   
   $$y(x)=a_0{\displaystyle \sum }_{k=0}^{\infty }\frac{1}{(2k)!}{x}^{2k}+a_1{\displaystyle \sum }_{k=0}^{\infty }\frac{1}{(2k+1)!}{x}^{2k+1}\text{(step 6).}$$
   
   Analysis for part a.
   As expected for a second-order differential equation, this solution depends on two arbitrary constants. However, note that our differential equation is a constant-coefficient differential equation, yet the power series solution does not appear to have the familiar form (containing exponential functions) that we are used to seeing. Furthermore, since $y(x)=c_1{e}^x+c_2{e}^{\text{−}x}$ is the general solution to this equation, we must be able to write any solution in this form, and it is not clear whether the power series solution we just found can, in fact, be written in that form.
   Fortunately, after writing the power series representations of $e^x$ and $e^{\text{−}x},$ and doing some algebra, we find that if we choose
   
   $$c_0=\frac{(a_0+a_1)}{2},c_1=\frac{(a_0-a_1)}{2},$$
   
   we then have $a_0=c_0+c_1$ and$a_1=c_0-c_1,$ and
   
   $$\begin{array}{cc}\hfill y(x)& =a_0+a_{1}x+\frac{a_0}{2}{x}^2+\frac{a_1}{3!}{x}^3+\frac{a_0}{4!}{x}^4+\frac{a_1}{5!}{x}^5+\text{⋯}\hfill \\ & =(c_0+c_1)+(c_0-c_1)x+\frac{(c_0+c_1)}{2}{x}^2+\frac{(c_0-c_1)}{3!}{x}^3+\frac{(c_0+c_1)}{4!}{x}^4+\frac{(c_0-c_1)}{5!}{x}^5+\text{⋯}\hfill \\ & =c_0{\displaystyle \sum _{n=0}^{\infty }\frac{x^n}{n!}}+c_1{\displaystyle \sum _{n=0}^{\infty }\frac{{(\text{−}x)}^n}{n!}}\hfill \\ & =c_0{e}^x+c_1{e}^{\text{−}x}.\hfill \end{array}$$
   
   So we have, in fact, found the same general solution. Note that this choice of $c_1$ and $c_2$ is not obvious. This is a case when we know what the answer should be, and have essentially “reverse-engineered” our choice of coefficients.
2. Assume $y(x)={\displaystyle \sum }_{n=0}^{\infty }{a}_n{x}^n$ (step 1). Then, $y^{\prime }(x)={\displaystyle \sum _{n=1}^{\infty }n}{a}_n{x}^{n-1}$ and $y\text{″}(x)={\displaystyle \sum _{n=2}^{\infty }n}(n-1)a_n{x}^{n-2}$ (step 2). We want to find values for the coefficients $a_n$ such that
   
   $$\begin{array}{}\\ \\ \hfill (x^2-1)y\text{″}+6x{y}^{\prime }+4y& =\hfill & \mathrm{-4}\hfill \\ \hfill \left(x^2-1\right){\displaystyle \sum _{n=2}^{\infty }n}(n-1)a_n{x}^{n-2}+6x{\displaystyle \sum _{n=1}^{\infty }n}{a}_n{x}^{n-1}+4{\displaystyle \sum _{n=0}^{\infty }{a}_n}{x}^n& =\hfill & \mathrm{-4}\hfill \\ \hfill x^2{\displaystyle \sum _{n=2}^{\infty }n}(n-1)a_n{x}^{n-2}-{\displaystyle \sum _{n=2}^{\infty }n}(n-1)a_n{x}^{n-2}+6x{\displaystyle \sum _{n=1}^{\infty }n}{a}_n{x}^{n-1}+4{\displaystyle \sum _{n=0}^{\infty }{a}_n}{x}^n& =\hfill & \mathrm{-4}.\hfill \end{array}$$
   
   Taking the external factors inside the summations, we get
   
   $${\displaystyle \sum _{n=2}^{\infty }n}(n-1)a_n{x}^n-{\displaystyle \sum _{n=2}^{\infty }n}(n-1)a_n{x}^{n-2}+{\displaystyle \sum _{n=1}^{\infty }6n}{a}_n{x}^n+{\displaystyle \sum _{n=0}^{\infty }4a_n}{x}^n=\mathrm{-4}\text{(step 3).}$$
   
   Now, in the first summation, we see that when $n=0$ or $n=1,$ the term evaluates to zero, so we can add these terms back into our sum to get
   
   $${\displaystyle \sum }_{n=2}^{\infty }n\left(n-1\right)a_n{x}^n={\displaystyle \sum }_{n=0}^{\infty }n\left(n-1\right)a_n{x}^n.$$
   
   Similarly, in the third term, we see that when $n=0,$ the expression evaluates to zero, so we can add that term back in as well. We have
   
   $${\displaystyle \sum _{n=1}^{\infty }6}n{a}_n{x}^n={\displaystyle \sum _{n=0}^{\infty }6}n{a}_n{x}^n.$$
   
   Then, we need only shift the indices in our second term. We get
   
   $${\displaystyle \sum _{n=2}^{\infty }n}(n-1)a_n{x}^{n-2}={\displaystyle \sum _{n=0}^{\infty }(n+2)}(n+1)a_{n+2}{x}^n.$$
   
   Thus, we have
   
   $$\begin{array}{ccc}\hfill {\displaystyle \sum _{n=0}^{\infty }n}(n-1)a_n{x}^n-{\displaystyle \sum _{n=0}^{\infty }(n+2)}(n+1)a_{n+2}{x}^n+{\displaystyle \sum _{n=0}^{\infty }6n}{a}_n{x}^n+{\displaystyle \sum _{n=0}^{\infty }4a_n}{x}^n& =\hfill & \mathrm{-4}\text{(step 4).}\hfill \\ \hfill {\displaystyle \sum }_{n=0}^{\infty }\left[n(n-1)a_n-(n+2)(n+1)a_{n+2}+6n{a}_n+4a_n\right]x^n& =\hfill & \mathrm{-4}\hfill \\ \hfill {\displaystyle \sum }_{n=0}^{\infty }\left[(n^2-n)a_n+6n{a}_n+4a_n-(n+2)(n+1)a_{n+2}\right]x^n& =\hfill & \mathrm{-4}\hfill \\ \hfill {\displaystyle \sum }_{n=0}^{\infty }\left[n^2{a}_n+5n{a}_n+4a_n-(n+2)(n+1)a_{n+2}\right]x^n& =\hfill & \mathrm{-4}\hfill \\ \hfill {\displaystyle \sum }_{n=0}^{\infty }\left[(n^2+5n+4)a_n-(n+2)(n+1)a_{n+2}\right]x^n& =\hfill & \mathrm{-4}\hfill \\ \hfill {\displaystyle \sum }_{n=0}^{\infty }\left[(n+4)(n+1)a_n-(n+2)(n+1)a_{n+2}\right]x^n& =\hfill & \mathrm{-4}\hfill \end{array}$$
   
   Looking at the coefficients of each power of x, we see that the constant term must be equal to $\mathrm{-4},$ and the coefficients of all other powers of x must be zero. Then, looking first at the constant term,
   
   $$\begin{array}{ccc}\hfill 4a_0-2a_2& =\hfill & \mathrm{-4}\hfill \\ \hfill a_2& =\hfill & 2a_0+2\text{(step 3).}\hfill \end{array}$$
   
   For $n\ge 1,$ we have
   
   $$\begin{array}{ccc}\hfill (n+4)(n+1)a_n-(n+2)(n+1)a_{n+2}& =\hfill & 0\hfill \\ \hfill (n+1)\left[(n+4)a_n-(n+2)a_{n+2}\right]& =\hfill & 0.\hfill \end{array}$$
   
   Since $n\ge 1,$ $n+1\ne 0,$ we see that
   
   $$(n+4)a_n-(n+2)a_{n+2}=0$$
   
   and thus
   
   $$a_{n+2}=\frac{n+4}{n+2}{a}_n.$$
   
   For even values of n, we have
   
   $$\begin{array}{ccc}\hfill a_4& =\hfill & \frac{6}{4}\left(2a_0+2\right)=3a_0+3\hfill \\ \hfill a_6& =\hfill & \frac{8}{6}\left(3a_0+3\right)=4a_0+4\hfill \\ & \text{⋮.}\hfill & \end{array}$$
   
   In general, $a_{2k}=\left(k+1\right)\left(a_0+1\right)$ (step 5).
   For odd values of n, we have
   
   $$\begin{array}{ccc}\hfill a_3& =\hfill & \frac{5}{3}{a}_1\hfill \\ \hfill a_5& =\hfill & \frac{7}{5}{a}_3=\frac{7}{3}{a}_1\hfill \\ \hfill a_7& =\hfill & \frac{9}{7}{a}_5=\frac{9}{3}{a}_1=3a_1\hfill \\ & \text{⋮.}\hfill & \end{array}$$
   
   In general, $a_{2k+1}=\frac{2k+3}{3}{a}_1$ (step 5 continued).
   Putting this together, we have
   
   $$y(x)={\displaystyle \sum _{k=0}^{\infty }(k+1)(a_0+1)}{x}^{2k}+{\displaystyle \sum _{k=0}^{\infty }\left(\frac{2k+3}{3}\right)a_1}{x}^{2k+1}\text{(step 6).}$$

The Bessel equation of order 0 is given by

We assume a solution of the form $y={\displaystyle \sum }_{n=0}^{\infty }{a}_n{x}^n.$ Then $y^{\prime }(x)={\displaystyle \sum _{n=1}^{\infty }n}{a}_n{x}^{n-1}$ and $y\text{″}(x)={\displaystyle \sum _{n=2}^{\infty }n}(n-1)a_n{x}^{n-2}.$ Substituting this into the differential equation, we get

Then, $a_1=0,$ and for $n\ge 2,$

Because $a_1=0,$ all odd terms are zero. Then, for even values of n, we have

In general,

Thus, we have

The graph appears below.