5.2 Series Solution to Differential Equations

1. Notice that ${\displaystyle P_1\left(x\right)=1}$ and thus the coefficients are analytic at every point. We assume ${\displaystyle x_0=0}$ and that the solution can be written as a power series

  ${\displaystyle y=\sum _{n=0}^{\infty } a_n{x}^n}$

2. First, we need to find ${\displaystyle y\prime \prime }$:

 ${\displaystyle y\prime =\frac{d}{dx}\sum _{n=0}^{\infty } a_n{x}^n}$ ${\displaystyle =\sum _{n=1}^{\infty } n{a}_n{x}^{n-1}}$

 ${\displaystyle y\prime \prime =\frac{d}{dx}\sum _{n=1}^{\infty } n{a}_n{x}^{n-1}}$ ${\displaystyle =\sum _{n=2}^{\infty } n(n-1)a_n{x}^{n-2}}$

3. Next, we substitute ${\displaystyle y}$ and ${\displaystyle y\prime \prime }$ into the equation:

${\displaystyle y\prime \prime +y=0}$

 ${\displaystyle \sum _{n=2}^{\infty } n(n-1)a_n{x}^{n-2}}$ ${\displaystyle +\sum _{n=0}^{\infty } a_n{x}^n=0}$

4. The next step is to align terms. To do this we need to shift the summation indices to start at the same value. Letting ${\displaystyle k=n-2}$ or equally ${\displaystyle n=k+2}$ in the first summation and ${\displaystyle n=k}$ in the second summation, we have

 ${\displaystyle \sum _{k=0}^{\infty } (k+2)(k+1)a_{k+2}{x}^k}$ ${\displaystyle +\sum _{k=0}^{\infty } a_k{x}^k=0}$

5. Adding the series yields

 ${\displaystyle \sum _{k=0}^{\infty } [(k+2)(k+1)a_{k+2}+a_k]x^k=0}$

6. From the Power Series Vanishing on an Interval property discussed in Section 5.1, we know that If a power series is zero for all ${\displaystyle x}$, then all its coefficients must be zero. Therefore, we conclude that

 ${\displaystyle (k+2)(k+1)a_{k+2}+a_k=0}$

or

 ${\displaystyle a_{k+2}=\frac{-a_k}{(k+2)(k+1)},}$   ${\displaystyle k\ge 0}$

This is called the recurrence relation for the values of ${\displaystyle k}$ for which the relation is true.

7. Next, we write a few terms of the series to see if we can determine the trend and hopefully the explicit formula of the series. Setting ${\displaystyle k=0,1,2,...,5}$, we get

Notice that the term with even indices can be written in terms of the previous term and eventually in terms of ${\displaystyle a_0}$ and so can be the odd indices in terms of ${\displaystyle a_1}$. Therefore, by writing the recurrence relation separately for odd (${\displaystyle k=2m+1}$) and even (${\displaystyle k=2m}$) indices, we get

 ${\displaystyle a_{2m}=\frac{{(-1)}^m{a}_0}{\left(2m\right)!}}$   ${\displaystyle , m\ge 0}$

 ${\displaystyle a_{2m+1}=\frac{{(-1)}^m{a}_1}{(2m+1)!}}$   ${\displaystyle , m\ge 0}$

8. Thus the general solution of the equation can be written as

  ${\displaystyle y=\sum _{n=0}^{\infty } a_n{x}^n}$

 ${\displaystyle =\sum _{m=0}^{\infty } a_{2m}{x}^{2m}}$ ${\displaystyle +\sum _{m=0}^{\infty } a_{2m+1}{x}^{2m+1}}$

 ${\displaystyle =a_0\sum _{m=0}^{\infty } {(-1)}^m\frac{x^{2m}}{\left(2m\right)!}}$ ${\displaystyle +a_1\sum _{m=0}^{\infty } {(-1)}^m\frac{x^{2m+1}}{(2m+1)!}}$

We recognize that the series in the solution are the Maclaurin series of ${\displaystyle \cos \left(x\right)}$ and ${\displaystyle \sin \left(x\right)}$, respectively.

 ${\displaystyle \cos \left(x\right)=\sum _{m=0}^{\infty } {(-1)}^m\frac{x^{2m}}{\left(2m\right)!}}$    and   ${\displaystyle \sin \left(x\right)=\sum _{m=0}^{\infty } {(-1)}^m\frac{x^{2m+1}}{(2m+1)!}}$

Therefore, the general solution can be expressed as

 ${\displaystyle y=a_0\cos \left(x\right)+a_1\sin \left(x\right)}$

for some arbitrary constant ${\displaystyle a_0}$  and ${\displaystyle a_1}$. This is the same solution we would obtain using the methods learned in previous sections.

The interval of convergence for both the cosine and sine series is all real numbers ${\displaystyle (-\infty ,\infty )}$.

For both series in the solution, the Ratio Test indicates that as ${\displaystyle m\to \infty }$ the limit ${\displaystyle L}$ approaches zero, which means the series converge for all real numbers. Therefore, without prior knowledge of the series representing sine and cosine, we would conclude that the interval of convergence for each series and hence the combined series solution is all real number ${\displaystyle (-\infty ,\infty )}$.

1. Note that ${\displaystyle P_1\left(x\right)=1+x^2}$ has no root and thus every point for this equation is an ordinary point. We assume that the solution can be written as a power series

  ${\displaystyle y=\sum _{n=0}^{\infty } a_n{x}^n}$

2. Next, we find ${\displaystyle y\prime }$ and ${\displaystyle y\prime \prime }$ :

 ${\displaystyle y\prime =\frac{d}{dx}\sum _{n=0}^{\infty } a_n{x}^n}$ ${\displaystyle =\sum _{n=1}^{\infty } n{a}_n{x}^{n-1}}$

 ${\displaystyle y\prime \prime =\frac{d}{dx}\sum _{n=1}^{\infty } n{a}_n{x}^{n-1}}$ ${\displaystyle =\sum _{n=2}^{\infty } n(n-1)a_n{x}^{n-2}}$

3. Next, we substitute ${\displaystyle y, y\prime ,}$ and ${\displaystyle y\prime \prime }$ into the equation:

 ${\displaystyle (1+x^2)\sum _{n=2}^{\infty } n(n-1)a_n{x}^{n-2}}$ ${\displaystyle +2x\sum _{n=1}^{\infty } n{a}_n{x}^{n-1}}$ ${\displaystyle -2\sum _{n=0}^{\infty } a_n{x}^n=0}$

Multiplying the coefficients by the series, we get

 ${\displaystyle \sum _{n=2}^{\infty } n(n-1)a_n{x}^{n-2}}$ ${\displaystyle +\sum _{n=2}^{\infty } n(n-1)a_n{x}^n}$ ${\displaystyle +2\sum _{n=1}^{\infty } n{a}_n{x}^n}$ ${\displaystyle -2\sum _{n=0}^{\infty } a_n{x}^n=0}$

4. Note that the exponent of ${\displaystyle x}$ is the same in all but the first series. Therefore, we only need to shift the index of the first summations by 2:

 ${\displaystyle \sum _{n=0}^{\infty } (n+2)(n+1)a_{n+2}{x}^n}$ ${\displaystyle +\sum _{n=2}^{\infty } n(n-1)a_n{x}^n}$ ${\displaystyle +2\sum _{n=1}^{\infty } n{a}_n{x}^n}$ ${\displaystyle -2\sum _{n=0}^{\infty } a_n{x}^n=0}$

Also, notice that the second series is zero at ${\displaystyle n=0,1}$ . So its index can start at ${\displaystyle n=0}$ . Likewise, the third series is zero at ${\displaystyle n=0}$ so it can too start at ${\displaystyle n=0}$ . Then we rewrite the indices and we have

 ${\displaystyle \sum _{n=0}^{\infty } (n+2)(n+1)a_{n+2}{x}^n}$ ${\displaystyle +\sum _{n=0}^{\infty } n(n-1)a_n{x}^n}$ ${\displaystyle +2\sum _{n=0}^{\infty } n{a}_n{x}^n}$ ${\displaystyle -2\sum _{n=0}^{\infty } a_n{x}^n=0}$

5. Combining the series yields

 ${\displaystyle \sum _{n=0}^{\infty } [(n+2)(n+1)a_{n+2}+(n(n-1)+2n-2)a_n]x^n=0}$

 ${\displaystyle \sum _{n=0}^{\infty } [(n+2)(n+1)a_{n+2}+(n^2+n-2)a_n]x^n=0}$

6. Now setting the coefficient to zero gives

 ${\displaystyle (n+2)(n+1)a_{n+2}+(n^2+n-2)a_n=0}$

 ${\displaystyle a_{n+2}=-\frac{n^2+n-2}{(n+2)(n+1)}{a}_n}$

 ${\displaystyle =-\frac{(n+2)(n-1)}{(n+2)(n+1)}{a}_n}$

7. So the recursive relation is simplified to

 ${\displaystyle a_{n+2}=-\frac{n-1}{n+1}{a}_n}$

Setting ${\displaystyle n=0,1,2,...,5}$ , we get

Notice that all the terms with odd indices are zero except ${\displaystyle a_1}$. Therefore, by writing the recurrence relation separately for odd (${\displaystyle n=2m+1}$) and even (${\displaystyle n=2m}$) indices, we obtain

 ${\displaystyle a_{2m}={(-1)}^{m+1}\frac{1}{2m-1}{a}_0}$   ${\displaystyle , m\ge 1}$

 ${\displaystyle a_{2m+1}=a_1}$   ${\displaystyle , m=0}$

8. Thus the general solution of the equation can be written as

  ${\displaystyle y=\sum _{n=0}^{\infty } a_n{x}^n}$

 ${\displaystyle =a_0+a_1{x}^{2\left(0\right)+1}+\sum _{m=1}^{\infty } a_{2m}{x}^{2m}}$

 ${\displaystyle =a_0+a_{1}x+a_0\sum _{m=1}^{\infty }{(-1)}^{m+1}\frac{x^{2m}}{2m-1}}$

In Example 5.2.2, we found the general series solution to this differential equation.

 ${\displaystyle =a_0+a_{1}x+a_0\sum _{m=1}^{\infty }{(-1)}^{m+1}\frac{x^{2m}}{2m-1}}$

To apply the initial conditions, we first recognize that ${\displaystyle a_0=y\left(0\right)=2}$ and ${\displaystyle a_1=y\prime \left(0\right)=3}$. Then, we substitute ${\displaystyle a_0}$  and ${\displaystyle a_1}$ into the general solution to compute the other terms.

${\displaystyle a_0=2}$

${\displaystyle a_1=3}$

 ${\displaystyle m=1 \to }$  ${\displaystyle a_2=}$ ${\displaystyle a_0\left[{(-1)}^{1+1}\frac{x^{2\left(1\right)}}{2\left(1\right)-1}\right]}$  ${\displaystyle \to a_2=2x^2}$

${\displaystyle a_3=0}$

 ${\displaystyle m=2 \to }$  ${\displaystyle a_4=a_0\left[{(-1)}^{2+1}\frac{x^{2\left(2\right)}}{2\left(2\right)-1}\right]}$  ${\displaystyle \to a_4=-\frac{2}{3}{x}^4}$

${\displaystyle a_5=0}$

 ${\displaystyle m=3 \to }$  ${\displaystyle a_6=a_0\left[{(-1)}^{3+1}\frac{x^{2\left(3\right)}}{2\left(3\right)-1}\right]}$  ${\displaystyle \to a_6=\frac{2}{5}{x}^6}$

${\displaystyle a_7=0}$

 ${\displaystyle m=4 \to }$  ${\displaystyle a_8=a_0\left[{(-1)}^{4+1}\frac{x^{2\left(4\right)}}{2\left(4\right)-1}\right]}$  ${\displaystyle \to a_8=-\frac{2}{7}{x}^8}$

Therefore, the solution to the initial value problem is

 ${\displaystyle y\left(x\right)=2+3x+2x^2-\frac{2}{3}{x}^4+\frac{2}{5}{x}^6-\frac{2}{7}{x}^8+...}$ .