# Series solution of linear DE

As was pointed before

• some DE are difficult to solve or can not be solved analytically;
• some of the solutions consist of functions which are not elementary functions.

Example: The purpose is to solve linear DE

$$y''-(x+1)y'+x^2y = x$$

by means of infinite series. In this case the solution is

$$y(x) = 1+x+\frac{x^2}{2}+\frac{x^3}{2}+\frac{x^4}{8}+ \dots \hskip2em \text{for} -\infty \lt x \lt \infty.$$

## Power series

A series of the form

$$\begin{equation} c_0 + c_1(x-a) + c_2 (x-a)^2 +c_3(x-a)^3 + \dots = \sum_{n=0}^{\infty}c_n(x-a)^n \label{ref:ser1} \end{equation}$$

where $c_0,\ c_1,\dots ,\ a$ are constants and $x$ is variable, is called a power series. Such power series is centered at $a$.

### Convergence of power series

We study whether the above sum approaches some given number. Then the series converges. A power series may

1. Converge only for a single value $x=a$.
2. Converge absolutely for values $x$ near value $a$, but not for other values. That means
• there is and interval $(a-h; a+h)$ within which $x$ converges
• and diverges for other values.
• At the end points $x\pm h$ may either converge or diverge.
3. Converge absolutely for all values of $x$, i.e. for $-\infty \lt x \lt \infty$.

### Ratio test

Convergence of power series can be determined by the ratio test. It states that the series $u_1+u_2+u_3+\dots+u_n + \dots$ converges absolutely if

$$\lim_{n\to \infty}|\frac{u_{n+1}}{u_n}| = L \hskip2em \text{and} \hskip1em L \lt 1.$$

If $L\gt1$ then the series diverges. If $L=1$ the test is inconclusive.

#### Example

Determine whether series $1 + 1!\ x + 2!\ x^2 + 3!\ x^3 + \dots +n!\ x^{n} + \dots$ converges.

Here

\begin{align} u_n &= n!\cdot x^n \nonumber \\ u_{n+1} &= (n+1)!\cdot x^{n+1} \nonumber \end{align}

Therefore

$$\big |\frac{u_{n+1}}{u_{n}}\big | = \big |\frac{(n+1)!\ x^{n+1}}{n!\ x^n} \big | = (n+1)|x|$$

The ratio between both members $u_{n+1}$ and $u_n$ grows with $n\to \infty$: $L = (n+1)|x| \to \infty$ as $n \to \infty$. Since it is not valid that $L\lt 1$, the series converges only for $x=0$.

#### Example

Determine whether series $1 + x + \frac 1 2 x^2 + \frac 1 3 x^3 + \dots \frac 1 n x^n$ converges.

Here

\begin{align} &u_n = \frac 1 n x^n, \nonumber \\ &u_{n+1} = \frac{1}{n+1} x^{n+1} \nonumber \\ &\lim_{n\to \infty}\big |\frac{x^{n+1}}{\color{red}{n+1}} \cdot \frac{\color{red}{n}}{x^n} \big | = |x| = L \nonumber \end{align}

A series converges if $L\lt 1$. In our case the series converges for $-1 \lt x \lt 1$.

Note: the test is not conclusive at endpoints $\pm 1$: in this case for $x=1$ diverges, for $x=-1$ converges (check the series itself).

### Shifting the summation index of power series

To solve DE by power series, it is crucial to be able to simplify two series into one.

#### Example

Simplify $\sum_{n=1}^{\infty}2nc_n x^{n-1} + \sum_{n=0}^{\infty}6c_n x^{n+1}$

The important observation is that both series are not expressed from $x^n$. Instead of that they use $x^{n-1},\ x^{n+1}$. It is desired to have both in the form of $x^n$.

\begin{align} \sum_{n=1}^{\infty}2nc_n x^{n-1} &= \sum_{n=0}^{\infty}2(n+1)c_{n+1} x^{n} \nonumber \\ \sum_{n=0}^{\infty}6c_n x^{n+1} &= \sum_{n=1}^{\infty}6c_{n-1} x^{n}\nonumber \end{align}

Then sum of above series is

\begin{align} \sum_{n=1}^{\infty}2nc_n x^{n-1} + \sum_{n=0}^{\infty}6c_n x^{n+1} &= \sum_{n=0}^{\infty}2(n+1)c_{n+1} \cdot x^{n} + \sum_{n=1}^{\infty}6c_{n-1} \cdot x^{n} = \nonumber \\ &= 2c_1 + \sum_{n=1}^{\infty}x^n(2(n+1)c_{n+1} + 6c_{n-1}) \nonumber \end{align}

### A power series defines a function

If power series converges on an interval $I$, then the power series $(\ref{ref:ser1})$ defines a function $f(x)$ which is continuous, differentiable and integrable on the interval:

\begin{align} y & = \sum_{n=0}^{\infty}c_n x^n = c_0 + c_1x + c_2 x^2 +c_3x^3 + \dots \nonumber \\ y' & = \sum_{n=1}^{\infty}c_n n x^{n-1} = c_1 + 2 c_2 x +3 c_3 x^2 + \dots \nonumber \\ y''& = \sum_{n=2}^{\infty}c_n n (n-1)x^{n-2} = 2c_2 +6 c_3 x + 12 c_4 x^2 + \cdots \nonumber \end{align}

### Analytic at point

Any function is said to be analytic at a point $a$ if it can be represented by a power series in $x-a$.

Example: function such as $e^x,\ \sin x,\ \log \ x$ can be represented by Taylor series. These functions are analytic.

If a function is analytic at point, it can be replaced either by Taylor series $(\ref{ref:ser3})$ or Maclaurin series $(\ref{ref:ser4})$, which are both power series.

\begin{align} \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n &= f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \dots \label{ref:ser3}\\ \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n &= f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \dots \label{ref:ser4} \end{align}

### Ordinary and singular points

A point $x=x_0$ is called ordinary point of the DE

$$y^{(n)} + F_{n-1}(x)y^{(n-1)} + \dots + F_1(x)y' + F_0(x)y = Q(x)$$

if each function $F_0,\ F_1,\ \dots,\ F_{n-1}$ and $Q$ is analytic at $x=x_0$. A point $x=x_0$ is called singular if one or more of the functions $F_0(x),\ \dots,\ F_{n-1}(x),\ Q(x)$ are not analytic at $x=x_0$.

#### Example

DE $(x-1)y'' + \frac 1 x y' - 2y = 0$ has two singularities: at $x=1$ and at $x=0$.

If we divide DE by $(x-1)$, we get

y''+\frac{1}{x(x-1)}y' - \frac 2 {x-1}y = 0 \implies \left\{ \begin{align} F_1(x) &= \frac{1}{x(x-1)} \nonumber \\ F_0(x) &= -\frac{2}{x-1} \nonumber \end{align} \right.

Both $F_0,\ F_1$ are singular, are not analytic and can not be described by power series.

### Existence of solution

If $x=x_0$ is an ordinary point of the DE, we can always find solutions in the form of power series centered at $x_0$. A power series solution converges at least on interval $|x-x_0| \lt R$, where $R$ is the distance from the closest singular point.

## Solving DE with power series

Note: in finding a series solution of a differential equation, it will usually not be easy to write the general form of the series. In fact, it will, in most cases, be very difficult if not impossible.

#### Example

Solve DE $(x-1)y'' - x y' +y = 0$, $y(0) = -2$, $y'(0) = 6$.

First, let us solve by successive differentiation using Maclaurin polynomial.

\begin{align} \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n &= f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \dots \nonumber \\ \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n &= -2 + 6x + \cdots \nonumber \end{align}

The first two members were taken from initial values. The next members will be computed by successive differentiations:

\begin{align} f''(0): \hskip1em &(x-1)y''-xy'+y =0 \implies \color{red}{y''}(x=0, y=-2, y'=6) = \color{red}{-2} \nonumber \\[2mm] f'''(0): \hskip1em &(x-1)y''' +y'' - xy''-y' +y' = 0 \nonumber \\ & (x-1)y'''+y''-xy'' = 0 \implies \color{red}{y'''}(x=0, y'=6, y''=-2) = \color{red}{-2} \nonumber \\[2mm] f^{(4)}(0): \hskip1em &(x-1)y^{(4)}+y'''+y'''-xy'''-y'' = 0 \nonumber \\ &(x-1)y^{(4)}+2y''' - xy'''-y''=0 \implies \color{red}{y^{(4)}}(x=0,y''=-2, y'''=-2) = \color{red}{-2} \nonumber \\[2mm] f^{(5)}(0): \hskip1em &(x-1)y^{(5)} + y^{(4)} + 2y^{(4)} - xy^{(4)} - y''' - y''' = 0 \nonumber \\ &(x-1)y^{(5)} + 3y^{(4)} - xy^{(4)} - 2y''' = 0 \implies \color{red}{y^{(5)}}(x=0, y'''=-2, y^{(4)}=-2) = \color{red}{-2} \nonumber \end{align}

From values above we may form the series for solution:

$$y = f(x) = -2 + 6x - 2\frac{x^2}{2!} -2\frac{x^3}{3!} -2\frac{x^4}{4!} -2\frac{x^5}{5!} + \dots$$

It can be noticed that the solution does not involve any arbitrary constant, because it is particular solution for given initial values.

#### Example

Solve DE $(x-1)y'' - x y' +y = 0$, $y(0) = -2$ and $y'(0) = 6$.

The same example as above but now solved by power series (undetermined coefficients approach).

Let us approximate solution by

$$y = \sum_{n=0}^{\infty}c_n x^n = c_0 + c_1x + c_2 x^2 +c_3x^3 + \dots \nonumber \\$$

Then

\begin{align} y' & = \sum_{n=1}^{\infty}c_n n x^{n-1} = c_1 + 2c_2 x +3 c_3 x^2 + \dots \nonumber \\ y''& = \sum_{n=2}^{\infty}c_n n (n-1)x^{n-2} = 2c_2 +6 c_3 x + 12 c_4 x^2 + \cdots \nonumber \end{align}

Let us substitute $y,\ y',\ y''$ into DE:

\begin{align} \color{red}{(x-1)}\big[\sum_{n=2}^{\infty} c_{n}n(n-1) x^\color{red}{n-2}\big] - \color{red}{x}\big[\sum_{n=1}^{\infty}c_{n}nx^\color{red}{n-1}\big] + \big[\sum_{n=0}^{\infty}c_nx^n\big] &= 0 \nonumber \\ \big[\sum_{n=2}^{\infty} c_{n}n(n-1) x^\color{red}{n-1} \color{red}{-} \sum_{n=2}^{\infty} c_{n}n(n-1) x^\color{red}{n-2}\big] - \big[\sum_{n=1}^{\infty}c_{n}nx^\color{red}{n}\big] + \big[\sum_{n=0}^{\infty}c_nx^n\big] &= 0 \nonumber \\ \end{align}

We need to shift the summation index in order to work with the series in a convenient way, i.e. all sums have to work with $x^n$:

\begin{align} \big[\sum_{n=\color{red}{1}}^{\infty} c_{\color{red}{n+1}}\color{red}{(n+1)n} x^\color{red}{n} - \sum_{n=\color{red}{0}}^{\infty} c_{n\color{red}{+2}}\color{red}{(n+2)(n+1)} x^\color{red}{n}\big] - \big[\sum_{n=1}^{\infty}c_{n}nx^{n}\big] + \big[\sum_{n=0}^{\infty}c_nx^n\big] &= 0 \nonumber \\ \end{align}

Let us collect coefficients $\boldsymbol{c_0,\ c_1,\ c_2,\ c_3,\ \dots}$ for $\boldsymbol{1,\ x, \ x^2,\ x^3, \dots}$. We are solving second order DE, which is linear. So we will have two independent solutions, which are distinguished by arbitrary constants $c_0,\ c_1$. The other constants have to be evaluated in terms of $c_0,\ c_1$.

\begin{align} n=0:\hskip1em & 1(-c_2\cdot 2\cdot 1 + c_0) = 0 \implies c_2 = \frac 1 2 c_0 \nonumber \\ n=1: \hskip1em &x(c_2\cdot 2 \cdot 1 - c_3\cdot 3 \cdot 2 - c_1 \cdot 1 + c_1) = 0 \implies c_3 = \frac 1 3 c_2 = \frac 1 6 c_0 \nonumber \\ n=2: \hskip1em &x^2(c_3\cdot 3 \cdot 2 - c_4 \cdot 4 \cdot 3 - c_2\cdot 2 + c_2) = 0 \implies c_4 = \frac{c_2 -6c_3}{-12} = \frac 1 {24} c_0 \nonumber \\ n=3: \hskip1em &x^3(c_4\cdot 4 \cdot 3 - c_5\cdot 5 \cdot 4 -c_3 \cdot 3 + c_3) = 0 \implies c_5 = \frac{2c_3-12c_4}{-20}= \frac{1}{120}c_0 \nonumber \end{align}

From coefficients found so far we can form power series of $y$:

$$y = c_0 + c_1x + \frac 1 2 c_0 x^2 + \frac 1 6 c_0 x^3 + \frac{1}{24} c_0x^4 + \frac 1 {120} c_0 x^5 + \dots$$

You can notice that we have two arbitrary constants ($c_0,\ c_1$) within the above general solution $y$. And that implies there are two solutions $y_1(x) = c_1f_1(x)$ and $y_2(x)=c_2f_2(x)$. And that could be expected since the DE to solve is not nonlinear. The first solution is an infinite series with $c_0$ and the second solution is a line $c_1x$. That complies with the analyticial solution of the given DE, which is $y = c_1x + c_2e^x$.

From initial values we will determinate constants to express particular solution. $y(0) = -2 \implies c_0 = -2$, $y'(0) = 6 \implies c_1 = 6$. Then

\begin{align} &y = -2 + 6x - x^2 -\frac 1 3 x^3 - \frac 1 {12} x^4 - \frac{1}{60} x^5 + \dots \hskip1em \text{or} \nonumber \\ &\underline{\underline{y = -2 + 6x - 2 \frac{x^2}{2!} - 2 \frac{x^3}{3!} - 2 \frac{x^4}{4!} -2\frac{x^5}{5!} + \dots }} \nonumber \end{align}

We have reached the same solution as within previous example.    Series (2, 3, 4, 5, 6 members of infinite series involved) around $x=0$ compared to analytical solution (in black). The second chart is a detail from the first one.

#### Example

Solve nonlinear DE $y''=xy-(y')^2$, $y(0) = 2$ and $y'(0) = 1$ by power series (undetermined coefficients approach).

We will by finding the general solution first and then the initial values will be used to evaluate $c_0, \ c_1$.

We can use the power series for $y,\ y',\ y''$ from the last example. $$y = \sum_{n=0}^{\infty}c_n x^n, \hskip1em y' = \sum_{n=1}^{\infty}c_n n x^{n-1}, \hskip1em y'' = \sum_{n=2}^{\infty}c_n n (n-1)x^{n-2}$$

and substitute them into DE.

$$\sum_{n=2}^{\infty}c_nn(n-1)x^{n-2} = x\sum_{n=0}^{\infty}c_nx^n - \big[\sum_{n=1}^{\infty}c_nnx^{n-1}\big]^2 \\$$

It would be quite difficult to work with the above series if they are not based on $x^n$. They can be rewritten to

$$\sum_{n=0}^{\infty}c_{n+2}(n+2)(n+1)x^{n} = \sum_{n=1}^{\infty}c_{n-1}x^n - \color{blue}{\big[\sum_{n=0}^{\infty}c_{n+1}(n+1)x^{n}\big]^2} \\$$

Note that the last series is squared, which brings us some more work. To handle the power, it is useful to list its members:

$$\color{blue}{\big[\sum_{n=0}^{\infty}c_{n+1}(n+1)x^{n}\big]^2} = (c_1 + 2c_2x + 3c_3x^2 + 4c_4x^3 + 5c_5x^4 + \dots)^2$$

Now we do not need to square the series itself. But we need to pick up coefficients for each $1,\ x,\ x^2,\ \dots$ For example we want to pick up coefficient of $x^2$. Now imagine there is the series multiplied by itself:

\begin{align} {\big[\sum_{n=0}^{\infty}c_{n+1}(n+1)x^{n}\big]^2} = & (c_1 + 2c_2x + 3c_3x^2 + 4c_4x^3 + \dots)\cdot \nonumber \\ & \cdot (c_1 + 2c_2x + 3c_3x^2 + 4c_4x^3+ \dots) \nonumber \\ \end{align}

Then for $x^2$: $(c_1\cdot 3c_3x^2) + (3c_3x^2\cdot c_1) + (2c_2x)^2.$

Now we can use method of undetermined coefficients for $\boldsymbol{1,\ x,\ x^2,\ \dots}$.

\begin{align} n=0:& \hskip1em 1(c_2\cdot 2 = -c_1^2) = 0\implies c_2 = \frac{c_1^2}{-2} \nonumber \\ n=1:& \hskip1em x(c_3\cdot 6 = c_0 -2c_1c_2\cdot2) = 0 \implies c_3 = \frac{c_0-4c_1c_2}{6} \nonumber \\ n=2:& \hskip1em x^2(c_4\cdot 12 = c_1 - (3c_1c_3\cdot 2 +4c_2^2)) = 0 \implies c_4 = \frac{c_1 - (6c_1c_3 + 4c_2^2)}{12} \nonumber \\ n=3:& \hskip1em x^3(c_5\cdot 20 = c_2 - (4c_1c_4\cdot 2 + c_2c_3\cdot 6\cdot 2)) = 0 \implies c_5 = \frac{c_2 - (8c_1 c_4 + 12c_2 c_3)}{20} \nonumber \end{align}

We have collected coefficients $c_2$ to $c_5$ to form a power series of solution. Since it is the second order DE, we should expect two arbitrary constants $c_0,\ c_1$. The others should be expressed in terms of $c_0,\ c_1$ and that is what can be confirmed by observing $c_2$ to $c_5$. If you observe $c_3$ you will notice that this time we can not separate solutions using $c_0$ and $c_1$ as in the previous example. The both solutions are somehow mixed. The reason is that the DE being solved is nonlinear.

Finally from given initial values $y(x=0) = 2$ and $y'(x=0) = 1$: $c_0=2, c_1=1$. Then the other constants can be evaluated as $c_2=-1/2$, $c_3=2/3$, $c_4=-1/3$, $c_5=37/120$.

$$\underline{\underline{y(x) = 2 + x - \frac 1 2 x^2 + \frac 2 3 x^3 - \frac 1 3 x^4 + \frac {37}{120}x^5 + \dots}}$$    Series (2, 3, 4, 5, 6 members of infinite series involved) around $x=0$. No analytical solution drawn. The second chart is a detail from the first one.

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