Lesson 20 - Power Series Solutions

Consider the second order equation: $y^{″}+p(x)y^{\prime }+q(x)y=0$. A point $x_0$ called an ordinary point of the above equation if both $p$ and $q$ are analytic at $x_0$, meaning that they both have infinite derivatives. If $x_0$ is not an ordinary point, it is called a singular point.

In order to differentiate a power series with respect to $x$, let $y(x)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$. Then, by power rule:

$y^{\prime }=\sum _{n=1}^{\mathrm{\infty }}n{a}_n{x}^{n-1}$

$y^{″}=\sum _{n=2}^{\mathrm{\infty }}n(n-1)a_n{x}^{n-2}$

Each derivative that is taken shifts the index up by $1$. By substituting for $n$, these summation derivatives can have their indexes shifted back to $n=0$. (See Lesson 19 - Taylor Polynomial, Power Series and Shifting Index)

Try it Yourself

Shift the starting index of the summation to $n=0$ for the following summations:

1. $y^{\prime }=\sum _{n=1}^{\mathrm{\infty }}n{a}_n{x}^{n-1}$
2. $y^{″}=\sum _{n=2}^{\mathrm{\infty }}n(n-1)a_n{x}^{n-2}$

We can use these derivatives of power solutions to find solutions to differential equations about a point.

Example

Find a power series solution about $x=0$ for $y^{\prime }+2xy=0$.

$y(x)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$

$y^{\prime }(x)=\sum _{n=1}^{\mathrm{\infty }}n{a}_n{x}^{n-1}$

Substitute $y(x)$ and $y^{\prime }(x)$ for the summations above.

$y^{\prime }+2xy=0$ $\to$ $\sum _{n=1}^{\mathrm{\infty }}n{a}_n{x}^{n-1}+2\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+1}=0$

Let's expand each summation.

Left side:

$\sum _{n=1}^{\mathrm{\infty }}n{a}_n{x}^{n-1}$ $=$ $a_1+2a_{2}x+3a_3{x}^2+\dots$

Right side:

$2\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+1}=2a_{0}x+2a_1{x}^2+2a_2{x}^3+\dots$

Since $a_1$ is the only coefficient that contains no $x$ term, and since the sum of the power series is equal to $0$, then we can say that $a_1$ is also equal to $0$.

$a_1=0$

Now let's take the coefficients of $x$ and determine the value of $a_2$.

$2a_2+2a_0=0$ $\to$ $2a_2=-2a_0$ $\to$ $a_2=-a_0$

Coefficients of $x^2$:

$3a_3+2a_1=0$ $\to$ $3a_3=-2a_1$ $\to$ $a_3=0$

Terms of $x^3$

$4a_4+2a_2=0$ $\to$ $4a_4=-2a_2$ $\to$ $a_4=-\frac{a_2}{2}$

Let's put the terms together:

$y(x)=a_0(1-a_2+\frac{a_4}{2!}+\dots )$

Recall the coefficients that each of the terms above represent: $a_2=x^2$, $a_4=x^3$, $\dots$
So,

$y(x)=a_0(1-x^2+\frac{x^3}{2!}+\dots )$

We can now deduce a general solution for the differential equation by looking at the pattern:

$a_{2n}=\frac{(-1{)}^n{a}_0}{n!}$

$y(x)=a_0\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{n!}{x}^{2n}$

Try for Yourself:

1. Find a power series solution about $x=0$ for $y^{\prime }-y=0$.
2. Find the general solution using power series about $x=0$ for $y^{″}+y=0$.

Was that confusing? That's okay. Let's go onto a much lighter topic that will make you wonder why we learned anything in the previous lessons.

Next Lesson: Lesson 21 - Definition of Laplace Transforms

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