# Introduction to Differential Equations

## Section4.2Series solutions of linear second order ODEs

In this video we introduce the idea of solving an ODE using a series solution. We will see how to form and solve a recurrence relation and then test for convergence of the resulting series.
Suppose we have a linear second order homogeneous ODE of the form
\begin{equation*} p(x) y'' + q(x) y' + r(x) y = 0 . \end{equation*}
Suppose that $$p(x)\text{,}$$ $$q(x)\text{,}$$ and $$r(x)$$ are polynomials. We will try a solution of the form
\begin{equation*} y = \sum_{k=0}^\infty a_k {(x-x_0)}^k \end{equation*}
and solve for the $$a_k$$ to try to obtain a solution defined in some interval around $$x_0\text{.}$$
The point $$x_0$$ is called an ordinary point if $$p(x_0) \not= 0\text{.}$$ That is, the functions
are defined for $$x$$ near $$x_0\text{.}$$ If $$p(x_0) = 0\text{,}$$ then we say $$x_0$$ is a singular point. Handling singular points is harder than ordinary points and so we now focus only on ordinary points.

### Example4.2.1.

\begin{equation*} y'' - y = 0 . \end{equation*}
Let us try a power series solution near $$x_0 = 0\text{,}$$ which is an ordinary point. Every point is an ordinary point in fact, as the equation is constant coefficient. We already know we should obtain exponentials or the hyperbolic sine and cosine, but let us pretend we do not know this.
We try
\begin{equation*} y = \sum_{k=0}^\infty a_k x^k . \end{equation*}
If we differentiate, the $$k=0$$ term is a constant and hence disappears. We therefore get
\begin{equation*} y' = \sum_{k=1}^\infty k a_k x^{k-1} . \end{equation*}
We differentiate yet again to obtain (now the $$k=1$$ term disappears)
\begin{equation*} y'' = \sum_{k=2}^\infty k(k-1) a_k x^{k-2} . \end{equation*}
We reindex the series (replace $$k$$ with $$k+2$$) to obtain
\begin{equation*} y'' = \sum_{k=0}^\infty (k+2)\,(k+1) \, a_{k+2} x^k . \end{equation*}
Now we plug $$y$$ and $$y''$$ into the differential equation
\begin{equation*} \begin{split} 0 = y''-y & = \Biggl( \sum_{k=0}^\infty (k+2)\,(k+1) \, a_{k+2} x^k \Biggr) - \Biggl( \sum_{k=0}^\infty a_k x^k \Biggr) \\ & = \sum_{k=0}^\infty \,\Bigl( (k+2)\,(k+1) \, a_{k+2} x^k - a_k x^k \Bigr) \\ & = \sum_{k=0}^\infty \,\bigl( (k+2)\,(k+1) \,a_{k+2} - a_k \bigr) \, x^k . \end{split} \end{equation*}
As $$y'' - y$$ is supposed to be equal to 0, we know that the coefficients of the resulting series must be equal to 0. Therefore,
\begin{equation*} (k+2)\,(k+1) \,a_{k+2} - a_k = 0 , \qquad \text{or} \qquad a_{k+2} = \frac{a_k}{(k+2)(k+1)} . \end{equation*}
The equation above is called a recurrence relation for the coefficients of the power series. It did not matter what $$a_0$$ or $$a_1$$ was. They can be arbitrary. But once we pick $$a_0$$ and $$a_1\text{,}$$ then all other coefficients are determined by the recurrence relation.
Let us see what the coefficients must be. First, $$a_0$$ and $$a_1$$ are arbitrary. Then,
\begin{equation*} a_2 = \frac{a_0}{2}, \quad a_3 = \frac{a_1}{(3)(2)}, \quad a_4 = \frac{a_2}{(4)(3)} = \frac{a_0}{(4)(3)(2)}, \quad a_5 = \frac{a_3}{(5)(4)} = \frac{a_1}{(5)(4)(3)(2)}, \quad \ldots \end{equation*}
So for even $$k\text{,}$$ that is $$k=2n\text{,}$$ we have
\begin{equation*} a_k = a_{2n} = \frac{a_0}{(2n)!} , \end{equation*}
and for odd $$k\text{,}$$ that is $$k=2n+1\text{,}$$ we have
\begin{equation*} a_k = a_{2n+1} = \frac{a_1}{(2n+1)!} . \end{equation*}
Let us write down the series
\begin{equation*} y = \sum_{k=0}^\infty a_k x^k = \sum_{n=0}^\infty \left( \frac{a_0}{(2n)!} \,x^{2n} + \frac{a_1}{(2n+1)!} \,x^{2n+1} \right) = a_0 \sum_{n=0}^\infty \frac{1}{(2n)!} \,x^{2n} + a_1 \sum_{n=0}^\infty \frac{1}{(2n+1)!} \,x^{2n+1} . \end{equation*}
We recognize the two series as the hyperbolic sine and cosine. Therefore,
\begin{equation*} y = a_0 \cosh x + a_1 \sinh x . \end{equation*}
Of course, in general we will not be able to recognize the series that appears, since usually there will not be any elementary function that matches it. In that case we will be content with the series.
We do a full example of a slightly more complicated example: Airy’s Equation

### Example4.2.2.

Let us do a more complex example. Consider Airy’s equation 1 :
\begin{equation*} y'' - xy = 0 , \end{equation*}
near the point $$x_0 = 0\text{.}$$ Note that $$x_0 = 0$$ is an ordinary point.
We try
\begin{equation*} y = \sum_{k=0}^\infty a_k x^k . \end{equation*}
We differentiate twice (as above) to obtain
\begin{equation*} y'' = \sum_{k=2}^\infty k\,(k-1) \, a_k x^{k-2} . \end{equation*}
We plug $$y$$ into the equation
\begin{equation*} \begin{split} 0 = y''-xy &= \Biggl( \sum_{k=2}^\infty k\,(k-1) \, a_k x^{k-2} \Biggr) - x \Biggl( \sum_{k=0}^\infty a_k x^k \Biggr) \\ &= \Biggl( \sum_{k=2}^\infty k\,(k-1) \, a_k x^{k-2} \Biggr) - \Biggl( \sum_{k=0}^\infty a_k x^{k+1} \Biggr) . \end{split} \end{equation*}
We reindex to make things easier to sum
\begin{equation*} \begin{split} 0 = y''-xy &= \Biggl( 2 a_2 + \sum_{k=1}^\infty (k+2)\,(k+1) \, a_{k+2} x^k \Biggr) - \Biggl( \sum_{k=1}^\infty a_{k-1} x^k \Biggr) \\ &= 2 a_2 + \sum_{k=1}^\infty \Bigl( (k+2)\,(k+1) \, a_{k+2} - a_{k-1} \Bigr) \, x^k . \end{split} \end{equation*}
Again $$y''-xy$$ is supposed to be 0, so $$a_2 = 0\text{,}$$ and
\begin{equation*} (k+2)\,(k+1) \,a_{k+2} - a_{k-1} = 0 , \qquad \text{or} \qquad a_{k+2} = \frac{a_{k-1}}{(k+2)(k+1)} . \end{equation*}
We jump in steps of three. First, since $$a_2 = 0$$ we must have , $$a_5 = 0\text{,}$$ $$a_8 = 0\text{,}$$ $$a_{11}=0\text{,}$$ etc. In general, $$a_{3n+2} = 0\text{.}$$
The constants $$a_0$$ and $$a_1$$ are arbitrary and we obtain
\begin{equation*} a_3 = \frac{a_0}{(3)(2)}, \quad a_4 = \frac{a_1}{(4)(3)}, \quad a_6 = \frac{a_3}{(6)(5)} = \frac{a_0}{(6)(5)(3)(2)}, \quad a_7 = \frac{a_4}{(7)(6)} = \frac{a_1}{(7)(6)(4)(3)}, \quad \ldots \end{equation*}
For $$a_k$$ where $$k$$ is a multiple of $$3\text{,}$$ that is $$k=3n$$ we notice that
\begin{equation*} a_{3n} = \frac{a_0}{(2)(3)(5)(6) \cdots (3n-1)(3n)} . \end{equation*}
For $$a_k$$ where $$k = 3n+1\text{,}$$ we notice
\begin{equation*} a_{3n+1} = \frac{a_1}{(3)(4)(6)(7) \cdots (3n)(3n+1)} . \end{equation*}
In other words, if we write down the series for $$y\text{,}$$ it has two parts
\begin{equation*} \begin{split} y &= \left( a_0 + \frac{a_0}{6} x^3 + \frac{a_0}{180} x^6 + \cdots + \frac{a_0}{(2)(3)(5)(6) \cdots (3n-1)(3n)} x^{3n} + \cdots \right) \\ &\phantom{=} + \left( a_1 x + \frac{a_1}{12} x^4 + \frac{a_1}{504} x^7 + \cdots + \frac{a_1}{(3)(4)(6)(7) \cdots (3n)(3n+1)} x^{3n+1} + \cdots \right) \\ & = a_0 \left( 1 + \frac{1}{6} x^3 + \frac{1}{180} x^6 + \cdots + \frac{1}{(2)(3)(5)(6) \cdots (3n-1)(3n)} x^{3n} + \cdots \right) \\ &\phantom{=} + a_1 \left( x + \frac{1}{12} x^4 + \frac{1}{504} x^7 + \cdots + \frac{1}{(3)(4)(6)(7) \cdots (3n)(3n+1)} x^{3n+1} + \cdots \right) . \end{split} \end{equation*}
We define
\begin{equation*} \begin{aligned} y_1(x) &= 1 + \frac{1}{6} x^3 + \frac{1}{180} x^6 + \cdots + \frac{1}{(2)(3)(5)(6) \cdots (3n-1)(3n)} x^{3n} + \cdots, \\ y_2(x) &= x + \frac{1}{12} x^4 + \frac{1}{504} x^7 + \cdots + \frac{1}{(3)(4)(6)(7) \cdots (3n)(3n+1)} x^{3n+1} + \cdots , \end{aligned} \end{equation*}
and write the general solution to the equation as $$y(x)= a_0 y_1(x) + a_1 y_2(x)\text{.}$$ If we plug in $$x=0$$ into the power series for $$y_1$$ and $$y_2\text{,}$$ we find $$y_1(0) = 1$$ and $$y_2(0) = 0\text{.}$$ Similarly, $$y_1'(0) = 0$$ and $$y_2'(0) = 1\text{.}$$ Therefore $$y = a_0 y_1 + a_1 y_2$$ is a solution that satisfies the initial conditions $$y(0) = a_0$$ and $$y'(0) = a_1\text{.}$$
The functions $$y_1$$ and $$y_2$$ cannot be written in terms of the elementary functions that you know. See Figure 4.3 for the plot of the solutions $$y_1$$ and $$y_2\text{.}$$ These functions have many interesting properties. For example, they are oscillatory for negative $$x$$ (like solutions to $$y''+y=0$$) and for positive $$x$$ they grow without bound (like solutions to $$y''-y=0$$).
Sometimes a solution may turn out to be a polynomial.

### Example4.2.3.

Let us find a solution to the so-called Hermite’s equation of order $$n$$ 3 :
\begin{equation*} y'' -2xy' + 2n y = 0 . \end{equation*}
Let us find a solution around the point $$x_0 = 0\text{.}$$ We try
\begin{equation*} y = \sum_{k=0}^\infty a_k x^k . \end{equation*}
We differentiate (as above) to obtain
\begin{equation*} \begin{aligned} y' &= \sum_{k=1}^\infty k a_k x^{k-1} , \\ y'' &= \sum_{k=2}^\infty k\,(k-1) \, a_k x^{k-2} . \end{aligned} \end{equation*}
Now we plug into the equation
\begin{equation*} \begin{split} 0 = y''-2xy'&+2ny \\ &= \Biggl( \sum_{k=2}^\infty k(k-1) a_k x^{k-2} \Biggr) - 2x \Biggl( \sum_{k=1}^\infty k a_k x^{k-1} \Biggr) + 2n \Biggl( \sum_{k=0}^\infty a_k x^k \Biggr) \\ &= \Biggl( \sum_{k=2}^\infty k(k-1) a_k x^{k-2} \Biggr) - \Biggl( \sum_{k=1}^\infty 2k a_k x^k \Biggr) + \Biggl( \sum_{k=0}^\infty 2n a_k x^k \Biggr) \\ &= \Biggl(2a_2+ \sum_{k=1}^\infty (k+2)(k+1) a_{k+2} x^k \Biggr) - \Biggl( \sum_{k=1}^\infty 2k a_k x^k \Biggr) + \Biggl( 2na_0 + \sum_{k=1}^\infty 2n a_k x^k \Biggr) \\ &= 2a_2+2na_0+ \sum_{k=1}^\infty \bigl( (k+2)(k+1) a_{k+2} - 2ka_k + 2n a_k \bigr) x^k . \end{split} \end{equation*}
As $$y''-2xy'+2ny = 0$$ we have
\begin{equation*} (k+2)(k+1) a_{k+2} + ( - 2k+ 2n) a_k = 0 , \qquad \text{or} \qquad a_{k+2} = \frac{(2k-2n)}{(k+2)(k+1)} a_k . \end{equation*}
This recurrence relation actually includes $$a_2 = -na_0$$ (which comes about from $$2a_2+2na_0 = 0$$). Again $$a_0$$ and $$a_1$$ are arbitrary.
\begin{equation*} \begin{aligned} & a_2 = \frac{-2n}{(2)(1)}a_0, \qquad a_3 = \frac{2(1-n)}{(3)(2)} a_1, \\ & a_4 = \frac{2(2-n)}{(4)(3)} a_2 = \frac{2^2(2-n)(-n)}{(4)(3)(2)(1)} a_0 , \\ & a_5 = \frac{2(3-n)}{(5)(4)} a_3 = \frac{2^2(3-n)(1-n)}{(5)(4)(3)(2)} a_1 , \quad \ldots \end{aligned} \end{equation*}
Let us separate the even and odd coefficients. We find that
\begin{equation*} \begin{aligned} a_{2m} &=\frac{2^m(-n)(2-n)\cdots(2m-2-n)}{(2m)!}a_0 , \\ a_{2m+1} &=\frac{2^m(1-n)(3-n)\cdots(2m-1-n)}{(2m+1)!}a_1 . \end{aligned} \end{equation*}
Let us write down the two series, one with the even powers and one with the odd.
\begin{equation*} \begin{aligned} y_1(x) & = 1+\frac{2(-n)}{2!} x^2 + \frac{2^2(-n)(2-n)}{4!} x^4 + \frac{2^3(-n)(2-n)(4-n)}{6!} x^6 + \cdots , \\ y_2(x) & = x+\frac{2(1-n)}{3!} x^3 + \frac{2^2(1-n)(3-n)}{5!} x^5 + \frac{2^3(1-n)(3-n)(5-n)}{7!} x^7 + \cdots . \end{aligned} \end{equation*}
We then write
\begin{equation*} y(x) = a_0 y_1(x) + a_1 y_2(x) . \end{equation*}
We remark that if $$n$$ is a positive even integer, then $$y_1(x)$$ is a polynomial as all the coefficients in the series beyond a certain degree are zero. If $$n$$ is a positive odd integer, then $$y_2(x)$$ is a polynomial. For example, if $$n=4\text{,}$$ then
\begin{equation*} y_1(x) = 1 + \frac{2(-4)}{2!} x^2 + \frac{2^2(-4)(2-4)}{4!} x^4 = 1 - 4x^2 + \frac{4}{3} x^4 . \end{equation*}
When do series solutions exist? In this video we develop a bit of the theory. This will be extended quite a bit if you take Math 342 in the future where we will talk about how to deal with singular points. For now, we’ll focus on ordinary points.

### Subsection4.2.1Exercises

In the following exercises, when asked to solve an equation using power series methods, you should find the first few terms of the series, and if possible find a general formula for the $$k^{\text{th}}$$ coefficient.

#### Exercise4.2.1.

Use power series methods to solve $$y''+y = 0$$ at the point $$x_0 = 1\text{.}$$
Solution.
\begin{equation*} \begin{aligned} y=& \sum_{k=0}^{\infty}a_k(x-1)^k \\ y''=& \sum_{k=2}^\infty k(k-1)a_k(x-1)^{k-2},\quad \rm (k\rightarrow k+2) \\ =& \sum_{k=0}^\infty (k+2)(k+1)a_{k+2}(x-1)^{k} \end{aligned} \end{equation*}
Plugging into the ODE
We can note that for even $$k\text{:}$$
\begin{equation*} \begin{aligned} a_k=a_{2n}=\frac{(-1)^na_0}{(2n)!} \end{aligned} \end{equation*}
And for odd $$k\text{:}$$
\begin{equation*} \begin{aligned} a_k=a_{2n+1}=\frac{(-1)^na_1}{(2n+1)!} \end{aligned} \end{equation*}
So the solution can be written as
\begin{equation*} \begin{aligned} y=a_0\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}(x-1)^{2n}+a_1\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(x-1)^{2n+1}=a_0\cos(x-1)+a_1\sin(x-1) \end{aligned} \end{equation*}

#### Exercise4.2.2.

Use power series methods to solve $$y''+4xy = 0$$ at the point $$x_0 = 0\text{.}$$
The solution can be written as $$y=a_0y_0+a_1y_1\text{,}$$ where
\begin{equation*} \begin{aligned} y_0=& 1+\frac{(-4)}{(2)(3)}x^3+\frac{(-4)^2}{(2)(3)(5)(6)}x^6+...+\frac{(-4)^n}{(2)(3)(5)(6)...(3n-1)(3n)}x^{3n}+... \\ y_1=& x+\frac{(-4)}{(3)(4)}x^4+\frac{(-4)^2}{(3)(4)(6)(7)}x^7+...+\frac{(-4)^n}{(3)(4)(6)(7)...(3n)(3n+1)}x^{3n+1}+... \end{aligned} \end{equation*}

#### Exercise4.2.3.

Use power series methods to solve $$y''-xy = 0$$ at the point $$x_0 = 1\text{.}$$ You may compute only terms up to 4th order.
Note: In contrast to Example 4.2.2, we need to expand in powers of $$(x-1)^n\text{.}$$
The solution can be written as $$y=a_0y_0+a_1y_1\text{,}$$ where
\begin{equation*} \begin{aligned} y_0=& 1+\frac{1}{2}(x-1)^2+\frac{1}{6}(x-1)^3+...+\frac{1}{24}(x-1)^4+... \\ y_1=& (x-1)+\frac{1}{6}(x-1)^3+\frac{1}{12}(x-1)^4+... \end{aligned} \end{equation*}

#### Exercise4.2.4.

Use power series methods to solve $$y''+x^2y = 0$$ at the point $$x_0 = 0\text{.}$$
The solution can be written as $$y=a_0y_0+a_1y_1\text{,}$$ where
\begin{equation*} \begin{aligned} y_0=& 1+\frac{(-1)}{(3)(4)}x^4+\frac{(-1)^2}{(3)(4)(7)(8)}x^8+...+\frac{(-1)^n}{(3)(4)(7)(8)...(4n-1)(4n)}x^{4n}+... \\ y_1=& x+\frac{(-1)}{(4)(5)}x^5+\frac{(-1)^2}{(4)(5)(8)(9)}x^9+...+\frac{(-1)^n}{(4)(5)(8)(9)...(4n)(4n+1)}x^{4n+1}+... \end{aligned} \end{equation*}

#### Exercise4.2.5.

Use power series methods to solve $$y'' + 2 x^3 y = 0$$ at the point $$x_0 = 0\text{.}$$
$$a_2 = 0\text{,}$$ $$a_3 = 0\text{,}$$ $$a_4 = 0\text{,}$$ recurrence relation (for $$k \geq 5$$): $$a_k = \frac{- 2 a_{k-5}}{k(k-1)}\text{,}$$ so:
$$y(x) = a_0 + a_1 x -\frac{a_0}{10} x^5 - \frac{a_1}{15} x^6 + \frac{a_0}{450} x^{10} + \frac{a_1}{825} x^{11} - \frac{a_0}{47250} x^{15} - \frac{a_1}{99000} x^{16} + \cdots$$

#### Exercise4.2.6.

The methods work for other orders than second order. Try the methods of this section to solve the first order system $$y'-xy = 0$$ at the point $$x_0 = 0\text{.}$$
Solution.
Letting $$y=\sum_{k=0}^\infty a_kx^k\text{,}$$ differentiating (and re-indexing) and plugging into the ODE we get
\begin{equation*} \begin{aligned} \sum_{k=0}^\infty(k+1)a_{k+1}x^k-\sum_{k=0}^\infty a_kx^{k+1}=&0 \\ a_1+\sum_{k=1}^\infty(k+1)a_{k+1}x^k-\sum_{k=1}^\infty a_{k-1}x^{k}=&0 \\ \end{aligned} \end{equation*}
This gives $$a_1=0$$ and
\begin{equation*} \begin{aligned} a_{k+1}=\frac{a_{k-1}}{k+1} \end{aligned} \end{equation*}
Giving
So the solution is
\begin{equation*} \begin{aligned} y=a_0\sum_{n=0}^\infty \frac{x^n}{(2n)!!}=a_0e^{x^2/2} \end{aligned} \end{equation*}

#### Exercise4.2.7.Chebyshev’s equation of order $$p$$.

1. Solve $$(1-x^2)y''-xy' + p^2y = 0$$ using power series methods at $$x_0=0\text{.}$$
2. For what $$p$$ is there a polynomial solution?
Solution.
(a) Setting $$y=\sum_{k=0}^\infty a_k x^k$$ and plugging into the ODE
\begin{equation*} \begin{aligned} &\sum_{k=2}^\infty k(k-1)a_kx^{k-2}-\sum_{k=2}^\infty k(k-1)a_kx^k-\sum_{k=1}^\infty ka_kx^k+\sum_{k=0}^\infty p^2a_kx^k=0 \\ & 2a_2+6a_3x-a_1x+p^2a_0+p^2a_1x+\sum_{k=2}^\infty \left[ (k+2)(k+1)a_{k+2}-k(k-1)a_k-ka_k+p^2a_k\right]x^k=0 \end{aligned} \end{equation*}
This gives $$a_2=-p^2a_0/2\text{,}$$ $$a_3=(1-p^2)a_1/6$$ and the relation
\begin{equation*} \begin{aligned} a_{k+2}=\frac{k^2-p^2}{(k+2)(k+1)}a_k \end{aligned} \end{equation*}
Giving
So for even $$k$$
\begin{equation*} \begin{aligned} a_{2n}=\frac{\left( (2n-2)^2-p^2\right)\left( (2n-4)^2-p^2\right)...\left( 0^2-p^2\right)}{(2n)!}a_0 \end{aligned} \end{equation*}
And for odd $$k$$
\begin{equation*} \begin{aligned} a_{2n+1}=\frac{\left( (2n-1)^2-p^2\right)\left( (2n-3)^2-p^2\right)...\left( 1^2-p^2\right)}{(2n+1)!}a_1 \end{aligned} \end{equation*}
So the solution can be written as $$y=a_0y_0+a_1y_1\text{,}$$ where
\begin{equation*} \begin{aligned} y_0 =& 1+\frac{(0^2-p^2)}{2}x^2+\frac{(2^2-p^2)(0^2-p^2)}{4!}x^4+...+\frac{\left( (2n-2)^2-p^2\right)\left( (2n-4)^2-p^2\right)...\left( 0^2-p^2\right)}{(2n)!}x^{2n}+... \\ y_1=& x+\frac{(1^2-p^2)}{3!}x^3+\frac{(3^2-p^2)(1^2-p^2)}{5!}x^5+...+\frac{\left( (2n-1)^2-p^2\right)\left( (2n-3)^2-p^2\right)...\left( 1^2-p^2\right)}{(2n+1)!}x^{2n+1}+... \end{aligned} \end{equation*}

(b) If $$p$$ is an even integer, the series for $$y_0$$ terminates, whereas if $$p$$ is an odd integer the series for $$y_1$$ terminates. Both cases giving a polynomial solution.

#### Exercise4.2.8.

Find a polynomial solution to $$(x^2+1) y''-2xy'+2y = 0$$ using power series methods.
$$y=a_0+a_1x-a_0x^2=a_0(1-x^2)+a_1x$$

#### Exercise4.2.9.

1. Use power series methods to solve $$(1-x)y''+y = 0$$ at the point $$x_0 = 0\text{.}$$
2. Use the solution to part a) to find a solution for $$xy''+y=0$$ around the point $$x_0=1\text{.}$$
Solution.
(a) Plugging the series solution into the ODE
\begin{equation*} \begin{aligned} &\sum_{k=2}^\infty a_k(k-1)kx^{k-2}-\sum_{k=2}^\infty a_k(k-1)kx^{k-1}+\sum_{k=0}^\infty a_kx^k=0 \\ & \sum_{k=0}^\infty a_{k+2}(k+1)(k+2)x^{k}-\sum_{k=1}^\infty a_{k+1}(k+1)kx^{k}+\sum_{k=0}^\infty a_kx^k=0 \\ & 2a_2+a_0 +\sum_{k=1}^\infty \left[ a_{k+2}(k+1)(k+2)-a_{k+1}k(k+1)+a_k\right] x^k=0 \end{aligned} \end{equation*}
This gives $$a_2=-a_0/2$$ and the relation
\begin{equation*} \begin{aligned} a_{k+2}=\frac{a_{k+1}k(k+1)-a_k}{(k+1)(k+2)} \end{aligned} \end{equation*}
Giving
The solution is then
\begin{equation*} \begin{aligned} y=&\sum_{k=0}^\infty a_kx^k= a_0+a_1x-\frac{a_0}{2 }x^2-\frac{a_0+a_1}{(2)(3)}x^3-\frac{a_0/2+a_1}{(3)(4)}x^4+... \\ y=& a_0\left[ 1-\frac{x^2}{2}-\frac{x^3}{6}-\frac{x^4}{24}-\frac{x^5}{60}+...\right] + a_1\left[x-\frac{x^3}{6}-\frac{x^4}{12}-\frac{x^5}{24}+...\right] \end{aligned} \end{equation*}

(b) Let $$x=z-1\text{,}$$ then the equation becomes
\begin{equation*} \begin{aligned} (z-1)\frac{d^2y}{dz^2}+y=0 \end{aligned} \end{equation*}
At $$x=1\rightarrow z=0\text{,}$$ so we’re solving this equation near $$z=0$$ which is the same as part (a), therefore the solution is
\begin{equation*} \begin{aligned} y=& a_0\left[ 1-\frac{z^2}{2}-\frac{z^3}{6}-\frac{z^4}{24}-\frac{z^5}{60}+...\right] + a_1\left[z-\frac{z^3}{6}-\frac{z^4}{12}-\frac{z^5}{24}+...\right] \\ y=& a_0\left[ 1-\frac{(x+1)^2}{2}-\frac{(x+1)^3}{6}-\frac{(x+1)^4}{24}-\frac{(x+1)^5}{60}+...\right] + \\ & a_1\left[(x+1)-\frac{(x+1)^3}{6}-\frac{(x+1)^4}{12}-\frac{(x+1)^5}{24}+...\right] \end{aligned} \end{equation*}

#### Exercise4.2.10.

(challenging)   Power series methods also work for nonhomogeneous equations.
1. Use power series methods to solve $$y'' - x y = \frac{1}{1-x}$$ at the point $$x_0 = 0\text{.}$$ Hint: Recall the geometric series.
2. Now solve for the initial condition $$y(0)=0\text{,}$$ $$y'(0) = 0\text{.}$$
a) $$a_2 = \frac{1}{2}\text{,}$$ and for $$k \geq 1$$ we have $$a_k = \frac{a_{k-3} + 1}{k(k-1)}\text{,}$$ so
$$y(x) = a_0 + a_1 x + \frac{1}{2} x^2 + \frac{a_0 + 1}{6} x^3 + \frac{a_1 + 1}{12} x^4 + \frac{3}{40} x^5 + \frac{a_0 + 2}{30} x^6 + \frac{a_1 + 2}{42} x^7 + \frac{5}{112} x^8 + \frac{a_0 + 3}{72} x^9 + \frac{a_1 + 3}{90} x^{10} + \cdots$$
b) $$y(x) = \frac{1}{2} x^2 + \frac{1}{6} x^3 + \frac{1}{12} x^4 + \frac{3}{40} x^5 + \frac{1}{15} x^6 + \frac{1}{21} x^7 + \frac{5}{112} x^8 + \frac{1}{24} x^9 + \frac{1}{30} x^{10} + \cdots$$

#### Exercise4.2.11.

Attempt to solve $$x^2 y'' - y = 0$$ at $$x_0 = 0$$ using the power series method of this section ($$x_0$$ is a singular point). Can you find at least one solution? Can you find more than one solution?
Applying the method of this section directly we obtain $$a_k = 0$$ for all $$k$$ and so $$y(x) = 0$$ is the only solution we find.
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