---
# System prepended metadata

title: Series Solutions of Linear Differential Equations
tags: [Mathematics]

---

# Notes
## Power Series
1. $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+...$
2. $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...$
3. $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$
4. $\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...$
## Power Series Solutions
Suppose the liner second-order differential equation
$a_2(x)y''+a_1(x)y'+a_0(x)y=0$
and the standard form is
$y''+P(x)y'+Q(x)y=0$
1. Introductions
    1. Ordinary and Sigular Points 
        A point $x_0$ is said to be an ordinary point of the differential equation if both $P(x)$ and $Q(x)$ in the standard form are analytic at $x_0$. A point that is not an ordinary point is said to be a singular point of the equaiton.
		* Regular and Irragular Singular Points
			A sigular point $x_0$ is said to be a regular singulr point of the differential equation if the functions $p(x)=(x-x_0)P(x)$ and $q(x)=(x-x_0)^2Q(x)$ are both analytic at $x_0$. A singular point that is not regular is said to be an irregular sigulr point of the equation.
            Moreover, observe that if $x=x_0$ is a regular singular point and
            $(x-x_0)^2y''+(x-x_0)p(x)y'+q(x)y=0$
            then $p(x)$ and $q(x)$ are both analytic at $x=x_0$
    2. $y=c_0y_1(x)+c_1y_2(x)$
2. Solution
	1. Oridinary Point
	    If $x=x_0$ is and ordinary point of the differential equation, we can always find two linearly indepentent solutions in the form of a power series centered at $x_0$; that is, $y=\sum_{n=0}^\infty c_n(x-x_0)^n$. A series solution converge at least on some interval defined bt $|x-x_0|<R$, where $R$ is the distance from $x_0$ to the closest singulat point.

	2. Sigular Point
		* Frobenius' Theorem
		    1. Definition
			    If $x=x_0$ is a regular singular point of a differential equation, then there exists at least one nonzero solution of the form 
			    $y=(x-x_0)^r\sum_{n=0}^\infty c_n(x-x_0)^n=\sum_{n=0}^\infty c_n(x-x_0)^{n+r},\ \text{where the number r is}\\a\  constant\ to\ be\ determined.$ The series will converge at least on some interval defined by $0<x-x_0<R$
            2. Indical Equation
            	If $x=0$ is a regular singular point,
                $Let\ y''+P(x)y'+Q(x)y=0\\\Rightarrow x^2y''+x[xP(x)]y'+[x^2Q(x)]y=0\\Let\ \begin{cases}p(x)=xP(x)=a_0+a_1x+a_2x^2+\cdots\\q(x)=x^2Q(x)=b_0+b_1x+b_2x^2+\cdots\end{cases}\\Substituting\ y=\sum_{n=0}^\infty c_nx^{n+r}\\\Rightarrow r(r-1)+a_0r+b_0=0$
                1. Case 1: If $r_1-r_2\not=integer$, there exist two linearly independent solutions of the form
                    $\begin{cases}y_1(x)=\sum_{n=0}^\infty c_nx^{n+r1}\\y_2(x)=\sum_{n=0}^\infty c_nx^{n+r2}\end{cases}$
                2. Case 2: If $r_1-r_2=N$, where $N$ is a positive integer, then there exist two linearly independent solutions of the form
                    $\begin{cases}y_1(x)=\sum_{n=0}^\infty c_nx^{n+r1},\ c_0\not=0\\y_2(x)=Cy_1(x)lnx+\sum_{n=0}^\infty b_nx^{n+r_2},\ b_0\not=0,\ where\ C\ is\ a\ constant\ that\\could\ be\ zero\end{cases}$
                3. Case 3: If $r_1=r_2$, then there always exist two lineraly independent solutions of the form
                    $\begin{cases}y_1(x)=\sum_{n=0}^\infty c_nx^{n+r_1},\ c_0\not=0\\y_2(x)=y_1(x)lnx+\sum_{n=0}^\infty b_nx^{n+r_2}\end{cases}$
                * ※In case 2 and case 3, the way of finding the second solution with knowing the first solution in advance is that
                    $y_2(x)=y_1(x)\int\frac{e^{-\int P(x)dx}}{y_1^2(x)}dx$
3. Special Functions
    1. Bessel Functions 
        1. Definition
            $x^2y''+xy'+(x^2-\nu^2)y=0$
            is called Bessel's equation of order $\nu$
        2. The Solutions
            1. Bessel Functions of the First and the Second Kind
                Because $x=0$ is a regular singular point of Bessel's equation,
                $Let\ y=\sum_{n=0}^\infty c_nx^{n+r}\\\Rightarrow x^2y''+xy'+(x^2-\nu^2)\\\begin{split}&=\sum_{n=0}^\infty c_n(n+r)(n+r-1)x^{n+r}+\sum_{n=0}^\infty c_n(n+r)x^{n+r}+\sum_{n=0}^\infty c_nx^{n+r+2}-\nu^2\sum_{n=0}^\infty c_nx^{n+r}\\&=c_0(r^2-r+r-\nu^2)x^r+x^r\sum_{n=0}^\infty c_n[(n+r)(n+r-1)+(n+r)-\nu^2]x^n+x^r\sum_{n=0}^\infty c_nx^{n+2}\\&=c_0(r^2-\nu^2)x^r+x^r\sum_{n=0}^\infty c_n[(n+r)^2-\nu^2]x^n+x^r\sum_{n=0}^\infty c_nx^{n+2}\end{split}\\r^2-\nu^2=0\\\Rightarrow\begin{cases}r_1=\nu\\r_2-\nu\end{cases}\\At\ r_1=\nu\\\Rightarrow x^\nu\sum_{n=1}^\infty c_nn(n+2\nu)x^n+x^\nu\sum_{n=0}^\infty c_nx^{n+2}\\\begin{split}&=x^\nu[(1+2\nu)c_1x+\sum_{n=2}^\infty c_nn(n+2\nu)x^n+\sum_{n=0}^\infty c_nx^{n+2}]\\&=x^\nu[(1+2\nu)c_1x+\sum_{k=0}^\infty [(k+2)(k+2+2\nu)c_{k+2}+c_k]x^{k+2}]=0\end{split}\\\Rightarrow\begin{cases}(1+2\nu)c_1=0\\(k+2)(k+2+2\nu)c_{k+2}+c_k=0\Rightarrow c_{k+2}=\frac{-c_k}{(k+2)(k+2+2\nu)},\ k=0,1,2,...\end{cases}\\\begin{split}(i)&We\ choose\ c_1=0\\&\Rightarrow c_3=c_5=c_7=\cdots=0\end{split}\\\begin{split}(ii)&For\ n=1,2,3,...,\\&c_{2n}=-\frac{c_{2_n-2}}{2^2n(n+\nu)}\\&\begin{split}\Rightarrow &c_2=\frac{c_0}{2^2\cdot1\cdot(1+\nu)}\\&c_4=-\frac{c_2}{2^2\cdot2(2+\nu)}=\frac{c_0}{2^4\cdot1\cdot2(1+\nu)(2+\nu)}\\&c_6=-\frac{c_4}{2^2\cdot3(3+\nu)}=-\frac{c_0}{2^6\cdot1\cdot2\cdot3(1+\nu)(2+\nu)(3+\nu)}\\&\vdots\\&c_{2n}=\frac{(-1)^nc_0}{2^{2n}n!(1+\nu)(2+\nu)\cdots(n+\nu)},\ n=1,2,3,...\end{split}\end{split}\\We\ choose\ c_0=\frac{1}{2^\nu\Gamma(1+\nu)},\ where\ \Gamma(1+\alpha)=\alpha\Gamma(\alpha)\\\Rightarrow c_{2n}=\frac{(-1)^n}{2^{2n+\nu}n!(1+\nu)(2+\nu)\cdots(n+\nu)\Gamma(1+\nu)}=\frac{(-1)^n}{2^{2n+\nu}n!\Gamma(1+\nu+n)},\ n=0,1,2,...$
                1. Bessel Functions fo the First Kind
                    $\begin{cases}J_\nu(x)=\sum_{n=0}^\infty\frac{(-1)^n}{n!\Gamma(1+\nu+n)}(\frac{x}{2})^{2n+\nu},\ the\ series\ converges\ on\ [0,\infty)\ if\ \nu\geq0 \\J_{-\nu}(x)=\sum_{n=0}^\infty\frac{(-1)^n}{n!\Gamma(1-\nu+n)}(\frac{x}{2})^{2n-\nu},\ the\ series\ converges\ on\ (0,\infty)\end{cases}$
                    1. Case 1: $r_1-r_2=2\nu\ is\ not\ a\ positive\ integer$
                        $J_\nu(x)$ and $J_{-\nu}(x)$ is linearly independent.
                    2. Case 2: $r_1-r_2=2\nu,\ where\ \nu=m\ is\ a\ positive\ integer$
                        $J_\nu(x)$ and $J_{-\nu}(x)$ is linearly dependent.
                    3. Case 3: $r_1-r_2=2\nu,\ where\ \nu\ is\ half\ an\ odd\ integer$
                        $J_\nu(x)$ and $J_{-\nu}(x)$ is linearly independent.
                    The solution on $(0,\infty)$ is
                    $y=c_1J_\nu(x)+c_2J_{-\nu}(x),\ \nu\not=integer$
                    
                2. Bessel Functions of the Second Kind
                    If $\nu\not=integer$
                    $Y_\nu(x)=\frac{cos\nu\pi J_\nu(x)-J_{-\nu}(x)}{sin\nu\pi},\ where\ Y_\nu(x)\ and\ J_\nu(x)\ are\ linearly\ independent$
                    Therefore the solution on $(0,\infty)$ can also be written as
                    $y=c_1J_\nu(x)+c_2Y_\nu(x)$
                    Noted that if $\nu\rightarrow m,\ where\ m=integer$, $Y_\nu(x)$ has the interdeterminate form $\frac{0}{0}$. By L'Hostipal's rule
                    $Y_m(x)=lim_{\nu\to m}Y_\nu(x)$
            2. Parametric Bessel equation
                $x^2y''+xy'+(\alpha^2x^2-\nu^2)y=0$
                $Let\ t=\alpha x\\\begin{cases}\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\alpha\frac{dy}{dt}\\\frac{d^2y}{dx^2}=\frac{d}{dt}(\frac{dy}{dx})\frac{dt}{dx}=\alpha^2\frac{d^2y}{dt^2}\end{cases}\\\Rightarrow t^2\frac{d^2y}{dt^2}+t\frac{dy}{dt}+(t^2-\nu^2)y=0$
                The solution on $(0,\infty)$ is
                $y=c_1J_\nu(\alpha x)+c_2Y_\nu(\alpha x)$
                * Modified Bessel Functions
                    1. The first kind
                        $x^2y''+xy'-(x^2+\nu^2)y=0$
                        $Let\ t=ix\Rightarrow i^2=-1\\\Rightarrow t^2\frac{d^2y}{dt^2}+t\frac{dy}{dt}+(t^2-\nu^2)y=0\\Let\ I_\nu(x)=i^{-\nu}J_\nu(ix)$
                        The solution on $(0,\infty)$ is
                        $y=c_1I_\nu(x)+c_2I_{-\nu}(x),\ \nu\not=integer$
                    2. The second kind
                        $K_\nu(x)=\frac{\pi}{2}\frac{I_{-\nu}(x)-I_\nu(x)}{sin\nu\pi}$
                        Therefore the solution on $(0,\infty)$ can also be written as
                        $y=c_1I_\nu(x)+c_2K_\nu(x)$
                        Noted that if $\nu\rightarrow m,\ where\ m=integer$, $I_\nu(x)$ has the interdeterminate form $\frac{0}{0}$. By L'Hostipal's rule
                    $K_m(x)=lim_{\nu\to m}K_\nu(x)$
                * ※Parametric form of the modified Bessel equation
                    1. Definition
                        $x^2y''+xy'-(\alpha^2x^2+\nu^2)y=0$
                    2. Solution on $(0,\infty)$
                        $y=c_1I_\nu(\alpha x)+c_2K_\nu(\alpha x)$
            3. Bessel Functions of Half-Integral Order
                We consider $\nu$ is half an odd integer.
                1. The First Kind
                    $J_{1/2}(x)=\sum_{n=0}^\infty \frac{(-1)^n}{n!\Gamma(1+\frac{1}{2}+n)}(\frac{x}{2})^{2n+1/2}\\Because\begin{cases}\Gamma(1+\alpha)=\alpha\Gamma(\alpha)\\\Gamma(\frac{1}{2})=\sqrt{\pi}\end{cases}\\\begin{split}\Rightarrow&\Gamma(\frac{3}{2})=\frac{1}{2}\Gamma(\frac{1}{2})=\frac{1}{2}\sqrt{\pi}\\&\Gamma(\frac{5}{2})=\frac{3}{2^2}\sqrt{\pi}\\&\Gamma(\frac{7}{2})=\frac{5!}{2^52!}\sqrt{\pi}\\&\Gamma(\frac{9}{2})=\frac{7!}{2^73!}\sqrt{\pi}\\&\vdots\\&\Gamma(1+\frac{1}{2}+n)=\frac{(2n+1)!}{2^{2n+1}n!}\sqrt{\pi}\end{split}\\\Rightarrow J_{1/2}=\sum_{n=0}^\infty \frac{(-1)^n}{n!\frac{(2n+1)!}{2^{2+1}n!}\sqrt{\pi}}(\frac{x}{2})^{2n+1/2}=\sqrt{\frac{2}{\pi x}}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}\\\begin{cases}J_{1/2}(x)=\sqrt{\frac{2}{\pi x}}sinx\\J_{-1/2}(x)\sqrt{\frac{2}{\pi x}}cosx\end{cases}$
                2. The Second Kind
                    $Y_{n+1/2}(x)=(-1)^{n+1}J_{-(n+1/2)}(x)\\\Rightarrow\begin{cases}Y_{1/2}(x)=-\sqrt{\frac{2}{\pi x}}cosx\\Y_{-1/2}(x)=\sqrt{\frac{2}{\pi x}}sinx\end{cases}$
                
                The sperical bessel functions of the first kind, $j_n()x)$ and the second kind, $y_n(x)$ is respectively
                $\begin{cases}j_n(x)=\sqrt{\frac{\pi}{2x}}J_{n+1/2}(x)\\y_n(x)=\sqrt{\frac{\pi}{2x}}Y_{n+1/2}(x)\end{cases}$
        3. Properties
            * Differential Recurrence Relation
                $xJ'_\nu(x)=\nu J_\nu(x)-xJ_{\nu+1}(x)$
    2. Legendre Functions
        1. Definition
            $(1-x^2)y''-2xy'+n(n+1)y=0$
        2. The Solutions
            Since $x=0$ is and ordinary point of Legendre's equation, 
            $Let\ y=\sum_{n=k}^\infty c_kx^k\\\begin{split}\Rightarrow &(1-x^2)y''-2xy'+n(n+1)y\\&=[n(n+1)c_0+2c_2]+[(n-1)(n+2)c_1+6c_3]x+\sum_{j=2}^\infty [(j+2)(j+1)c_{j+2}+(n-j)(n+j+1)c_j]x^j=0\end{split}\\\begin{cases}n(n+1)c_0+2c_2=0\\(n-1)(n+2)c_1+6c_3=0\\(j+2)(j+1)c_{j+2}+(n-j)(n+j+1)c_j=0\end{cases}\\\Rightarrow\begin{cases}c_2=-\frac{n(n+1)}{2!}c_0\\c_3=-\frac{(n-1)(n+2)}{3!}c_1\\c_{j+2}=-\frac{(n-j)(n+j+1)}{(j+2)(j+1)}c_j,\ j=2,3,4,...\end{cases}\\\begin{split}\Rightarrow&j=2,\ c_4=-\frac{(n-2)(n+3)}{4\cdot3}c_2=\frac{(n-2)n(n+1)(n+3)}{4!}c_0\\&j=3,\ c_5=-\frac{(n-3)(n+4)}{5\cdot4}c_3=\frac{(n-3)(n-1)(n+2)(n+4)}{5!}c_1\\&j=4,\ c_6=-\frac{(n-4)(n+5)}{6\cdot5}=-\frac{(n-4)(n-2)n(n+1)(n+3)(n+5)}{6!}c_0\\&j=5,\ c_7=-\frac{(n-5)(n+6)}{7\cdot5}c_5=-\frac{(n-5)(n-3)(n-1)(n+2)(n+4)(n+6)}{7!}c_1\\&\vdots\end{split}\\For\ |x|<1\\\begin{cases}y_1(x)=c_0[1-\frac{n(n+1)}{2!}x^2+\frac{(n-2)n(n+1)(n+3)}{4!}x^4-\frac{(n-4)(n-2)n(n+1)(n+3)(n+5)}{6!}x^6+\cdots]\\y_2(x)=c_1[x-\frac{(n-1)(n+3)}{3!}x^3+\frac{(n-3)(n-1)(n+2)(n+4)}{5!}x^5-\frac{(n-5)(n-3)(n-1)(n+2)(n+4)(n+6)}{7!}x^7+\cdots]\end{cases}\\\Rightarrow\begin{cases}For\ n=0,\ we\ choose\ c_0=1,\ and\ for\ n=2,4,6,...\ c_0=(-1)^{n/2}\frac{1\cdot3\cdots(n-1)}{2\cdot4\cdots n}\\For\ n=1,\ we\ choose\ c_1=1,\ and\ for\ n=3,5,7,...\ c_0=(-1)^{(n-1)/2}\frac{1\cdot3\cdots n}{2\cdot4\cdots(n-1)}\end{cases}$
        3. Properties
            1. $P_n(-x)=(-1)^nP_n(x)$
            2. $P_n(1)=1$
            3. $P_n(-1)=(-1)^n$
            4. $P_n(0)=0,\ n\ is\ an\ odd\ integer$
            5. $P'_n(0)=0,\ n\ is\ an\ even\ integer$
            6. Recurrence Relation
                $(k+1)P_{k+1}(x)-(2k+1)xP_k(x)+kP_{-1}(x)=0$
            7. Rodrigues' Formula
                $P_n(x)=\frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^n,\ n=0,1,2,...$
                
                        
        
# Exercises
## Advanced engineering mathematics, 6th edition, By Dennis G. Zill
### P.268 Chapter 5.1 Example 4
Solve $y''-(1+x)y=0$ at $x=0$
![](https://i.imgur.com/ng193mO.jpg)
![](https://i.imgur.com/YAT3DDd.jpg)
### P.269 Chapter 5.1 Example 5
Solve $y''+(cosx)y=0$ at $x=0$
![](https://i.imgur.com/M4ow0vN.jpg)
### P.271 Chapter 5.2 Example 2
Solve $3xy''+y'-y=0$ at $x=0$
![](https://i.imgur.com/l9AbIdf.png)
![](https://i.imgur.com/LHHRmCu.png)
![](https://i.imgur.com/IpzTUET.png)
### P.276 Chapter 5.2 Example 4
Solve $xy''+y=0$ at $x=0$
![](https://i.imgur.com/YyfsvP8.png)
![](https://i.imgur.com/ypnvzDS.png)
![](https://i.imgur.com/1cRtj87.png)
![](https://i.imgur.com/tb9J3IH.png)


**Reference**
NCKUCSIE 1101 Engineering Mathematics
Advanced engineering mathematics, 6th edition, By Dennis G. Zill