Higher-order Linear ODE

Definition

Second-Order Liner ODE

$y^{\prime \prime }+p(x)y^{\prime }+q(x)y=r(x)$

Higher-Order Liner ODE

$L(y)=\underset{=\sum _{k=0}^n{p}_k(x)y^{(k)},p_n(x)=1}{\underbrace{y^{(n)}+p_{n-1}(x)y^{(n-1)}+\cdots +p_1(x)y^{\prime }+p_0(x)y}}=r(x)$

Solution

Second-Order Homogeneous Linear ODE with Constant Coefficients

Definition

$y^{\prime \prime }=a{y}^{\prime }+by=0,a,b\in \mathbb{R}$

Solution

Assume the solutions will be $y=e^{\lambda x}$ form then ${\lambda }^2{e}^{\lambda x}+a\lambda e^{\lambda X}+b{e}^{\lambda x}=({\lambda }^2+a\lambda +b)\underset{>0}{\underbrace{e^{\lambda x}}}=0$ $\Rightarrow {\lambda }^2+a\lambda +b=0$^[characteristic equation]

${\lambda }_{1,2}=\frac{-a\pm \sqrt{a^2-4b}}{2}\Rightarrow y_1=e^{{\lambda }_{1}x}$, $y_2=e^{{\lambda }_{2}x}$ are solutions

Two Distinct Real Roots ${\lambda }_1\ne {\lambda }_2$

General Solution

$y=c_1{e}^{{\lambda }_{1}x}+c_2{e}^{{\lambda }_{2}x},c_1,c_2\in \mathbb{R}$

Multiple Real Roots $\lambda =-\frac{a}{2}$

$y_1=e^{\lambda x}=e^{-\frac{a}{2}x}$

by using the Reduction of Order, we can get $y_2$.

$U=\frac{1}{y_1^2}{e}^{-\int pdx}=\frac{1}{e^{-ax}}{e}^{-ax}=1$ $\Rightarrow y_2=u{y}_1=y_1\int Udx=e^{-\frac{a}{2}x}x=x{e}^{\lambda x}$

General Solution

$y=c_1{e}^{\lambda x}+c_{2}x{e}^{\lambda x}=(c_1+c_{2}x)e^{\lambda x},c_1,c_2\in \mathbb{R}$

Complex Roots ${\lambda }_1=-\frac{a}{2}+i\omega ,{\lambda }_2=-\frac{a}{2}-i\omega$

$\begin{array}{rl}\Rightarrow & {\hat{y}}_1=e^{-\frac{a}{2}x}{e}^{+i\omega x}=e^{-\frac{a}{2}x}(\cos \omega x+i\sin \omega x)\\ & {\hat{y}}_2=e^{-\frac{a}{2}x}{e}^{-i\omega x}=e^{-\frac{a}{2}x}(\cos \omega x-i\sin \omega x)\\ \Rightarrow & y_1=\frac{{\hat{y}}_1+{\hat{y}}_2}{2}=e^{-\frac{a}{2}x}\cos \omega x\\ & y_2=\frac{{\hat{y}}_1-{\hat{y}}_2}{2i}=e^{-\frac{a}{2}x}\sin \omega x\end{array}$

General Solution

$y=A{e}^{-\frac{a}{2}x}\cos \omega x+B{e}^{-\frac{a}{2}x}\sin \omega x,A,B\in \mathbb{R}$

Examples

$$\begin{array}{rl}& y^{\prime \prime }+5y^{\prime }+6y=0\\ & \text{assume}y=e^{\lambda x}\\ & {\lambda }^2+5\lambda +6=0\Rightarrow (\lambda +3)(\lambda +2)=0\Rightarrow \lambda =-3\text{or}-2\\ & y=c_1{e}^{-3x}+c_2{e}^{-2x}\end{array}$$Link to original

Transclude of Higher-Order-Homogeneous-Linear-ODE-with-Constant-Coefficients

Euler-Cauchy Equations

Definition

$x^2{y}^{\prime \prime }+ax{y}^{\prime }+by=0(x\ne 0)$

Solution

Assume the solutions $y=x^m$ form
then $x^{2}m(m-1)x^{m-2}+axm{x}^{m-1}+b{x}^m=m(m-1)+am+b=0$

$\Rightarrow m^2+(a-1)m+b=0$^[auxiliary equation]

$m=\frac{(1-a)\pm \sqrt{(1-a{)}^2-4b}}{2}=\frac{1-a}{2}\pm \sqrt{\frac{(1-a{)}^2}{4}-b}$

Two Distinct Real Roots $m_1,m_2$

$y_1=x^{m_1},y_2=x^{m_2}$

General Solution

$y=c_1{x}^{m_1}+c_2{x}^{m_2},c_1,c_2\in \mathbb{R}$

Multiple Real Roots $m=\frac{1-a}{2}$

$y_1=x^m=x^{\frac{1-a}{2}}$

By using the Reduction of Order, we can get $y_2$.

$U=\frac{1}{y_1^2}{e}^{-\int pdx}=\frac{1}{y_1^2}{e}^{-\int \frac{a}{x}dx}=\frac{1}{x^{1-a}}{e}^{-a\ln x}=\frac{1}{x^{1-a}}{x}^{-a}=\frac{1}{x},p=\frac{ax}{x^2}=\frac{a}{x}$ $\Rightarrow y_2=u{y}_1=y_1\int Udx=x^m\ln x(x>0)$

General Solution

$y=c_1{x}^m+c_2{x}^m\ln x,c_1,c_2\in \mathbb{R}$

Complex Roots $m=\mu \pm i\nu ,\mu ,\nu \in \mathbb{R}$

$\Rightarrow {\hat{y}}_1=x^{\mu +i\nu }=x^{\mu }{x}^{i\nu }=x^{\mu }(e^{\ln x}{)}^{i\nu }=x^{\mu }{e}^{i\nu \ln x}=x^{\mu }(\cos (\nu \ln x)+i\sin (\nu \ln x))$ ${\hat{y}}_2=x^{\mu -i\nu }=x^{\mu }{x}^{-i\nu }=x^{\mu }(e^{\ln x}{)}^{-i\nu }=x^{\mu }{e}^{-i\nu \ln x}=x^{\mu }(\cos (\nu \ln x)-i\sin (\nu \ln x))$ $\Rightarrow y_1=\frac{{\hat{y}}_1+{\hat{y}}_2}{2}=x^{\mu }\cos (\nu \ln x)$ $y_2=\frac{{\hat{y}}_1-{\hat{y}}_2}{2i}=x^{\mu }\sin (\nu \ln x)$

General Solution

$y=A{x}^{\mu }\cos (\nu \ln x)+B{x}^{\mu }\sin (\nu \ln x),A,B\in \mathbb{R}$

Examples

$$\begin{array}{rl}& x^2{y}^{\prime \prime }-3x{y}^{\prime }-12y=0\\ & \text{substitute}{x}^m:=y\\ & x^2(x^m{)}^{\prime \prime }-3x(x^m{)}^{\prime }-12(x^m)=0\\ & m(m-1)x^m-3m{x}^m-12x^m=0\\ & m^2-4m-12=0\Rightarrow (m-6)(m+2)=0\Rightarrow m=6\text{or}-2\\ & y=C_1{x}^6+C_2{x}^{-2}\end{array}$$Link to original

Higher-Order Non-Homegeneous Linear ODE

Definition

Second-order

$y^{\prime \prime }+p(x)y^{\prime }+q(x)y=r(x)\ne 0⑴$^[Non-homogeneous ODE]

$y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0⑵$^[Corresponding homogeneous ODE]

Higher-order

$L(y)=y^{(n)}+a_{n-1}{y}^{(n-1)}+\cdots +a_0{y}^{(0)}=r(x)\ne 0$^[Non-homogeneous ODE]

$y^{(n)}+a_{n-1}{y}^{(n-1)}+\cdots +a_0{y}^{(0)}=0$^[Corresponding homogeneous ODE]

Facts

Thm 1

The sum of solution $y$ of ⑴ on some open interval $I$ and a solution $\tilde{y}$ of ⑵ on $I$ $y+\tilde{y}$ is a solution of ⑴ on $I$

The difference of two solutions of ⑴ on $I$ is a solution of ⑵ on $I$

Thm 2

If $p(x),q(x),r(x)$ in Non-homogeneous ODE ⑴ are continuous on $I$, then every solution of Non-homogeneous ODE ⑴ on $I$ can be obtained from ⑶

Solution

General Solutions of Non-homogeneous Linear ODE

$y(x)=y_h(x)+y_p(x)⑶$, where $y_h(x)$ is a general solution of the corresponding homogeneous ODE ⑵ $y_p(x)$ is any solution (particular solution) of the Non-homogeneous ODE ⑴

By Thm 1 $y(x)$ is a solution of Non-homogeneous ODE ⑴

Variation of parameters

Second-order

for $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=r(x)\Rightarrow y=y_h+y_p$

If $p,q,r$ are continuous on $I$ and if $y_1,y_2$ form a basis of solutions of the corresponding homogeneous ODE, then $y_p(x)=-y_1\int \frac{y_{2}r}{W}dx+y_2\int \frac{y_{1}r}{W}dx$, $W$ is Wronskian Determinant

Higher-order

If $y_1,\dots ,y_n$ are linearly independent solutions of the corresponding homogeneous equation of $\sum _{k=0}^n{p}_k(x)y^{(k)}=r(x),p_n(x)=1$

Then, $y_p$ becomes $y_p(x)=\sum _{k=1}^n{y}_k(x)\int \frac{W_k(x)}{W(x)}r(x)dx$, where Wronskian Determinant $W(x)=\left|\begin{array}{ccc}{y}_1& \dots & y_n\\ y_1^{\prime }& \dots & y_n^{\prime }\\ ⋮& \ddots & ⋮\\ y_1^{(n-1)}& \dots & y_n^{(n-1)}\end{array}\right|$, $W_k(x)=\left|\begin{array}{ccccc}{y}_1& \dots & 0& \dots & y_n\\ y_1^{\prime }& \dots & 0& \dots & y_n^{\prime }\\ ⋮& \ddots & ⋮& \ddots & ⋮\\ y_1^{(n-1)}& \dots & \underset{\text{k-th column}}{\underbrace{1}}& \dots & y_n^{(n-1)}\end{array}\right|$

Method of Undetermined Coefficients

Definition

To solve $y^{\prime \prime }+a{y}^{\prime }+by=r(x)$, we have to solve corresponding homogeneous ODE $y^{\prime \prime }+a{y}^{\prime }+by=0$ and find a particular solution $y_p$

Table

Terms in $r(x)$	Choice for $y_p(x)$
$k{e}^{\gamma x}$	$C{e}^{\gamma x}$
k x^n,\quad _{n=0, 1, 2}	$K_n{x}^n+K_{n-1}{x}^{n-1}+\cdots +K_{1}x+K_0$
$k\cos \omega x$ $k\sin \omega x$	$K\cos \omega x+M\sin \omega x$
$k{e}^{\alpha x}\cos \omega x$ $k{e}^{\alpha x}\sin \omega x$	$e^{\alpha x}(K\cos \omega x+M\sin \omega x)$

Choice Rules

1. Basic Rule: If $r(x)$ belongs a column in the left, choose corresponding right $y_p$
2. Modification Rule: If $y_p$ is a solution of a corresponding homogeneous ODE, multiply the selected $y_p$ by $x$ or $x^2$
3. Sum Rule: If $r(x)$ is a sum of several functions, $y_p$ is selected as the sum of functions corresponding to each.

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Examples

&y'' + 4y = x^{2}+ 3e^{x}\\ &\text{find the solution of the corresponding homogene ODE:}\\ &y'' + 4y = 0\\ &\text{by the solution of separable ODE}\\ &y_{h} = C_{1}\cos(2x) + C_{2}\sin(2x)\\ \\ &\text{by method of undetermined coefficients, assume}\\ &y_{p} = ax^{2}+bx+c+de^{x}\\ &y_{p}'' + 4y_{p} = (2a+de^{x}) + 4(ax^{2}+bx+c+de^{x}) = 5de^{x}+4ax^{2}+4bx+(2a+4c) = x^{2}+ 3e^{x}\\ &\Rightarrow a=\frac{1}{4}, b=0, c=-\frac{1}{8},d=\frac{3}{5}\\ &y_{p}= \frac{1}{4}x^{2} + \frac{3}{5}e^{x} -\frac{1}{8}\\ \\ &y = y_{h}+y_{p} = C_{1}\cos(2x) + C_{2}\sin(2x) +\frac{1}{4}x^{2} + \frac{3}{5}e^{x} -\frac{1}{8} \end{aligned}$$ $$\begin{aligned} &y''+4y=3\sin(2x)\\ &\text{find the solution of the corresponding homogene ODE:}\\ &y''+4y=0\\ &\text{by the solution of separable ODE}\\ &y_{h} = C_{1}\cos(2x)+C_{2}\sin(2x)\\ \\ &\text{by method of undetermined coefficients (modification rule), assume}\\ &y_{p}=ax\cos(2x)+bx\sin(2x)\\ &y_{p}''=-4a\sin(2x)+4b\cos(2x)-4ax\cos(2x)-4bx\sin(2x)\\ &y_{p}''+4y_{p}=-4a\sin(2x)+4b\cos(2x)=3\sin(2x)\\ &\Rightarrow a=-\frac{3}{4},b=0\\ &y_{p}= -\frac{3}{4}x\cos(2x)\\ \\ &y = y_{h}+y_{p} = C_{1}\cos(2x)+C_{2}\sin(2x)-\frac{3}{4}x\cos(2x) \end{aligned}$$

\begin{aligned} &y”-2y’+y=xe^{x}+4\ &\text{find the solution of the corresponding homogene ODE:}\ &y”-2y’+y=0\ &\text{by the solution of separable ODE}\ &y_{h}=C_{1}e^{x}+C_{2}xe^{x}\ \ &\text{by method of undetermined coefficients (modification rule), assume}\ &y_{p}=ax^{3}e^{x}+b\ &y_{p}‘=3ax^{2}e^{x}+ax^{3}e^{x}\ &y_{p}”=6axe^{x} + 6ax^{2}e^{x} + ax^{3}e^{x}\ &y_{p}”-2y_{p}‘+y_{p} = 6axe^{x} + b = xe^{x}+4\ &\Rightarrow a=\frac{1}{6}, b=4\ &y_{p} = \frac{1}{6}x^{3}e^{x}+4\ \ &y=y_{h}+y_{p} = C_{1}e^{x}+C_{2}xe^{x} + \frac{1}{6}x^{3}e^{x}+4 \end{aligned}

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