【笔记】高等数学复习笔记-微分方程

基于 William E, Boyce, Richard C 所著的《Elementary Differential Equations and Boundary Value Problems》以及课堂笔记；

加上网上资料结合进行整理；

对于部分基础多有忽略，主要用于知识点查找；

First Order Differential Equations

The general first order linear equation is

$$\frac{dy}{dt}+p(t)y=g(t)\qquad y(t_0)=y_0$$

If we multiply the equation above by $\mu(t)$

$$\mu(t)\frac{dy}{dt}+p(t)\mu(t)y=\mu(t)g(t)$$

In order to construct the equation

$$(uv)'=uv'+u'v$$

we set

$$\frac{d\mu(t)}{dt}=p(t)\mu(t)$$

then we have

$$\frac{d\mu(t)/dt}{\mu(t)}=p(t)$$

and consequently

$$\ln\mu(t)=\int_{t_0}^t p(s)ds+k$$

By choosing the arbitrary constant $k$ to be zero

$$\mu(t)=\exp\int_{t_0}^t p(s)ds$$

Returning to the first equation, we have

$$\frac d{dt}[\mu(t)y]=\mu(t)g(t)$$

To satisfy the initial condition (2) we must choose $c = y_0$, Hence

$$y=\frac{\int_{t_0}^t \mu(s)g(s)ds+y_0}{\mu(t)}$$

Separable Equations

The general first order equation is

$$\frac{dy}{dx}=f(x,y)$$

rewrite it in the form

$$M(x,y)+N(x,y)\frac{dy}{dx}=0$$

If $M $ is a function of $x$ only and $N$ is a function of $y$ only (separable) , set

$$H_1'(x)=M(x),\qquad H_2'(y)=N(y)$$

the equation becomes

$$H_1'(x)+H_2'(y)\frac{dy}{dx}=0$$

Consequently,

$$\frac{d}{dx}[H_1(x)+H_2(y)]=0$$

we obtain

$$H_1(x)+H_2(y)=c$$

Exact Equations and Integrating Factors

let the differential equation

$$M(x,y)+N(x,y)y'=0$$

If

$$M_y(x,y)=N_x(x,y)$$

there exists a function $\psi$

$$\psi_x=M(x,y),\qquad \psi_y=N(x,y)$$

We define $y = \phi(x)$, Then

$$\frac{\partial \psi}{\partial x}+\frac{\partial \psi}{\partial y}\frac{dy}{dx}=\frac d{dx}\psi[x,\phi(x)]=0$$

Solutions are given implicitly by

$$\psi(x,y)=c$$

where $c$ is an arbitrary constant.

If

$$M_y(x,y)\ne N_x(x,y)$$

we need an integrating factor $\mu(x,y)$, and satisfies

$$\begin{aligned} (\mu M)_y&=(\mu N)_x\\ M\mu_y+M_y\mu&=N\mu_x+N_x\mu \end{aligned}$$

Unfortunately, this function is ordinarily at least as difficult to solve as the original equation. The most important situations in which simple integrating factors can be found occur when $\mu$ is a function of only one of the variables $x$ or $y$, instead of both.

$$(\mu M)_y=\mu M_y,\qquad(\mu N)_x=\mu N_x+N\frac{d\mu}{dx}$$

Thus

$$\frac{d\mu}{dx}=\frac{M_y-N_x}{N}\mu$$

Homogeneous Equations with Constant Coefficients

$$ay''+by'+cy=0$$

we set

$$y_1(t)=e^{r_1t},\qquad y_2(t)=e^{r_2t}$$

The characteristic equation for the differential equation is

$$ar^2+br+c=0$$

if $r_1\ne r_2$

$$y=c_1y_1(t)+c_2y_2(t)$$

if $r_1,r_2=\lambda\pm i\mu$

$$y=c_1e^{\lambda t}\cos\mu t+c_2e^{\lambda t}\sin\mu t$$

if $r_1 = r_2$

$$y=c_1y(t)+c_2ty(t)$$

Fundamental Solutions of Linear Homogeneous Equations

$$y''+p(t)y'+q(t)y=0$$

Wronskian determinant

$$W(y_1,y_2)=\begin{vmatrix}y_1&y_2\\y_1'&y_2'\end{vmatrix}=y_1y_2'-y_1'y_2$$ $$W'(y_1,y_2)=y_1'y_2'+y_1y_2''-y_1'y_2'-y_1''y_2=y_1y_2''-y_1''y_2=\begin{vmatrix}y_1&y_2\\y_1''&y_2''\end{vmatrix}$$

扩展

$$\begin{aligned} y_1''+p(t)y_1'+q(t)y_1=0\\ y_2''+p(t)y_2'+q(t)y_2=0 \end{aligned}$$

If we multiply the first equation by $−y_2$, the second by $y_1$, and add the resulting
equations, we obtain

$$(y_1y_2''-y_1''y_2)+p(t)(y_1y_2'-y_1'y_2) = 0$$

Then we can write in the form

$$W'+p(t)W=0$$

It is a first order linear equation

$$W(t)=c\exp[-\int p(t)dt]$$

Euler Equations

$$L[y]=x^2y''+\alpha xy'+\beta y=0$$

we assume that

$$y=x^r$$

The roots of equation are

$$r_1,r_2=\frac{-(\alpha-1)\pm\sqrt{(\alpha-1)^2-4\beta}}2$$

If the roots are real and different, then

$$y=c_{1}|x|^{r_{1}}+c_{2}|x|^{r_{2}}$$

If the roots are real and equal, then

$$y=\left(c_{1}+c_{2} \ln |x|\right)|x|^{r_{1}}$$

If the roots are complex, then

$$y=|x|^{\lambda}\left[c_{1} \cos (\mu \ln |x|)+c_{2} \sin (\mu \ln |x|)\right]$$

where $r_{1}, r_{2}=\lambda \pm i \mu$

Nonhomogeneous Equations

$$ay''+by'+cy=g(t)$$ $$\begin{array}{l|l} \hline g(t) & Y(t) \\ \hline P_{n}(t)=a_{0} t^{n}+a_{1} t^{n-1}+\cdots+a_{n} & t^{s}\left(A_{0} t^{n}+A_{1} t^{n-1}+\cdots+A_{n}\right) \\ P_{n}(t) e^{\alpha t} & t^{s}\left(A_{0} t^{n}+A_{1} t^{n-1}+\cdots+A_{n}\right) e^{\alpha t} \\ P_{n}(t) e^{\alpha t}\left\{\begin{array}{l}\sin \beta t \\ \cos \beta t\end{array}\right. & \begin{array}{l}t^{s}[\left(A_{0} t^{n}+A_{1} t^{n-1}+\cdots+A_{n}\right) e^{\alpha t} \cos \beta t\\+\left(B_{0} t^{n}+B_{1} t^{n-1}+\cdots+B_{n}\right) e^{\alpha t} \sin \beta t]\end{array} \end{array}$$ $$y''+p(t)y'+q(t)y=g(x)$$

then a particular solution is

$$Y(t)=-y_{1}(t) \int \frac{y_{2}(t) g(t)}{W\left(y_{1}, y_{2}\right)(t)} d t+y_{2}(t) \int \frac{y_{1}(t) g(t)}{W\left(y_{1}, y_{2}\right)(t)} d t$$

and the general solution is

$$y=c_{1} y_{1}(t)+c_{2} y_{2}(t)+Y(t)$$

Series Solutions near an Ordinary Point

Find a series solution of the equation

$$y''+y=0$$

We look for a solution in the form of a power series about $x_0 = 0$

$$y=\sum_{n=0}^\infty a_nx^n$$

Differentiating term by term yields

$$y''=\sum_{n=2}^\infty n(n-1)a_nx^{n-2}$$

Substituting the series and y’’ gives

$$\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum_{n=0}^\infty a_nx^n=0$$

We obtain

$$\sum_{n=0}^\infty[(n+2)(n+1)a_{n+2}+a_n]x^n=0$$

hence

$$(n+2)(n+1)a_{n+2}+a_n=0$$

by this recurrence relation we get

$$a_n=a_{2k}=\frac{(-1)^k}{(2k)!}a_0,\qquad k=1,2,3...$$

and

$$a_n=a_{2k}=\frac{(-1)^k}{(2k+1)!}a_1,\qquad k=1,2,3...$$

Substituting these coefficients into equation

$$y=a_0+a_1x-\frac{a_0}{2!}x^2-\frac{a_1}{3!}x^3+\frac{a_0}{4!}x^4+\frac{a_1}{5!}x^5+...=a_0\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}x^{2n}+a_1\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{2n+1}$$

Series Solutions near a Regular Singular Point

if $x = 0$ is a regular singular point of the equation

$$P(x)y''+Q(x)y'+R(x)y=0$$ $$p_0=\lim_{x\to0}x\frac{Q(x)}{P(x)},\qquad q_0=\lim_{x\to0}x^2\frac{R(x)}{P(x)}$$

Indicial equation

$$F(r)=r(r-1)+p_0r+q_0$$ $$y_1(x)=|x|^{r_1}[1+\sum_{n=1}^\infty a_n(r_1)x^n]$$

If $r_1\ge r_2$

$$y_2(x)=ay_1(x)\ln|x|+|x|^{r_2}[1+\sum_{n=1}^\infty c_n(r_2)x^n]$$

If $r_1=r_2$

$$y_2(x)=y_1(x)\ln|x|+|x|^{r_1}\sum_{n=1}^\infty b_n(r_1)x^n$$

Homogeneous Linear Systems with Constant Coefficients

$$x'=Ax$$

Set

$$x=\xi e^{rt}$$

Substituting for $x$ in equation

$$r\xi e^{rt}=A\xi e^{rt}$$

we obtain

$$(A-rI)\xi=0$$

Nonhomogeneous Linear Systems

$$x'=P(t)x+g(t)$$

Let $T$ be the matrix whose columns are the eigenvectors $\xi^{(1)},…,\xi^{(n)}$ of $A$, and
define a new dependent variable $y$ by

$$x=Ty$$

Then substituting for $x$ in equation

$$Ty'=ATy+g(t)$$

By multiplying by $T^{-1}$

$$y'=(T^{-1}AT)y+T^{-1}g(t)=Dy+h(t)$$

the equations can be solved separately

$$y_j(t)=e^{r_jt}\int_{t_0}^te^{-r_js}h_j(s)ds+c_je^{r_jt},\qquad j=1,...,n,$$