微积分A(1)期末复习笔记(仅包括下半学期相关内容)，仅供参考。 PDF版本链接

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定理

黎曼积分

$\int_a^b f(x) \;\text{d}x = \lim_{\lambda(P) \to 0} \sigma(f;P,\xi) = \lim_{\lambda(P) \to 0} \sum_{i=1}^n f(\xi_i) \Delta x_i$

$f$可积:

IFF $\forall \varepsilon \gt 0, U(f;P)-L(f;P) \lt \varepsilon$.

IFF $\underline{\int}_a^b f(x) \;\text{d}x = \overline{\int}_a^b f(x) \;\text{d}x$.

IFF $\lim_{\lambda(P) \to 0} \sum_{i=1}^n \omega(f;\Delta_i) \Delta x_i = 0$.

一致连续

$\forall \varepsilon \gt 0, \exists \delta \gt 0, \forall x,y \in X, \vert x-y \vert \lt \delta$, 都有$\vert f(x) - f(y) \vert \lt \varepsilon$, 则$f$在$X$上一致连续.

$f$一致连续:

IFF $\forall \lbrace x_n \rbrace, \lbrace y_n \rbrace, \lim_{n \to \infty} (x_n-y_n) = 0$, 都有$\lim_{n \to \infty} (f(x_n)-f(y_n)) = 0$.

$L$-Lipschitz函数一致连续.

闭区间上连续函数一致连续.

闭区间上连续函数可积.

闭区间上单调函数可积.

有界函数可积当且仅当的间断点集零测度.

积分中值定理

积分第一中值定理: $f \in \mathscr C[a,b], \exists \xi \in [a,b], \int_a^b f(x) \;\text{d}x = f(\xi)(b-a)$.

加强积分第一中值定理: $f \in \mathscr R[a,b], f \in \mathscr C(a,b), \exists \xi \in (a,b), \int_a^b f(x) \;\text{d}x = f(\xi)(b-a)$.

广义积分第一中值定理: $f \in \mathscr C[a,b], g \in \mathscr R[a,b], \exists \xi \in [a,b], \int_a^b f(x)g(x) \;\text{d}x = f(\xi)\int_a^b g(x) \;\text{d}x$.

积分第二中值定理: $f \in \mathscr R[a,b]$, $g$在$[a,b]$上单调, $\exists \xi \in [a,b], \int_a^b f(x)g(x) \;\text{d}x = g(a) \int_a^\xi f(x) \;\text{d}x + g(b)\int_\xi^b f(x) \;\text{d}x$.

广义积分收敛性判断

Cauchy判别准则: $\int_a^\omega f(x) \;\text{d}x$收敛:

IFF $\forall \varepsilon \gt 0, \exists c \in (a,\omega), \forall A_1,A_2 \in (c,\omega), \vert \int_{A_1}^{A_2} f(x) \;\text{d}x \vert \lt \varepsilon$.

Abel判别准则: $\int_a^\omega f(x) \;\text{d}x$收敛且$g$单调有界则$\int_a^\omega f(x)g(x) \;\text{d}x$收敛.

Dirichlet判别准则: $F(A) = \int_a^A f(x) \;\text{d}x$有界且$g$单调趋于$0$, 则$\int_a^\omega f(x)g(x) \;\text{d}x$收敛.

比较判敛法.

测试函数.

公式

高阶导数公式

1. $(x^\alpha)^{(n)} = \alpha^{\underline{n}}x^{\alpha-n}$.

2. $(e^{\alpha x})^{(n)} = \alpha^n e^{\alpha x}$ ($\alpha \in \mathbb C$).

3. $(\ln(1+x))^{(n)} = \frac{(-1)^{n-1}(n-1)!}{(1+x)^n}$.

4. $\sin^{(n)}(x) = \sin(x+\frac{n\pi}{2})$, $\cos^{(n)}(x) = \cos(x+\frac{n\pi}{2})$.

$1 \Rightarrow 3$:

$(\ln(1+x))' = \frac{1}{1+x} = (1+x)^{-1} \\ [(\ln(1+x))']^{(n-1)} = -1^{\underline{n-1}}(1+x)^{-1-(n-1)} = \frac{(-1)^{n-1}(n-1)!}{(1+x)^n} \\$

$2 \Rightarrow 4$:

$\cos^{(n)}(x) + i\sin^{(n)}(x) = (\cos x + i\sin x)^{(n)} = (e^{ix})^{(n)} = \\ i^ne^{ix} = (e^\frac{i\pi}{2})^ne^{ix} = e^{i(x+\frac{n\pi}{2})} = \cos(x+\frac{n\pi}{2}) + i\sin(x+\frac{n\pi}{2}) \\$

$(\lambda f + \mu g)^{(n)} = \lambda f^{(n)} + \mu g^{(n)} \\ (fg)^{(n)} = \sum_{k=0}^n {n \choose k} f^{(k)}g^{(n-k)} \\$

常用泰勒展开

$\begin{aligned} e^x &= 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + ... \\ \sin x &= x - \frac{x^3}{6} + \frac{x^5}{120} - ... \\ \cos x &= 1 - \frac{x^2}{2} + \frac{x^4}{24} - ... \\ \ln(1+x) &= x - \frac{x}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ... \\ \ln(\frac{1}{1-x}) &= x + \frac{x}{2} + \frac{x^3}{3} + \frac{x^4}{4} + ... \\ \frac{1}{1-x} &= 1 + x + x^2 + x^3 + x^4 + ... \\ \end{aligned}$

变限积分求导

$\left( \int_{\psi(x)}^{\varphi(x)} f(t) \;\text{d}t \right)' = f(\varphi(x)) \varphi'(x) - f(\psi(x))\psi'(x)$

$\left( \int_{\psi(x)}^{\varphi(x)} f(x,t) \;\text{d}t \right)' = f(x,\varphi(x))\varphi'(x) - f(x,\psi(x))\psi'(x) + \int_{\psi(x)}^{\varphi(x)} \frac{\text{d}f(x,t)}{\text{d}x} \;\text{d}t$

定积分与数列极限

$\lim_{n \to \infty}\frac{b-a}{n}\sum_{k=1}^n f(\xi_{n,k}) = \int_a^b f(x) \;\text{d}x$

不等式

均值不等式:

$\frac{n}{\sum_{i=1}^n\frac{1}{x_i}} \le \sqrt[n]{\prod_{i=1}^nx_i} \le \frac{1}{n} \sum_{i=1}^n x_i \le \sqrt{\frac{\sum_{i=1}^n x_i^2}{n}}$

Young不等式($\frac{1}{p}+\frac{1}{q} = 1$):

$x^\frac{1}{p}y^{\frac{1}{q}} \le \frac{1}{p}x + \frac{1}{q}y$

Holder不等式($\frac{1}{p}+\frac{1}{q} = 1$):

$\sum_{k=1}^n x_ky_k \le \left( \sum_{k=1}^nx_k^p \right)^{\frac{1}{p}} \left( \sum_{k=1}^ny_k^q \right)^{\frac{1}{q}}$

积分Cauchy不等式:

$\left( \int_a^b f(x)g(x) \;\text{d}x \right)^2 \le \left( \int_a^b f^2(x) \;\text{d}x \right) \left( \int_a^b g^2(x) \;\text{d}x \right)$

积分Holder不等式($\frac{1}{p}+\frac{1}{q} = 1$):

$\left\vert \int_a^b f(x)g(x) \;\text{d}x \right\vert \le \left( \int_a^b \vert f(x) \vert^p \;\text{d}x \right)^\frac{1}{p} \left( \int_a^b \vert f(x) \vert^q \;\text{d}x \right)^\frac{1}{q}$

积分Jensen不等式($\varphi$为凸函数):

$\varphi \left( \frac{1}{b-a} \int_a^b f(x) \;\text{d}x \right) \le \frac{1}{b-a} \int_a^b \varphi(f(x)) \;\text{d}x$

积分Minkowski不等式($p \gt 1$):

$\left( \int_a^b (\vert f(x) \vert + \vert g(x) \vert)^p \;\text{d}x \right)^{\frac{1}{p}} \le \left( \int_a^b \vert f(x) \vert^p \;\text{d}x \right)^{\frac{1}{p}} + \left( \int_a^b \vert g(x) \vert^p \;\text{d}x \right)^{\frac{1}{p}}$

几何图形计算

直角坐标系下面积(矩形逼近):

$S = \int_a^b \vert f(x) \vert \;\text{d}x$

极坐标系下面积(扇形逼近):

$S = \frac{1}{2} \int_\alpha^\beta \rho^2(\theta) \;\text{d}\theta$

参数方程面积:

若$x(t)$单调, 反函数$t(x)$存在, $a = x(\alpha)$, $b = x(\beta)$, 则:

$S = \int_a^b \vert y(t(x)) \vert \;\text{d}x \stackrel{x=x(t)}{=} \int_\alpha^\beta \vert y(t)x'(t) \vert \;\text{d}t$

直角坐标系弧长:

$L = \int_a^b \sqrt{1+(f'(x))^2} \;\text{d}t$

极坐标系弧长:

$L = \int_\alpha^\beta \sqrt{\rho^2(\theta) + \rho'^2(\theta)} \;\text{d}\theta$

参数方程曲线弧长:

$L = \int_a^b \sqrt{(x'(t))^2+(y'(t))^2} \;\text{d}t$

曲线的曲率: $\kappa = \frac{\text{d}\alpha}{\text{d}L}$. 曲线的曲率半径$R = \frac{1}{\kappa}$

参数方程曲线曲率:

$\kappa = \left \vert \frac{\text{d}\alpha}{\text{d}L} \right \vert = \left \vert \frac{\text{d}\arctan\frac{y'(t)}{x'(t)}}{\sqrt{x'^2(t)+y'^2(t)}} \right \vert = \frac{\vert x'(t)y''(t) - x''(t)y'(t) \vert}{(x'^2(t)+y'^2(t))^\frac{3}{2}}$

直角坐标系曲线曲率:

$\kappa = \frac{\vert y''(x) \vert}{(1+y'^2(x))^\frac{3}{2}}$

极坐标系曲线曲率:

$\kappa = \frac{\vert \rho^2(\theta)+2\rho'^2(\theta)-\rho(\theta)\rho''(\theta) \vert}{(\rho^2(\theta)+\rho'^2(\theta))^{\frac{3}{2}}}$

$y = f(x)$绕$x$轴旋转体体积:

$V = \pi \int_a^b f^2(x) \;\text{d}x$

$y = f(x)$绕$y$轴旋转体体积:

$\frac{\text{d}V}{\text{d}x}(x) = \lim_{\Delta x \to 0}\frac{(\pi(x+\Delta x)^2-\pi x^2)\vert y(x) \vert}{\Delta x} = 2\pi x \vert y(x) \vert = 2\pi x f(x) \\ V = 2\pi \int_a^b \vert xf(x) \vert \;\text{d}x$

极坐标曲线与原点连线所成图形绕极轴旋转体积(一般情况下适用):

$V = \frac{2\pi}{3}\int_\alpha^\beta r^3(\theta) \sin \theta \;\text{d}\theta$

$y = f(x)$绕$x$轴旋转体侧面积微元: $\text{d}S = 2\pi \vert y \vert\;\text{d} L$.

参数方程旋转体侧面积($x$轴):

$S = 2 \pi \int_a^b \vert y(t) \vert \sqrt{x'^2(y)+y'^2(t)} \;\text{d}t$

参数方程旋转体侧面积($y$轴):

$S = 2 \pi \int_a^b \vert x(t) \vert \sqrt{x'^2(y)+y'^2(t)} \;\text{d}t$

直角坐标系旋转体侧面积($x$轴):

$S = 2 \pi \int_a^b \vert y(x) \vert \sqrt{1+y'^2(x)} \;\text{d}x$

直角坐标系旋转体侧面积($y$轴):

$S = 2 \pi \int_a^b \vert x \vert \sqrt{1+y'^2(x)} \;\text{d}x$

极坐标系旋转体侧面积($x$轴):

$S = 2 \pi \int_\alpha^\beta \vert \rho(\theta)\sin\theta \vert \sqrt{\rho^2(\theta)+\rho'^2(\theta)} \;\text{d}\theta$

极坐标系旋转体侧面积($y$轴):

$S = 2 \pi \int_\alpha^\beta \vert \rho(\theta)\cos\theta \vert \sqrt{\rho^2(\theta)+\rho'^2(\theta)} \;\text{d}\theta$

曲线质心: 力矩平衡. 设线密度$\mu$为常数. 则一小段线段的质量为$\text{d} m = \mu \text{d} L$. 线段总质量为$m = \int \mu \;\text{d}L$.

沿$y$轴方向的静力矩为: $M_y = \int x\mu \;\text{d}L$. 类似地, $M_x = \int y\mu \;\text{d}L$.

设质心坐标为$(\bar x,\bar y)$, 则:

$\bar x = \frac{M_y}{m} = \frac{\int x \mu \;\text{d}L}{\int \mu \;\text{d}L} = \frac{\int x \;\text{d}L}{L}$

类似地,

$\bar y = \frac{\int y \;\text{d}L}{L}$

若线密度不为常数，将$\mu$替换为$\mu(x)$即可，此时分子分母积分不再能消去$\mu(x)$.

ODE

直接积分型ODE:

$\frac{\text{d}y}{\text{d}x} = f(x)$

则直接积分. 通解为:

$y = \int f(x) \;\text{d}x$

一阶线性齐次ODE

$\frac{\text{d}y}{\text{d}x} + P(x)y = 0 \\$

求导后能和自身抵消, 猜测函数为$Ce^{f(x)}$形式, 解出. 通解为:

$y = C e^{-\int P(x) \;\text{d}x} \\$

注意这里的积分符号表示任意一个原函数, 因此积分结果本身不需要$+C$.

一阶线性非齐次ODE:

$\frac{\text{d}y}{\text{d}x} + P(x)y = Q(x) \\$

常数变易法, 猜测函数为$y = C(x)e^{f(x)}$, 解出. 通解为:

$y = e^{-\int P(x) \;\text{d}x} (C + \int Q(x)e^{\int P(x) \;\text{d}x} \;\text{d}x) \\$

分离变量型一阶ODE:

$\frac{\text{d}y}{\text{d}x} = f(x)g(y) \\$

移项: $\frac{\text{d}y}{g(y)} = f(x) \;\text{d}x$, 因此可以直接积分. 通解为(隐函数):

$\int \frac{1}{g(y)} \;\text{d}y = \int f(x) \;\text{d}x + C \\$

若存在$g(y_0) = 0$, 则$y_0$为该ODE的一个奇解.

ax+by+c型ODE

$\frac{\text{d}y}{\text{d}x} = f(ax+by+c) \\$

若$b = 0$, 则为直接积分型ODE. 通解为:

$y = \int f(ax+c) \;\text{d}x + C \\$

令$u = ax+by+c$, 则$\frac{\text{d}u}{\text{d}x} = a+b \frac{\text{d}y}{\text{d}x} = a+bf(u)$, 为分离变量型ODE. 通解为(隐函数):

$\int \frac{1}{a+bf(u)} \;\text{d}u = x + C \\$

若存在$a+bf(u_0) = 0$, 则$y = \frac{1}{b}(u_0-ax-c)$为该ODE的一个奇解.

y/x型ODE

$\frac{\text{d}y}{\text{d}x} = F\left(\frac{y}{x}\right)$

令$u = \frac{y}{x}$, 则$\frac{\text{d}y}{\text{d}x} = \frac{\text{d}ux}{\text{d}x} = x\frac{\text{d}u}{\text{d}x} + u = F(u)$, 为分离变量型ODE: $x\frac{\text{d}u}{\text{d}x} = F(u)-u$. 通解为(隐函数):

$\int \frac{1}{F(u)-u} \;\text{d}u = \ln \vert x \vert + C \\$

若存在$F(u_0)-u_0 = 0$, 则$y = u_0x$为该ODE的一个奇解.

x/y型ODE

$\frac{\text{d}y}{\text{d}x} = F\left(\frac{x}{y}\right)$

令$u = \frac{x}{y}$, 则$\frac{\text{d}y}{\text{d}x} = \frac{1}{\frac{\text{d}uy}{\text{d}y}} = \frac{1}{y\frac{\text{d}u}{\text{d}y} + u} = F(u)$, 为分离变量型ODE: $y\frac{\text{d}u}{\text{d}y} = \frac{1}{F(u)}-u$. 通解为(隐函数):

$\int \frac{1}{\frac{1}{F(u)}-u} \;\text{d}u = \ln \vert y \vert + C \\$

若存在$\frac{1}{F(u_0)}-u_0 = 0$, 则$y = \frac{1}{u_0}x$为该ODE的一个奇解.

直线交点型ODE

$\frac{\text{d}y}{\text{d}x} = f\left(\frac{a_1x+b_1y+c_1}{a_2x+b_2y+c_2}\right)$

若直线$a_1x+b_1y+c_1=0$与$a_2x+b_2y+c_2=0$有唯一交点$(x_0,y_0)$, 则$X = x-x_0, Y = y-y_0$:

$\frac{\text{d}Y}{\text{d}X} = f\left(\frac{a_1X+b_1Y}{a_2X+b_2Y}\right) = f\left(\frac{a_1+b_1\frac{Y}{X}}{a_2+b_2\frac{Y}{X}}\right) = F(\frac{Y}{X})$

否则两直线平行, $a_1b_2 = a_2b_1$. 则:

$f\left(\frac{a_{1} x+b_{1} y+c_{1}}{a_{2} x+b_{2} y+c_{2}}\right) = f\left(k+\frac{c_{1}-k c_{2}}{a_{2} x+b_{2} y+c_{2}}\right) = F\left(a_{2} x+b_{2} y+c_{2}\right)$

伯努利方程

$\frac{\text{d}y}{\text{d}x} + p(x)y = q(x)y^\alpha$

若$\alpha = 0$则为一阶线性非齐次ODE, 若$\alpha = 1$则为分离变量型一阶ODE, 否则令$z = y^{1-\alpha}$, 则$\frac{\text{d}z}{\text{d}x} = (1-\alpha)y^{-\alpha}\frac{\text{d}y}{\text{d}x}$, 为一阶线性非齐次ODE: $\frac{\text{d}z}{\text{d}x} + (1-\alpha)p(x)z = (1-\alpha)q(x)$. 通解为:

$y = \left(e^{-(1-\alpha) \int p(x) \;\text{d}x} (C + (1-\alpha)\int q(x)e^{(1-\alpha) \int p(x) \;\text{d}x} \;\text{d}x)\right)^{\frac{1}{1-\alpha}} \\$

另外, 若$\alpha \gt 0$, $y = 0$为该ODE的奇解.

高阶ODE与基本解组

$n$阶ODE: $y^{(n)} + \sum_{i=0}^{n-1} a_i(x)y^{(i)} = f(x)$. 对于任意一个柯西初值问题, 区间$I$上存在唯一解.

高阶齐次ODE解集为$n$维线性空间. 可以找到$n$个线性无关的函数作为基底(基本解组).

Wronsky行列式:

$W(f_1,f_2,\cdots,f_n)(x) = \\ \det \begin{pmatrix} f_1(x) & f_2(x) & f_3(x) & \cdots & f_n(x) \\ f_1'(x) & f_2'(x) & f_3'(x) & \cdots & f_n'(x) \\ f_1''(x) & f_2''(x) & f_3''(x) & \cdots & f_n''(x) \\ & \vdots & & \ddots & \vdots \\ f_1^{(n-1)}(x) & f_2^{(n-1)}(x) & f_3^{(n-1)}(x) & \cdots & f_n^{(n-1)}(x) \\ \end{pmatrix}$

THEOREM

$f$线性相关 $\Leftrightarrow$ 在区间$I$上$W(f_1,f_2,\cdots,f_n)(x) \equiv 0$. 且$f$线性无关 $\Leftrightarrow$ 在区间$I$上$W(f_1,f_2,\cdots,f_n)(x)$恒不为$0$.

给定一个基本解组, 构造以这个基本解组为解的ODE:

$\det \left( \begin{matrix} f_1(x) & f_2(x) & f_3(x) & \cdots & f_n(x) \\ f_1'(x) & f_2'(x) & f_3'(x) & \cdots & f_n'(x) \\ f_1''(x) & f_2''(x) & f_3''(x) & \cdots & f_n''(x) \\ & \vdots & & \ddots & \vdots \\ f_1^{(n-1)}(x) & f_2^{(n-1)}(x) & f_3^{(n-1)}(x) & \cdots & f_n^{(n-1)}(x) \\ f_1^{(n)}(x) & f_2^{(n)}(x) & f_3^{(n)}(x) & \cdots & f_n^{(n)}(x) \\ \end{matrix} \right\vert \left. \begin{matrix} y \\ y' \\ y'' \\ \vdots \\ y^{(n-1)} \\ y^{(n)} \\ \end{matrix} \right) = 0$

高阶积分型ODE

$y^{(n)} = f(x)$

求$n$次原函数.

降阶型ODE

$y^{(n)} = F(x,y^{(k)}, \cdots, y^{(n-1)})$

其中$k \ge 1$, 则令$p(x) = y^{(k)}(x)$, 阶数降低, 解出$p(x)$后为高阶积分型ODE, $y(x)$为$p(x)$求$k$次原函数.

不显含x型二阶ODE

$F(y,\frac{\text{d}y}{\text{d}x},\frac{\text{d}^2y}{\text{d}x^2}) = 0$

令$p = \frac{\text{d}y}{\text{d}x}$, 则$\frac{\text{d}^2y}{\text{d}x^2} = p\frac{\text{d}p}{\text{d}y}$, 转换为$F(y,p,p\frac{\text{d}p}{\text{d}y}) = 0$, 为分离变量型ODE: $\frac{\text{d}y}{\text{d}x} = p(y)$.

二阶线性齐次ODE

$y'' + py' + qy = 0$

求导后能和自身抵消, 猜测函数为$e^{\lambda x}$形式, 解出$(\lambda^2 + p\lambda +q)e^{\lambda x} = 0$. 判别式$\Delta = p^2-4q$

若$\Delta \gt 0$, 通解为: $y = C_1e^{\lambda_1 x} + C_2 e^{\lambda_2 x}$.

若$\Delta = 0$, 通解为: $y = (C_1+C_2x)e^\lambda = (C_1+C_2x)e^{-\frac{p}{2}x}$.

若$\Delta \lt 0$, $\lambda_1,\lambda_2$可能为复数$\alpha \pm \beta i$. 对于一个复值函数解$y(x)$, $u(x) = \text{Re}(x)$和$v(x) = \text{Im(x)}$分别满足:

$u'' + pu' + qu = \text{Re} 0 = 0 \\ v'' + pv' + qv = \text{Im} 0 = 0 \\$

故复值函数解$y = Ce^{(\alpha + \beta i)x}$可以构造两个实值函数解:

$y_1 = e^{\alpha x} \cos (\beta x) y_2 = e^{\alpha x} \sin (\beta x)$

可以检验两个解线性无关, 则实值通解为:

$y = e^{\alpha x}(C_1 \cos (\beta x) + C_2 \sin (\beta x))$

事实上, 直接考虑复值通解:

$y = C_1e^{(\alpha + \beta i)x} + C_2e^{(\alpha - \beta i)x}$

当$C_1 = \overline{C_2}$时即可得实值通解.

高阶线性齐次ODE

$y^{(n)} + \sum_{i=0}^{n-1} a_iy^{(i)} = f(x)$

类似于二阶求其特征多项式$P(\lambda) = \lambda^n + \sum_{i=0}^{n-1} a_i\lambda^i$. 设其有$\lambda_1,\cdots,\lambda_k$共$k$个特征根, 其中$\lambda_i$重数为$m_i$, 则其复值通解为:

$y(x) = \sum_{j=1}^k \sum_{t=0}^{m_i-1} C_{j,t} \cdot x^t e^{\lambda_j x}$

特殊二阶线性非齐次ODE

$y'' + py' + qy = P_n(x)e^{\mu x}$

考虑$\mu$的重数$m$, 假设一个特解$z_0 = Q_n(x)x^me^{\mu x}$. 解出$Q_n(x)$后用特解加上其次ODE通解得到非齐次ODE通解.

因此, 任意$y'' + py' + qy = P_n(x)f(x)$, 其中$f(x) = e^{ax}$或$e^{ax}\cos(bx) = \text{Re}(e^{(a+bi)x})$或$e^{ax}\sin(bx) = \text{Im}(e^{(a+bi)x})$都可以解出.

一般二阶线性非齐次ODE

$y'' + py' + qy = f(x)$

首先求出二阶线性齐次ODE的两个线性无关解$y_1,y_2$, 那么一个特解为:

$z_0(x) = \int_{x_0}^x \frac{y_1(t)y_2(x) - y_1(x)y_2(t)}{W(y_1,y_2)(t)}f(t) \; \text{d}t$

此公式一般很难积分计算, 因此通常计算上一部分的特殊二阶线性非齐次ODE.

欧拉方程

$x^ny^{(n)} + \sum_{i=0}^{n-1} a_ix^iy^{(i)} = 0$

令$t = \ln \vert x \vert$, 则:

$\begin{aligned} \frac{\text{d}y}{\text{d}x} &= \frac{1}{x} \cdot \frac{\text{d}y}{\text{d}t} \\ \frac{\text{d}^2y}{\text{d}x^2} &= \frac{\text{d}}{\text{d}t}\left(\frac{\text{d}y}{\text{d}x}\right) \cdot \frac{1}{\frac{\text{d}x}{\text{d}t}} \\ &= \left(-\frac{1}{x^2} \cdot x \cdot \frac{\text{d}y}{\text{d}t} + \frac{1}{x} \cdot \frac{\text{d}^2y}{\text{d}t^2}\right) \cdot \frac{1}{x} \\ &= \frac{1}{x^2} \left( \frac{\text{d}^2y}{\text{d}t^2} - \frac{\text{d}y}{\text{d}t} \right) \\ \frac{\text{d}^3y}{\text{d}x^3} &= \frac{\text{d}}{\text{d}t}\left(\frac{\text{d}^2y}{\text{d}x^2}\right) \cdot \frac{1}{\frac{\text{d}x}{\text{d}t}} \\ &= \left( -\frac{2}{x^3} \cdot x \cdot \left( \frac{\text{d}^2y}{\text{d}t^2} - \frac{\text{d}y}{\text{d}t} \right) + \frac{1}{x^2} \cdot \left( \frac{\text{d}^3y}{\text{d}t^3} - \frac{\text{d}^2y}{\text{d}t^2} \right) \right) \cdot \frac{1}{x} \\ &= \frac{1}{x^3}\left( \frac{\text{d}^3y}{\text{d}t^3} - 3 \frac{\text{d}^2y}{\text{d}t^2}+ 2 \frac{\text{d}y}{\text{d}t} \right) \\ \end{aligned}$

一阶线性ODEg

$\begin{matrix} \frac{\text{d}y_i}{\text{d}x} = f_i(x) + \sum_{j=1}^n a_{ij}(x)y_j(x) & (i \in \lbrace 1,2, \cdots, n \rbrace) \\ y_j(x_0) = \xi_j & (j \in \lbrace 1,2, \cdots, n \rbrace) \\ \end{matrix}$

用向量形式可以表示为:

$\frac{\text{d}\mathbf Y}{\text{d} x} = \mathbf A(x) \mathbf Y + \mathbf F(x)$

如果该齐次方程存在$n$个线性无关的解, 奇解矩阵为$\mathbf \Phi = (\mathbf Y_1, \mathbf Y_2, \mathbf Y_3, \cdots, \mathbf Y_n)$, 则通解为$\mathbf{\Phi C}$, 其中$\mathbf C = (C_1,C_2,C_3, \cdots C_n)^\top$. 一个特解为$\mathbf Z(x) =\mathbf \Phi(x) \int_{x_0}^x (\mathbf \Phi (t))^{-1} \mathbf F(t) \;\text{d}t$. 因此, 原非齐次方程的通解为:

$\mathbf Y(x) = \mathbf \Phi(x) \mathbf C + \mathbf \Phi(x) \int_{x_0}^x (\mathbf \Phi (t))^{-1} \mathbf F(t) \;\text{d}t$

如何求$\mathbf \Phi$? 类比一阶线性齐次ODE, 考虑形如$\mathbf Y = e^{\lambda x}\mathbf r$的形式: 若$\mathbf A$有$k$个不同的特征值, 特征值$\lambda_i$重数为$m_i$, 则对于每个$\lambda_i$存在$m_i$个线性无关的解$e^{\lambda_ix}\mathbf P_i(x)$, 其中$\mathbf P_i$为系数为向量的多项式.

向量方程的Wronsky行列式$W(x) = \det(\mathbf \Phi(x))$. $n$个解线性相关当且仅当$W(x) \equiv 0$. 进一步地, 可以观察到(对行列式求到即对每一行分别求导后求行列式相加):

$W'(x) = (\det(\mathbf \Phi)) = -a_{n-1}(x)\det(\mathbf \Phi) = -a_{n-1}W(x)$

故$W(x) = W(x_0)e^{-\int_{x_0}^xa_{n-1}(t)\;\text{d}t}$.

事实上, 如果令$y_i = y^{(i)}$, 一个$n$阶线性ODE可以表示为一个一阶$n$元线性ODEg:

$\begin{matrix} \frac{\text{d}y_i}{\text{d}x} = y_{i+1} & (i \in \lbrace 0,2, \cdots, n-2 \rbrace) \\ \frac{\text{d}y_{n-1}}{\text{d}x} = f(x) - \sum_i a_i(x)y_i \\ y_j(x_0) = \xi_j & (j \in \lbrace 0,2, \cdots, n-1 \rbrace) \\ \end{matrix}$

求解一阶线性ODEg的常用方法事实上是转为ODE.

技巧

$\ln$求导法

$(\ln \vert f(x) \vert)' = \frac{f'(x)}{f(x)}\;\;\;\;(f(x) \ne 0) \\$

$f'(x) = f(x) \cdot (\ln \vert f(x) \vert)' \\$

分部积分

1. 对数函数: $\ln x$, …

2. 反三角函数: $\text{arcsin} x$, $\text{arctan} x$, …

3. 幂函数: $x^n$, $P(x)$, …

4. 三角函数: $\sin x$, $\cos x$, …

5. 指数函数: $e^x$, …

上述函数结合时, 排名靠前的适合求导, 留在原地准备分部积分后求导; 排名靠后的适合积分, 积分放入$\text{d}$后.

换元

$\sqrt{ax^2+bx+c} = \pm \sqrt{a}x + t$

$\sqrt{ax^2+bx+c} = tx \pm \sqrt{c}$

$\sqrt{a(x-b)(x-c)} = t(x-b)$

$\sqrt[n]{\frac{ax+b}{cx+d}} = t$

被积函数关于$\sin x$奇函数: $\cos x = t$.

被积函数关于$\cos x$奇函数: $\sin x = t$.

被积函数关于$\sin x$, $\cos x$都为偶函数: $\tan x = t$.

$\sqrt{x^2+a^2}$: $x = a \text{sh} t$或$x = a \tan t$.

$\sqrt{x^2-a^2}$: $x = a \text{ch} t$或$x = \pm a \sec t$.

$\sqrt{a^2-x^2}$: $x = a \sin t$.

三角有理函数积分

$\int \frac{a \sin x + b \cos x}{a \sin x + b \cos x} \;\text{d}x = \int \text{d}x = x + C \\$

$\int \frac{-b \sin x + a \cos x}{a \sin x + b \cos x} \;\text{d}x = \ln \vert a \sin x + b \cos x \vert + C \\$

而$a\sin x + b \cos x$与$-b \sin x + a \cos x$线性无关, 因此可以求出所有形如下方的积分:

$\int \frac{c \sin x + d \cos x}{a \sin x + b \cos x} \;\text{d}x$

鸣谢

感谢何昊天学长的微积分复习讲座(手动@微信公众号:乐学)。

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