一阶线性微分方程的初等积分法

变量分离方程与变量变换可分离变量的微分方程齐次方程可化为齐次方程的类型线性方程与常数变易法一阶线性齐次微分方程一阶线性非齐次微分方程伯努利方程黎卡提方程恰当方程与积分因子恰当方程积分因子一阶隐式方程与参数表示

变量分离方程与变量变换

可分离变量的微分方程

dydx=f(x)φ(y)(1.1)\frac{{dy}}{{dx}} = f(x)\varphi(y)\tag{1.1}dxdy​=f(x)φ(y)(1.1)

其中f(x),φ(y)f(x),\varphi (y)f(x),φ(y)分别是x与y的已知连续函数

如果φ(y)≠0{\varphi (y) \ne 0}φ(y)​=0

原式可化为：

dyφ(y)=f(x)dx\frac{{dy}}{{\varphi (y)}} = f(x)dxφ(y)dy​=f(x)dx

两边分别求积分，得：

∫dyφ(y)=∫f(x)dx\int {\frac{{dy}}{{\varphi (y)}} = \int {f(x)dx} } ∫φ(y)dy​=∫f(x)dx

用G(y)G(y)G(y)，F(x)F(x)F(x)分别表示1ϕ(y)\frac{1}{{\phi (y)}}ϕ(y)1​和f(x)f(x)f(x)的某一个原函数

则方程(1.1)的通解为

G(y)=F(x)+CG(y)=F(x)+CG(y)=F(x)+C

如果存在yi{y_i}yi​使得φ(yi)=0{\varphi (y_i) = 0}φ(yi​)=0，那么yi{y_i}yi​就是方程(1.1)的解

齐次方程

dydx=g(yx)(1.2)\frac{{dy}}{{dx}} = g(\frac{y}{x})\tag{1.2} dxdy​=g(xy​)(1.2)

其中g(u)g(u)g(u)为uuu的连续函数

做变量变换

yx=u\frac{y}{x} = uxy​=u

即y=uxy=uxy=ux

将上式带入原方程，得：

xdudx+u=g(u)x\frac{du}{{dx}} + u = g(u)xdxdu​+u=g(u)

整理，得：

dudx=1x⋅(g(u)−u)\frac{{du}}{{dx}} = \frac{1}{x} \cdot (g(u) - u)dxdu​=x1​⋅(g(u)−u)至此，问题转化为(1.1)

可化为齐次方程的类型

dydx=a1x+b1y+c1a2x+b2y+c2(1.3)\frac{{dy}}{{dx}} = \frac{{{a_1}x + {b_1}y + {c_1}}}{{{a_2}x + {b_2}y + {c_2}}} \tag{1.3}dxdy​=a2​x+b2​y+c2​a1​x+b1​y+c1​​(1.3)

其中ai,bi,ci,i=1,2{a_i},{b_i},{c_i},i = 1,2ai​,bi​,ci​,i=1,2均为常数，且c1,c2{c_1},{c_2}c1​,c2​不同时为零.

如果

∣a1b1a2b2∣=0\begin{vmatrix} {{a_1}}&{{b_1}}\\ {{a_2}}&{{b_2}}\end{vmatrix} =0 ∣∣∣∣​a1​a2​​b1​b2​​∣∣∣∣​=0

设 a1=ka2,b1=kb2{a_1} = k{a_2}, {b_1} = k{b_2}a1​=ka2​,b1​=kb2​

则原方程可化为:

dydx=k(a2x+b2y)+c1a2x+b2y+c2=f(a2x+b2y)\frac{{dy}}{{dx}} = \frac{{k({a_2}x + {b_2}y) + {c_1}}}{{{a_2}x + {b_2}y + {c_2}}} = f({\kern 1pt} {a_2}x + {b_2}y)dxdy​=a2​x+b2​y+c2​k(a2​x+b2​y)+c1​​=f(a2​x+b2​y)

令

u=a2x+b2yu = {a_2}x + {b_2}yu=a2​x+b2​y

带入原方程，整理，得：dudx=a2+b2f(u)\frac{{du}}{{dx}} = {a_2} + {b_2}f(u)dxdu​=a2​+b2​f(u)

至此，问题转化为(1.1)

如果

∣a1b1a2b2∣≠0\begin{vmatrix} {{a_1}}&{{b_1}}\\ {{a_2}}&{{b_2}}\end{vmatrix} \ne 0 ∣∣∣∣​a1​a2​​b1​b2​​∣∣∣∣​​=0

那么方程组

{a1x+b1y+c1=0a2x+b2y+c2=0\left\{ \begin{array}{l} {a_1}x + {b_1}y + {c_1} = 0\\ {a_2}x + {b_2}y + {c_2} = 0 \end{array} \right.{a1​x+b1​y+c1​=0a2​x+b2​y+c2​=0​

存在唯一解(α,β)(\alpha ,\beta )(α,β)

令{X=x−αY=y−β\left\{ \begin{array}{l} X = x - \alpha \\ Y = y - \beta \end{array} \right.{X=x−αY=y−β​

则原方程可化为

dYdX=dydx=a1(X+α)+b1(Y+β)+c1a2(X+α)+b2(Y+β)+c2=a1X+b1Y+(a1α+b1β+c1)a2X+b2Y+(a2α+b2β+c2)=a1X+b1Ya2X+b2Y\begin{aligned} \frac{d Y}{d X}=\frac{d y}{d x} &=\frac{a_{1}(X+\alpha)+b_{1}(Y+\beta)+c_{1}}{a_{2}(X+\alpha)+b_{2}(Y+\beta)+c_{2}} \\ &=\frac{a_{1} X+b_{1} Y+\left(a_{1} \alpha+b_{1} \beta+c_{1}\right)}{a_{2} X+b_{2} Y+\left(a_{2} \alpha+b_{2} \beta+c_{2}\right)}\\ &=\frac{a_{1} X+b_{1} Y}{a_{2} X+b_{2} Y} \end{aligned}dXdY​=dxdy​​=a2​(X+α)+b2​(Y+β)+c2​a1​(X+α)+b1​(Y+β)+c1​​=a2​X+b2​Y+(a2​α+b2​β+c2​)a1​X+b1​Y+(a1​α+b1​β+c1​)​=a2​X+b2​Ya1​X+b1​Y​​

至此，问题转化为(1.2)

线性方程与常数变易法

一阶线性齐次微分方程

y′=p(x)y(2.1)y' = p(x)y \tag{2.1}y′=p(x)y(2.1)

该问题与问题(1.1)同理，此处不再赘述

方程的通解为：

y=ce∫p(x)dxy = c{e^{\int {p(x)dx} }}y=ce∫p(x)dx

一阶线性非齐次微分方程

y′=P(x)y+Q(x)(2.2)y' = P(x)y + Q(x)\tag{2.2}y′=P(x)y+Q(x)(2.2)

假设y=c(x)e∫P(x)dxy = c(x){e^{\int {P(x)dx} }}y=c(x)e∫P(x)dx

是方程的解

将上式带入原方程，得：

c′(x)e∫P(x)dx+c(x)e∫P(x)dxP(x)=P(x)c(x)e∫P(x)dx+Q(x)c'(x){e^{\int {P(x)dx} }} + c(x){e^{\int {P(x)dx} }}P(x) = P(x)c(x){e^{\int {P(x)dx} }} + Q(x)c′(x)e∫P(x)dx+c(x)e∫P(x)dxP(x)=P(x)c(x)e∫P(x)dx+Q(x)

即：

c′(x)=Q(x)e−∫P(x)dxc'(x) = Q(x){e^{ - \int {P(x)dx} }}c′(x)=Q(x)e−∫P(x)dx

积分，得

c(x)=∫Q(x)e−∫p(x)dxdx+cc(x) = \int {Q(x){e^{ - \int {p(x)dx} }}dx + c} c(x)=∫Q(x)e−∫p(x)dxdx+c

将c(x)c(x)c(x)代入原式，得方程的通解

y=e∫P(x)dx[∫Q(x)e−∫P(x)dxdx+c]y = {e^{\int {P(x)dx} }}[\int {Q(x){e^{ - \int {P(x)dx} }}dx + c} ]y=e∫P(x)dx[∫Q(x)e−∫P(x)dxdx+c]

另外，我们注意到：

非齐次线性方程的通解等于其对应齐次方程通解与自身的一个特解之和。

伯努利方程

y′=P(x)y+Q(x)yn(2.3)y' = P(x)y + Q(x){y^n} \tag{2.3}y′=P(x)y+Q(x)yn(2.3)

将方程左右两边同乘以y−ny^{- n}y−n，得：

y−ny′=P(x)y1−n+Q(x){y^{ - n}}y' = P(x){y^{1 - n}} + Q(x)y−ny′=P(x)y1−n+Q(x)

令z=y1−nz = {y^{1 - n}}z=y1−n，带入上式，得：

11−ndzdx=P(x)z+Q(x)\frac{1}{{1 - n}}\frac{{dz}}{{dx}} = P(x)z + Q(x)1−n1​dxdz​=P(x)z+Q(x)

整理，得：

dzdx=(1−n)P(x)z+(1−n)Q(x)\frac{{dz}}{{dx}} = (1 - n)P(x)z + (1 - n)Q(x)dxdz​=(1−n)P(x)z+(1−n)Q(x)

至此，问题转化为(2.2)

通解为：

y1−n=e∫(1−n)P(x)dx[∫(1−n)Q(x)e−∫(1−n)P(x)dxdx+c]{y^{1 - n}} = {e^{\int {(1 - n)P(x)dx} }}[\int {(1 - n)Q(x){e^{ - \int {(1 - n)P(x)dx} }}dx + c} ]y1−n=e∫(1−n)P(x)dx[∫(1−n)Q(x)e−∫(1−n)P(x)dxdx+c]

黎卡提方程

dydx=P(x)y2+Q(x)y+R(x)(2.4)\frac{{dy}}{{dx}} = P(x){y^2} + Q(x)y + R(x)\tag{2.4}dxdy​=P(x)y2+Q(x)y+R(x)(2.4)

首先找出方程一特y(x)=y~(x)y(x) = \tilde y(x)y(x)=y~​(x)

则y′=z′+y~′(x)=P(x)(z+y~)2+Q(x)(z+y~)+R(x)=P(x)z2+(2y~P(x)+Q(x))z+P(x)y~2+Q(x)y~+R(x)\begin{aligned} y'=z' + \tilde y'(x) &=P(x){(z + \tilde y)^2} + Q(x)(z + \tilde y) + R(x){\rm{ }}\\ &=P(x){z^2} + (2\tilde yP(x) + Q(x))z + P(x){\tilde y^2} + Q(x)\tilde y + R(x)\end{aligned} y′=z′+y~​′(x)​=P(x)(z+y~​)2+Q(x)(z+y~​)+R(x)=P(x)z2+(2y~​P(x)+Q(x))z+P(x)y~​2+Q(x)y~​+R(x)​

考虑到y~′(x)=P(x)y~2+Q(x)y~+R(x)\tilde y'(x)=P(x){\tilde y^2} + Q(x)\tilde y + R(x)y~​′(x)=P(x)y~​2+Q(x)y~​+R(x)，上式可化为

z′=P(x)z2+(2y~P(x)+Q(x))zz' = P(x){z^2} + (2\tilde yP(x) + Q(x))zz′=P(x)z2+(2y~​P(x)+Q(x))z

至此，问题求解z′z'z′转化为(2.3)

求得z′z'z′后，可自然求得y(x)y(x)y(x)

恰当方程与积分因子

恰当方程

M(x,y)dx+N(x,y)dy=0(3.1)M(x,y)dx + N(x,y)dy = 0\tag{3.1} M(x,y)dx+N(x,y)dy=0(3.1)

其中，∂M(x,y)∂y=∂N(x,y)∂x\frac{{\partial M(x,y)}}{{\partial y}} = \frac{{\partial N(x,y)}}{{\partial x}} ∂y∂M(x,y)​=∂x∂N(x,y)​

解法1：不定积分法

求u(x,y)=∫M(x,y)dx+ϕ(y)u(x,y) = \int {M(x,y)dx + \phi (y)} u(x,y)=∫M(x,y)dx+ϕ(y)

由∂u∂y=N(x,y)\frac{{\partial u}}{{\partial y}} = N(x,y)∂y∂u​=N(x,y)求得ϕ(y)\phi (y)ϕ(y)解法2：分组凑微法

采用“分项组合”的方法,把本身已构成全微分的项分出来,再把余的项凑成全微分.

需要熟记的简单二元函数的全微分：

ydx+xdy=d(xy)ydx + xdy = d(xy)ydx+xdy=d(xy)

ydx−xdyy2=d(xy)\frac{{ydx - xdy}}{{{y^2}}} = d(\frac{x}{y})y2ydx−xdy​=d(yx​)

−ydx+xdyx2=d(yx)\frac{{ - ydx + xdy}}{{{x^2}}} = d(\frac{y}{x})x2−ydx+xdy​=d(xy​)

ydx−xdyxy=d(ln⁡∣xy∣)\frac{{ydx - xdy}}{{xy}} = d(\ln |\frac{x}{y}|)xyydx−xdy​=d(ln∣yx​∣)

ydx−xdyx2+y2=d(arctan⁡xy)\frac{{ydx - xdy}}{{{x^2} + {y^2}}} = d(\arctan \frac{x}{y})x2+y2ydx−xdy​=d(arctanyx​)

ydx−xdyx2−y2=12d(ln⁡∣x−yx+y∣)\frac{{ydx - xdy}}{{{x^2} - {y^2}}}=\frac{1}{2}d(\ln \left| {\frac{{x - y}}{{x + y}}} \right|)x2−y2ydx−xdy​=21​d(ln∣∣∣∣​x+yx−y​∣∣∣∣​)

xdx+ydy=d(x2+y22)xdx + ydy = d\left( {\frac{{{x^2} + {y^2}}}{2}} \right)xdx+ydy=d(2x2+y2​)

xdy+ydxxy=d(ln⁡∣xy∣)\frac{{xdy + ydx}}{{xy}} = d\left( {\ln \left| {xy} \right|} \right)xyxdy+ydx​=d(ln∣xy∣)

xdx+ydyx2+y2=d(12ln⁡(x2+y2))\frac{{xdx + ydy}}{{{x^2} + {y^2}}} = d\left( {\frac{1}{2}\ln ({x^2} + {y^2})} \right)x2+y2xdx+ydy​=d(21​ln(x2+y2))解法3：线积分法

通解为：

∫x0xM(x,y0)dx+∫y0yN(x,y)dy=c\begin{aligned} \int_{{x_0}}^x {M(x,{y_0})dx}+ \int_{{y_0}}^y {N(x,y)dy} =c \end{aligned} ∫x0​x​M(x,y0​)dx+∫y0​y​N(x,y)dy=c​

积分因子

定义

M(x,y)dx+N(x,y)dy=0M(x,y)dx+N(x,y)dy=0 M(x,y)dx+N(x,y)dy=0

如果存在连续可微函数μ(x,yv)≤0\mu(x, yv)\leq 0μ(x,yv)≤0,使得

μ(x,y)M(x,y)dx+μ(x,y)N(x,y)dy=0\mu (x, y)M(x, y)dx+ \mu (x, y)N(x, y)dy= 0 μ(x,y)M(x,y)dx+μ(x,y)N(x,y)dy=0

为恰当方程，则μ(x,y\mu(x, yμ(x,y)是方程(1)的一个积分因子

定理

微分方程(3.2)有一个仅依赖于x的积分因子的充要条件是

ψ(x)=(∂M∂y−∂N∂x)N\psi(x)=\frac{(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}) }{N} ψ(x)=N(∂y∂M​−∂x∂N​)​

仅与xxx有关，这时(3.2)的积分因子为

μ(x)=e∫ψ(x)dx\mu(x)=e^{\int\psi(x)dx} μ(x)=e∫ψ(x)dx

同理

微分方程(3.2)有一个仅依赖于x的积分因子的充要条件是

ψ(x)=(∂M∂y−∂N∂x)−M\psi(x)=\frac{(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}) }{-M} ψ(x)=−M(∂y∂M​−∂x∂N​)​

仅与xxx有关，这时(3.2)的积分因子为

μ(y)=e∫ψ(y)dy\mu(y)=e^{\int\psi(y)dy} μ(y)=e∫ψ(y)dy

一阶隐式方程与参数表示