数学常微分方程第二章

1、任课：林上海交通大学博士复旦大学博士后副教授导师林：办公室：数学系404：65904529办公室电子课件常微分方程Ordinary differentialequation王高雄周之铭王（第三版）第二章一阶微分方程的初等法Integrated Method of First Order ODECh.2 Integrated Method of First Order ODE方程类型/Classifications/：y¢ = f (x, y)F (x, y, y¢) = 0初等法/Integrated Method/：通过求解常微分方程的法，其特点是微分方

2、程的用初等函数以及初等函数的形式来表示。本章内容/Main Contents/Ø §2.1Ø §2.2Ø §2.3Ø §2.4变量分离方程与变量变换线性方程与常数变易法恰当方程与因子一阶隐式方程与参数表示Ch.2 Integrated Method of First Order ODE本章要求/Requirements/熟练掌握一些重要的常见的一阶方程的类型及其求解方法。 注意：正确方程的类型本章目录/Main Contents/Ø §2.1 变量分离方程与变量变换Ø §2.

3、2 线性方程与常数变易法 Ø §2.3 恰当方程与因子 Ø §2.4 一阶隐式方程与参数表示 § 2.2线性方程与常数变易法/Linear ODE and variation of constants Method/内容提要/Constant Abstract/ìì齐次线性方程: 特点× 解法× 举例ïïì常数变易法(ï线性方程í因子方法）ï非齐次线性方程íïïïî求解步骤× 举例

4、15;随堂练习方程ì特点îì线性方程与常数变易法íïï可化为线性方程的方程í提方程íï其他可化为线性方程的方程î解法ïïîï重点与难点ïïî思考本节要求/Requirements/ØØ熟练掌握线性方程和方程的求解方法。了解提方程的简单性质及其求解方法。§ 2.2 Linear ODE and variation of constants Method 、一阶线性微分方程/ First-Order

5、Linear ODE/a(x) dy + b(x) y + c(x) = 0一般形式dxy¢ = P(x) y + Q(x)形如(2.2.1)的方程称为一阶线性微分方程(即关于 y, y¢是线性的)其中 P(x),Q(x) 为 x 的已知函数。当Q(x) º 0时，y¢ = P(x) y(2.2.2)称为齐次线性方程;Q(x) ¹ 0当时，称为非齐次线性方程。§ 2.2 Linear ODE and variation of constants Method假设 P(x), Q(x) 函数在区间a<x<b上连续,则根据解的

6、存在性及唯一性定理可知，在区域D : a < x < b- ¥ < y < +¥方程(2.2.1)的初值问题的存在唯一的。y¢ = P(x) y + Q(x)(2.2.1)§ 2.2 Linear ODE and variation of constants Method（1）齐次线性方程/Homogenous Linear ODE/y¢ = p(x) y.(2.2.2)解法:分离变量，得:dy = p(x)dxydy =òòp(x)dx + C，得:1yln y = ò p(x)dx +

7、C1y = ±eC1 eò p( x)dxy = eC1 eò p( x)dxc = ±eC1§ 2.2 Linear ODE and variation of constants Methody = ceò p( x)dx得y º 0因为为(2.2.2)的解,所以其通解为:y = ceò p( x)dx其中c为任意常数。.(2.2.3)y(x0 ) = y0满足初始条件的òxp(t )dty = y ex0.(2.2.3)0§ 2.2 Linear ODE and variation of c

8、onstants Methody¢ + y sin x = 0试求微分方程例1p2的通解,并求满足条件的= 2 特解y()p(x) = -sin x解由公式(2.2.3)得，所求通解为：y = ce -òsin xdx= cecos x由公式(2.2.3)得，所求特解为:y = 2ecos x§ 2.2 Linear ODE and variation of constants Method（2）非齐次线性方程/Non-Homogenous Linear ODE/采用常数变易法求解y¢ = P(x) y + Q(x)设想方程有形如(2.2.3)的解，但其

9、中的常数c变易为x的待定函数y = ceò P( x)dx(2.2.3)y = c(x)eò P( x)dx即设.(2.2.4)方程的解。§ 2.2 Linear ODE and variation of constants Methody = c(x)eò P( x)dx上式代入方程(2.2.1)，得:y¢ = P(x) y + Q(x)c¢(x)eòP( x)dx+ c(x)eòP( x)dx P(x) = P(x)c(x)eò P( x)dx+ Q(x)c¢(x)eò P( x)

10、dx即:= Q(x)c¢(x) = Q(x)e-òP( x)dx得:c(x) = òQ(x)e òdx + c- p( x)dx§ 2.2 Linear ODE and variation of constants Methodc(x) = òQ(x)e òdx + c- p( x)dxy = c(x)eò P( x)dx代入(2.2.4)得:.(2.2.5)同时，方程满足初始条件 j (x0 ) = y0的特解为 :xxy = eòx P(t )dt y+x Q(x)e-òxP(t )dt d

11、x0òx000y = eò P( x)dx òQ(x)e-ò P( x)dxdx + c§ 2.2 Linear ODE and variation of constants Method由(2.2.5)得:y = ceò P( x)dx + eò P( x)dx òQ(x)e-ò P( x)dxdx其中第一项是线性齐次方程的通解，第二项是线性非齐次方程特解。非齐次线性方程通解的结构：通解等于其对应齐次方程通解与自身的一个特解之和。§ 2.2 Linear ODE and variation o

12、f constants Methodcos x dy = y sin x + cos2 x例2dx先求对应的齐次方程通解解1)cos x dy = y sin xdy = sin x dxdx= -ln cos x + ln cycos xln ycy =(c为任意常数)cos x2) 用常数变易法求方程通解c(x)y =设是方程的解，代入原方程，得cos x§ 2.2 Linear ODE and variation of constants Methodcos x dy = y sin x + cos2c(x)y =xcos xdxcos x( c¢(x) cos x

13、+ c(x) sin x ) =c(xcos2 xcosc¢(x) = cos2c(x) = ò cos2x+ c24+ c)1y =(c为任意常数)cos说明：对于一阶线性方程，也可直接用通解公式计算得出。§ 2.2 Linear ODE and variation of constants Methoddyy=dx2x - y2例3解1)转换变量位置dx = 2x - y2= 2x - ydyyy2)用公式求方程通解1 dx y-21 dyòò2-ò yedy + c = eln y (-ò ye-2 dy + c)2l

14、n yx = eyx = y2 (-ò 1 dy + c) = - y2 ln y + cy 2yx = - y2 ln y + cy2y = eò P( x)dx òQ(x)e-ò P( x)dxdx + c§ 2.2 Linear ODE and variation of constants Method注意:y, dy不是线性的，但如果视有时方程关于dxx, dx是线性的，x 为y 的函数，方程关于dy于是仍可以根据上面的方法求解。例 求值问题dy = 3 y + 4x2+1,y(1) = 1dxx的解.先求原方程的通解解:ò&

15、#242;- p( x)dxp( x)dxòy = edx + c)( Q(x)e3x3xòò-dxdx(ò (4x2 +1)e= edx + c)= x3 (ò (4+ c)x3 (ò (4+ c)=+ c)2=+ c x3232将初始条件y(1) =1代入后得故所给初值问题的通解为c =y = x3 ln22§ 2.2 Linear ODE and variation of constants Methodxy'+(1+ x) y = exdy + 2xy = 4x(1)练习(2)dxdy =y(3)x + y3

16、dx§ 2.2 Linear ODE and variation of constants Methodxy'+(1+ x) y = ex练习解(1)xy'+(1 + x) y = 0dy + 1 + x dx = 01)先解齐次方程yx，得：ln y + ln x + x = c1e- xy = cxe- xy = c(x)2)设,代入原方程，得:xe- xe- x'+(1 + x)c(x)= exxc(x)xx§ 2.2 Linear ODE and variation of constants Methode- x- xe¢+ c(x

17、) + (1+ x)c(x)= exxc (x)x2xxc¢(x)e-x = exc¢(x) = e2x化简得:c(x) = ò e2x dx = 1 e2x+ c2e- x1所以，通解为:y =+ c)2 x(ex2ce- xexy =+2xx§ 2.2 Linear ODE and variation of constants Methoddy + 2xy = 4x练习(2)dxp(x) = -2x, Q(x) = 4x解用公式求解,y = e-ò 2 xdx (ò 4xeò 2 xdx dx + c)(ò 4

18、xex2 dx + c)2(2ex+ c)2= e- x2= e- x2y = 2 + ce -x即:§ 2.2 Linear ODE and variation of constants Methoddy =y练习(3)x + y 3dx解方程可以改写为:dx = 1 x + y2p( y) = 1 ,yQ( y) = y2dyy故通解为:11yòò12-dydy(ò y 2e+ c) = y(+ c)y 2x = eyx = 1 y3 + cy2或 y = cx - 1 y32即:§ 2.2 Linear ODE and variation

19、 of constants Method二、 可化为线性方程的方程1方程/Bernoulli ODE/2*提方程/ Riccati ODE/§ 2.2 Linear ODE and variation of constants Method方程/Bernoulli ODE/1y¢ = P(x) y + Q(x) yn(2.2.6)形如的方程称为n ¹ 0, n ¹ 1方程，其中它通过变量代换可化为线性方程。将方程(2.2.6)的各项同乘以 y - n解法：y-n y¢ = P(x) y1-n + Q(x)z = y1-n得:令1dz = y-n

20、dydz = (1- n) y-ndy则1- n dxdxdxdx§ 2.2 Linear ODE and variation of constants Method1dz = P(x)z + Q(x)1- n dxdz = (1- n)P(x)z + (1- n)Q(x)dx用上式求解后，代入原变量z = y1-n,便得原方程的通解。y1-n = eò(1-n) P( x)dx ò (1- n)Q(x)e-ò(1-n) P( x)dxdx + c§ 2.2 Linear ODE and variation of constants Metho

21、dxy'+ y = xy2将方程改写为:ln xy -2 y'+ 1 y -1 = ln xxdz - 1 z = -ln x例4解z = y-1dxx11xòò-dxdx(ò (- ln x)ez = edx + c)x= x(ò - ln x dx + c)x1= x-(ln x)2 + c2 112= xc -(ln x)2 故y例求方程x2dyy=+dx2x2 y的通解.这是Bernoulli 方程, n = -1,dz = 1 z + x2令z = y2 , 代入方程得解:dxx解以上线性方程得1x1xòò1

22、2-dxdx(ò x2e= cx +z = edx + c)3x= cx + 1 x32y2将z = y 代入得所给方程的通解为:2§ 2.2 Linear ODE and variation of constants Method提方程 / Riccati ODE/2dy = P(x) y 2 + Q(x) y + R(x)l 形如(2.2.7)dx的方程称为l 特点:提方程。在一般情况下，此类方程的解不能用初等函数及其形示表示，如果先由观察法或其他方法知道它的一个特解时，才可以通过初等法，求出它的通解。§ 2.2 Linear ODE and variatio

23、n of constants Methoddy = P(x) y 2 + Q(x) y + R(x)(2.2.7)dx若方程有一特解为y(x) = y (x)y¢ = z¢ + y ¢(x)解法y = z + y (x)设则z¢ + y ¢(x) =P(x)(z + y )2 + Q(x)(z + y ) + R(x)z¢ + y ¢(x) =P(x)(z2 + 2yz + y 2 ) + Q(x)(z + y ) + R(x)z¢ + y ¢(x) =P(x)z2+ (2yP(x) + Q(x)z +

24、P(x)y 2 + Q(x)y + R(x)z¢ = P(x)z2 + (2yP(x) + Q(x)z化为方程。§ 2.2 Linear ODE and variation of constants Methody¢ = y2 - x2 +1由观察看出y = x 是方程的一个特解，于是例5解1+ u¢ = (u + x)2 - x2+1y = x + u令，则得1 u¢ = 1+ 2xu-1u¢ = u2 + 2xu- (u-1)¢ = 1+ 2xu-1u2z = e-x (ò - ex dx + C)22z

25、62; = -2xz -1(C - ò ex2 dx)-12故原方程的通解为 y = x + ex§ 2.2 Linear ODE and variation of constants Methodaxx2 y¢ = x2 y2 + xy +1解此微分方程。形如例6试求的特解,y = a ，y¢ = - a ,代入方程得:解设x2x(a +1)2 = 0- a = a2 + a +1y = - 1a = -1所以是方程的一个特解。故xy = - 1 + u令xx2 ( 1 + u¢) = x2 (- 1 + u)2 + x(- 1 + u) +

26、1x2于是方程化为x方程xu¢ = x2u 2 - xu§ 2.2 Linear ODE and variation of constants Method1u¢ = -xu-1 + x2- (u-1)¢ = -xu-1 + x2u2æöx22- x2çdx + C ÷z¢ = xz - x2òz = e-2x e2ç÷èøö-1æx2x22-u = e 2-çdx + C ÷ò- x e2ç&#

27、247;èø故原方程的通解为ö-1æ- x22- x221x1xçdx + C ÷òy = -+ u = -+ e- x e2ç÷èø§ 2.2 Linear ODE and variation of constants Method练习(1)xdy - y += 0(2)y¢e-§ 2.2 Linear ODE and variation of constants Methodxdy - y += 0(1)练习解xy3dx方程各项同除以dy = 1 y-

28、2y-3+1+ ln xdz = -2 y -3得:dxxdydxz = y -2 ,令dxdz = - 2 z - 2(1+ ln x)于是方程化为:dxx1 1(- 49x2òz =- 2(1+ ln+ c=+ c)x 23c - 4y-2=即x293§ 2.2 Linear ODE and variation of constants Method(2)y¢e-练习y¢ = - y2ex + 2 ye2解y = ex经观察，方程有一个特解y = ex + u令u¢ = -u 2ex - 2ue2x1y = ex +c + ex§

29、 2.2 Linear ODE and variation of constants Method思考题(1)æödy- y+1= xexç dx÷eè- exø+ ex+ y= 0y'(2)y ln ydx + (x - ln y)dy = 0(3)§ 2.2 Linear ODE and variation of constants Method提示: 1dy + e- yæödye- y= xexe- y+ 1= xexç dx÷dxèøde- y= e- y- xe（x线性方程）dx(ey )¢ - exey+ ex