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1、常微分方程OrdinaryDifferentialEquations第二章§2.2线性方程与常数变易法(LinearEquationandMethodofConstantVariation)OrdinaryDifferentialEquations,SchoolAppl.Maths.一阶线性微分方程dya(x)+b(x)y+c(x)=0dx在a(x)≠0的区间上可写成dy=P(x)y+Q(x))1(dx这里假设P(x),Q(x)在考虑的区间上是x的连续函数若Q(x)=,0则)1(变为dy=P(x)y)2(dx)2(称为一阶齐次线性方程若Q(x)≠,0则)1(
2、称为一阶非齐线性方程OrdinaryDifferentialEquations,SchoolAppl.Maths.一、一阶线性微分方程的解法--常数变易法01解对应的齐次方程dy=pxy()(2)dx得对应齐次方程解∫pxdx()yc=e,c为任意常数0dy2常数变易法求解=P(x)y+Q(x))1(dx(将常数c变为x的待定函数c(x),使它为)1(的解)∫p(x)dx令y=c(x)e为)1(的解,则OrdinaryDifferentialEquations,SchoolAppl.Maths.dydc(x)∫p(x)dx∫p(x)dx=e+c(x)p(x)
3、edxdxdc(x)−∫p(x)dx代入(1)得=Q(x)edx~−∫p(x)dx积分得c(x)=∫Q(x)edx+c03故)1(的通解为~∫p(x)dx−∫p(x)dxy=e(∫Q(x)edx+c))3(注求(1)的通解可直接用公式(3)(formular)OrdinaryDifferentialEquations,SchoolAppl.Maths.例1求方程dyxn+1(x+)1−ny=e(x+)1dx通解,其中n为常数.dynxn解:将方程改写为=y+e(x+)1dxx+1dyn首先,求齐次方程=y的通解dxx+1dyndyn从=y分离变量得=dxdx
4、x+1yx+1两边积分得lny=nlnx+1+c1OrdinaryDifferentialEquations,SchoolAppl.Maths.n故对应齐次方程通解为y=c(x+)1n∫pxdx()∫xdxn或yc===ece+1cx(1+)其次应用常数变易法求非齐线性方程的通解,n令y=c(x)(x+)1为原方程的通解,代入得dc(x)nn−1n−1xn(x+)1+nc(x)(x+)1=nc(x)(x+)1+e(x+)1dx~dc(x)xc(x)=ex+c即=e积分得dx~~nx故通解为y=(x+)1(e+c),c为任意常数OrdinaryDifferen
5、tialEquations,SchoolAppl.Maths.dyy=例2求方程dx2x−y2通解.解:原方程不是未知函数y的线性方程，但将它改写为2dx2dx2x−y=即=x−ydyydyy它是以x为未知函数,y为自变量的线性方程，故其通解为~∫p(y)dy−∫p(y)dyx=e(∫Q(y)edy+c)22∫dy−∫dy~yy=e(∫(−y)edy+c)~2=y(−lny+c),c为任意常数。OrdinaryDifferentialEquations,SchoolAppl.Maths.dy32例3求初值问题=y+4x+,1y)1(=1的解.dxx解:先求原
6、方程的通解~∫p(x)dx−∫p(x)dxy=e(∫Q(x)edx+c)33∫dx−∫dx~=ex(∫4(x2+)1exdx+c)~323=++xxx((∫41)/dxc)~x~=−+xxxc32(ln41/())2343=xlnx−+cx~2将初始条件y)1(=1代入后得c=32/故所给初值问题的通解为3433xy=xlnx+x−22OrdinaryDifferentialEquations,SchoolAppl.Maths.二、伯努利()Bernoulli方程dyn形如=+pxyQxyn()()，≠0,1dx的方程,称为伯努利方程.这里P(x),Q(x)
7、为x的连续函数。01−n解法:1引入变量变换z=y,方程变为dz=1(−n)P(x)z+1(−n)Q(x)dx02求以上线性方程的通解03变量还原OrdinaryDifferentialEquations,SchoolAppl.Maths.JacobBernoulli(alsoknownasJamesorJacques)(27dec.1654–16Aug.1705)wasoneofthemanyprominentmathematiciansintheBernoullifamily.J.BernoulliwasborninBasel,Switzerland.F
8、ollowinghisfather'swish,he
