大一上高數(shù)課件線性微分方程解結(jié)構(gòu)

第4 線性微分方程解的結(jié)1/2/3/y′′+p(x)y′+q(x)=疊加原理:設(shè)y1,y2都是解,c1,c2為常數(shù),y=c1y1+

7.1節(jié),yφ(xC1C22獨(dú)立的常數(shù).(4.2)中,λ,y1λy2,y4/定義nk1,k2,···,kn,k1y1(x)+k2y2(x)+···+knyn(x)≡0,?x∈則稱函數(shù)y1(x),y2(x),···,yn(x)在區(qū)間I上線性相關(guān).若上述常數(shù)ki全為零,則稱···線性無關(guān).例如:1,x,x2在,+∞)上線性無關(guān)1,x+1,x?1在,+∞)上線性相關(guān)cosx,sinx在,+∞)上線性無關(guān)5/注.

y2(xI

y1 )

?=0′y12y′?=1,∈I.解的結(jié)構(gòu):y′′p(x)y′q(x0y=C1y1+y1,y2是兩個線性無關(guān)的解,C1,C2為任意常數(shù)6/y′′+p(x)y′+q(x)=設(shè)y1,y2是方程的解,則Yy1?y2Y′′+p(x)Y′+q(x)Y=

即Y7/疊加原理:設(shè)y1,y2y′′+p(x)y′+ =f y′′+p(x)y′+ =f 則yy1y2y′′+p(x)y′+q(x)y=f(x)+f 例如,y′′py′qyx2excosx的特解,y′′py′qyx2和y′′py′qyexcosy?和y?.于是y?

為原方程的特解 8/y′′+p(x)y′+q(x)=(一)令y(x)=y1(x)u(x), :1:::::::1:::::::y1u′′+(2y′+py)u′+(y′′+ :1:::::::1:::::::

1+ u′ 例(1?x2)y′′+2xy′?2y=?2.9/y′′+p(x)y′+q(x)= y=

1 1(二)觀察出齊次方程的一個特解再求與之線性無關(guān)的另一個特解.根據(jù)方程系數(shù)的特點,用較簡單的函數(shù)如1(或任何常數(shù)),x,x2,1/x,1/x2,ex,lnx,或sinx,cosx等代入試算.xy′′+3y′=xy′′?(x+1)y′+y=10/y′′+p(x)y′+q(x)= (三)y1(x),Y=C1y1(x)+y?=u1(x)y(x)+u2(x)y 則 (y?)′=(u′y+ ::: :: ::::::

)+u1y′+u2y′ 2::: 2 (y?)′′=(u′ +u′y)′+u′y′+u′ ::: :: :::::: ::: ::::::: :: ::::: :::11/f(x)=(y?)′′+p(y?)′+ =(u′ +u′y)′+u′y′+u ::: :: :::::: :: ::::::: ::: :::::: :: :::::::: ::: ::::: ::: +u1·(y′′+py′+qy)+u2·(y′′+py′+qy =(u′y+u′y)′+p(u′y+u′y)+u′y′+u′y′1 2 1 2 1 2令u′ +u′ =

+u′y′= 因y, 線性無關(guān),y′y?yy?′0,上述方程組有解u,′ 1 1 12/y′′+p(x)y′+q(x)= y1,y2.y?=u1(x)y1(x)+u2(x)y2(x).u′y+u′ =1 2u′

+u′y′=例

2y′′+y=0y1=cosy2=sinx.y′′+y=2sec3x的通解13/7-4:A-6(1)EX1.設(shè)p,q,α,β,A,B為常數(shù),y′′+py′+ycosβx·F(xsinβx·G(x),F(x),G(x)二階