Question

微分方程题目？

Answer 1

(3)

y''+2ny'+k^2y=0
The aux. equation :
p^2+2np+k^2 = 0
△= 4n^2 -4k^2
case 1: 4n^2 -4k^2 >0
p= -n±√(n^2-k^2)
通解 : y = Ae^{ [-n+√(n^2-k^2)]x } +Be^{ [-n-√(n^2-k^2)]x }
case 2: n^2-k^2 =0
p=-n
通解 : y = (Ax+B).e^(-nx)
case 3: n^2- k^2 <0
p= -n±√(n^2-k^2)i
通解 : y = e^(-nx) .【 Acos{√(n^2-k^2)]x } +Bsin{√(n^2-k^2)]x 】
(4)
y''+y=e^x +cosx
The aux. equation
p^2 +1=0
p=i or -i
let

yg=Acosx +Bsinx
yp= Ce^x +x(Dcosx + Esinx)
yp'=Ce^x +(Dcosx + Esinx) +x(-Dsinx + Ecosx)
yp''
=Ce^x +(-Dsinx + Ecosx) +(-Dsinx + Ecosx) +x(-Dcosx - Esinx)
=Ce^x +2(-Dsinx + Ecosx) +x(-Dcosx - Esinx)
yp''+yp=e^x +cosx

Ce^x +2(-Dsinx + Ecosx) +x(-Dcosx - Esinx) + Ce^x +x(Dcosx + Esinx) =e^x +cosx
2Ce^x +2(-Dsinx + Ecosx) =e^x +cosx
=>
2C =1 and -2D =0 and 2E =1
C=1/2 and D=0 and E =1/2
yp=Ce^x +x(Dcosx + Esinx) = (1/2)e^x + (1/2)xsinx
通解
y=yg+yp=Acosx +Bsinx + (1/2)e^x + (1/2)xsinx