Question

第一大题 求解 求详细过程

Answer 1

（1）
y1=x =>y1'=1, y1''=0
y1''+(1/x)y1' -(1/x^2)y1
=0+(1/x)(1) -(1/x^2)(x)
=1/x -1/x
=0
//
y2=2x =>y2'=2 , y2''=0
y2''+(1/x)y2' -(1/x^2)y2
=0+(1/x)(2) -(1/x^2)(2x)
=2/x -2/x
=0
y1=x, y2=2x 是 y''+(1/x)y' -(1/x^2)y=0 的特解
(2)
y''-y'-2y=0
The aux. equation
p^2-p-2=0
(p-2)(p+1)=0
p=2 or -1
y= Ae^(2x)+Be^(-x)
(3)
4y''+4y'+y=0
The aux. equation
4p^2+4p+1=0
(2p+1)=0
p=-1/2
y= (Ax+B)e^[(-1/2)x]
(4)
4y''-12y'+y=0
The aux. equation
4p^2-12p+1=0
p=(3+2√2)/2 or (3-2√2)/2
y= Ae^((3+2√2)/2)+Be^((3-2√2)/2)

(5)
y''-2y'=3x+1
let
yp=Ax^2 +Bx
yp'=2Ax +B
yp''=2A
yp''-2yp'=3x+1
2A -2(2Ax +B) =3x+1
-4Ax +(2A-2B)=3x+1
-4A =3 and 2A-2B=1
A=-3/4 and B=-5/4
特解
yp=-(3/4)x^2 -(5/4)x

Answer 2

直接带进去不就好了。