Question

求该方程的求解详细过程

Answer 1

2x+2λx=0 (1)
-2y+(λ/2)y=0 (2)
x^2+(1/4)y^2-1 =0 (3)
from (1)
2x+2λx=0
x(1+λ)=0
x=0 or λ=-1
from (2)
-2y+(λ/2)y=0
4y -λy=0
y(4-λ)=0
y=0 or λ=4
case 1: x=0
from (3)
x^2+(1/4)y^2-1 =0
(1/4)y^2-1 =0
y^2 =4
y=2 or -2
(x,y)=(0,2) or (0,-2)
case 2: λ=-1
from (2)
=> y=0
from (3)
x^2+(1/4)y^2-1 =0
x^2-1 =0
x=1 or -1
(x,y)=(1,0) or (-1,0)
case 3: y=0 和case 2 一样
case 4 : λ=4
from (1) => x=0 和case 1 一样
ie
(x,y)=(0,2) or (0,-2) or (1,0) or (-1,0)