✔ 最佳答案
參考一下:
(1)x"=x- 1.5x^2, then x'x"= xx'- 1.5x^2 x'
integrate w.r.t. time t, then 0.5(x')^2 = 0.5x^2 - 0.5x^3+C
thus, 0.5[(x')^2-x^2+x^3]= C independent on time t
(2) (a)phase diagram is the famliy of y^2-x^2+x^3= 2C, C is any real number
(b) critical point (node)
x"=x-1.5x^3 then
{ x'=y
{ y'=x- 1.5x^2 = -1.5(x- 2/3)^2 - (x- 2/3)
thus, critical point are (x,y)=(0,0), (2/3, 0)
case 1: node(0,0)
the linear form of the system is
{ x'=y
{ y'=x
eigenvalue of the system are 1,-1, so, the node(0,0) is a saddle point
(unstable)
case 2: node(2/3, 0)
the linear form of the system is
{ X'= Y
{ Y'= -X , where X=x-2/3, Y=y
eigenvalue are i, -i, so, the node (2/3, 0) is a center point(stable)
(3)Separatrix
the separatrix of the system is y^2-x^2+x^3=0, shown as:
圖片參考：
http://imgcld.yimg.com/8/n/AD04686329/o/161002110070113872639760.jpg
