Khan exercises Maths Tools 1
The vector field F:R2→R2,(y1y2)↦(10y1+9y25y12+6y1y22)F: \mathbb R^2 \to \mathbb R^2, \begin{pmatrix} y_1 \\ y_2 \end{pmatrix}\mapsto \begin{pmatrix} 10y_1 + 9y_2 \\ 5y_1^2 + 6y_1 y_2^2 \end{pmatrix} F:R2→R2,(y1y2)↦(10y1+9y25y12+6y1y22) defines a system y′=F(y)y' = F(y)y′=F(y).
F:R2→R2,(y1y2)↦(10y1+9y25y12+6y1y22)F: \mathbb R^2 \to \mathbb R^2, \begin{pmatrix} y_1 \\ y_2 \end{pmatrix}\mapsto \begin{pmatrix} 10y_1 + 9y_2 \\ 5y_1^2 + 6y_1 y_2^2 \end{pmatrix} F:R2→R2,(y1y2)↦(10y1+9y25y12+6y1y22)
y′=F(y)y' = F(y)y′=F(y)
Find the fixed point y∞=(y∞,1,y∞,2)≠(0,0)y_{\infty} = \left({\color{red}y_{\infty,1}},{\color{blue}y_{\infty,2}}\right) \neq (0,0)y∞=(y∞,1,y∞,2)=(0,0).
y∞=(y∞,1,y∞,2)≠(0,0)y_{\infty} = \left({\color{red}y_{\infty,1}},{\color{blue}y_{\infty,2}}\right) \neq (0,0)y∞=(y∞,1,y∞,2)=(0,0)
y∞,1={\color{red}y_{\infty,1}} =y∞,1=
y∞,2={\color{blue}y_{\infty,2}} =y∞,2=