# System of equations: (y_1 y_2)=(2 0, 4 −1)(y_1 y_2)

System of equations:
$\left(\begin{array}{c}{\stackrel{˙}{y}}_{1}\\ {\stackrel{˙}{y}}_{2}\end{array}\right)=\left(\begin{array}{cc}2& 0\\ 4& -1\end{array}\right)\left(\begin{array}{c}{y}_{1}\\ {y}_{2}\end{array}\right)$
Looking at matrix $A$ I can see a value not on the diagonal equal to zero. So the eigenvalues here are $2$ and $-1$, meaning I have a saddle node. Next I look at the eigenvectors:
$\lambda =2,\left[\begin{array}{ccc}0& 0& 0\\ 4& -3& 0\end{array}\right]$
$\lambda =-1,\left[\begin{array}{ccc}3& 0& 0\\ 4& 0& 0\end{array}\right]$
$y=A{e}^{2x}\left(\begin{array}{c}3\\ 4\end{array}\right)+B{e}^{-x}\left(\begin{array}{c}0\\ 1\end{array}\right)$
So I have an attractive trajectory on the $y$-axis, and a repulsive trajectory on the line $y=\frac{3}{4}x$
With the slope at origin being
$\frac{{\stackrel{˙}{y}}_{2}}{{\stackrel{˙}{y}}_{1}}=\frac{4{y}_{1}-{y}_{2}}{2{y}_{1}}=\frac{0}{0}=±\mathrm{\infty }$
Does all of this seem correct?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

domwaheights0m
It is indeed correct, but the slope you computed is the slope of the orbit on the phase space, because each coordinate ${y}_{1}$ and ${y}_{2}$ have an exponencial growth/decay rate. also the origin is not a node, but a saddle equilibria