Assuming I have
$\frac{dx}{dt}=5x^2+2xy+x$
$\frac{dy}{dt}=xy-y$
which leads to a jacobian matrix
$$\begin{pmatrix}
10x+2y & 2y \\
y & x-1
\end{pmatrix}$$
one of the fixed points is $(0,0)$, how do I find the form of the linearized system
at that fixed point so that it is at the form of example: $\frac{dx}{dt}=5 \cdot x$
Best Answer
First off: the correct Jacobian would be: $$ J(x,y)=\left(\matrix{f_x(x,y) & f_y(x,y) \\ g_x(x,y) & g_y(x,y)}\right), $$ where your system is: $$ \frac{d}{dt}\left(\matrix{x\\y}\right)=\left(\matrix{f(x,y)\\g(x,y)}\right). $$ then you have to find a point $P(x^*,y^*)$ such that $f(P)=g(P)=0$. $(0,0)$ satisfies this condition.
Moreover, now you can proxy, in a "small" interval cointaining $P$, the dynamic in the following way: $$ \frac{d}{dt}\left(\matrix{x\\y}\right)=\left(\matrix{f(x,y)\\g(x,y)}\right)\approx\left(\matrix{f(P)\\g(P)}\right)+\left(\matrix{f_x(P) & f_y(P) \\ g_x(P) & g_y(P)}\right)\cdot\left(\matrix{x\\y}\right), $$ so, reminding that $\left(\matrix{f(P)\\g(P)}\right)=\left(\matrix{0\\0}\right)$ you then have: $$ \frac{d}{dt}\left(\matrix{x\\y}\right)=\approx\left(\matrix{f_x(P) & f_y(P) \\ g_x(P) & g_y(P)}\right)\cdot\left(\matrix{x\\y}\right)=J(P)\cdot\left(\matrix{x\\y}\right). $$ In your case: $$ J(x,y)=\left(\matrix{10x+2y+1 & 2x \\ y & x-1}\right);\ P(0,0). $$ Therefore: $$ J(P)=\left(\matrix{1 & 0 \\ 0 & -1}\right), $$ which leads to: $$ \frac{dx}{dt}=x, $$ $$ \frac{dy}{dt}=-y. $$ Or, rather: $$ x(t)=x_0e^t, $$ $$ y(t)=y_0e^{-t}, $$ But be careful that this ODE is ill-posed, as one eigenvalue has real part greater than 0.