Computing the index of fixed point for the system of equations $x'=x^2$ and $y'=-y$. Consider the system of equations $$x' = x^2$$ $$y'=-y.$$ Compute the index of fixed point at the origin both visually and computationally, or just write down the integral.
$\textbf{Solution:}$ To find the fixed point, integrate both equations $$\int f'(x) = \int x^2 dx \text{ and } \int f'(y) = \int -y dy.$$ Therefore, $f(x) = \frac{x^3}{3} + C \text{ and } f(y) = -\frac{y^2}{2} + C.$ Next, to find the fixed point index, let $x=0, \text{ and } y=0$ so $f(0) = C.$ Therefore, $f(x) = \frac{x^3}{3} + 0$ and $f(y) =-\frac{y^2}{2} + 0.$ So the fixed point of $f(x) = 0$ and $f(y) = 0.$
So the index for the first function is $I(f(x),x_0) = (\frac{x^3}{3}, 0)$ and for our second function it is $I(f(y),y_0) = (-\frac{y^2}{2},0).$ Can someone please help me complete the problem? 
 A: To get fixed points, we need $x'=0$ and $y'=0$. So, the fixed points are $x^2 = 0$ and $-y=0$ so $x=0$ and $y=0$. Thus, $(0,0)$ is a fixed point. 
Let $f=x^2$ and $g=-y$, then $f_x = 2x, f_y=0, g_x = 0,$ and $g_y=-1.$ Therefore the Jacobian is $$J = \begin{pmatrix} f_x & g_x \\ f_y & g_y \end{pmatrix}$$ $$J_{(0,0)} = \begin{pmatrix} 2x & 0 \\ 0 & -1 \end{pmatrix}_{(0,0)} = \begin{pmatrix} 0 & 0 \\ 0 & -1 \end{pmatrix}.$$
The eigenvalues at $(0,0)$ are $|J-I\lambda| = 0.$ $$\begin{pmatrix} 0-\lambda & 0 \\ 0 & -1-\lambda \end{pmatrix} = 0 \text{ so } \lambda = 0, -1.$$ The system is marginally stable if it has one or more distinct poles on the imaginary axis and any remaining poles have negative real part. 
Now, taking the integrals, $x' = x^2, \frac{dx}{x^2} = dt$ $$\implies \int x^{-2}dx = t + C$$ $$\implies -\frac{1}{x} = t + C.$$ Now, $y' = -y , \frac{dy}{dt} = -y \implies \frac{dy}{y} = \int -dt = \log y = -t + \log C$. So, $\frac{y}{C} = e^{-t}$ and $y= Ce^{-t}.$
A: You cant treat function as variable. Solving then this system of equations would look completely different. Treat these functions like functions dependable from paramter t.
Thus, as a simple differentional equation solving we get:
$.\\y(t)=C*e^{-t} \\ x(t)=-\frac{1}{t}+C $
I don't know what do to furhter, but hope this will help a bit.
