# Determine the flow of the differential equation $\dot{y}=Ay$

Determine the flow of the differential equation $$\dot{y}=Ay$$, where $$A=\Big(\begin{matrix} 2&1\\0&2 \end{matrix}\Big)$$

The solution to the differential equation would be $$y(t)=e^{(t-t_0)A}y_0$$, but there are no initial conditions given, so I don't know if I can just assume that.

If you just write everything out I get the differential equations $$\frac{dy_1}{dt}=2y_1+y_2$$ and $$\frac{dy_2}{dt}=2y_2$$.Which would give the solutions $$y_1(t)=c_1e^{2t}+c_2e^{2t}t$$ and $$y_2(t)=c_2e^{2t}$$.

I don't know how to get the flow from this.

• Welcome to MSE. Do you mean by stream what is called flow, that is the mapping assigning to any point $x_0 \in \mathbb{R}^2$ and any time $t$ the value at time $t$ of the solution of the ODE satisfying the initial condition $x(0)=x_0$? Then just calculate $c_1$ and $c_2$ in terms of $x_0$ and substitute to the formula for the solution. Or perhaps you mean something else? Feb 18 '19 at 9:53

Just consider parametric curves of the form $$(y_1(t),y_2(t)), t \in \mathbb{R}$$
For each set of initial conditions you will obtain a stream line. The picture below was obtained with mathematica using the StreamPlot[] command. In red one of the curves I mentioned before. 