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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.

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  • $\begingroup$ 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? $\endgroup$ – user539887 Feb 18 at 9:53
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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.enter image description here

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