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I have $y_1=\left[\begin{array}{rr}e^t\\-e^t\end{array}\right]$ $y'=\left[\begin{array}{rr}2-t & 1-t\\t-2 & t-1\end{array}\right]$.

Using Liouville formula $W(t)=W(t_0)\exp(\int_{t_0}^t\mathrm{Tr}(A(s))\mathrm{d}s)$

find the solution with initial condition $y(0)=\left[\begin{array}{rr}1\\0\end{array}\right]$.

The supposed result is $y_2=\left[\begin{array}{cc}t\\1-t\end{array}\right]$

I have no idea where to start at all.

Well $\exp(\int_{t_0}^t\mathrm{Tr}(A(s))\mathrm{d}s)=\exp(\int_{0}^t(2-s)+(s-1)\mathrm{d}s)=\exp(\left.s\right|_{0}^t)=\exp(t)=e^t$

But that doesn't help me at all.

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  • $\begingroup$ is $W_0=\left|begin{array}{cc}1 & 1 \\ -1 & 0\right|=1$? $\endgroup$ – mtr Jun 18 '15 at 0:35
  • $\begingroup$ OK, I guess I got it. Will present my solution later. $\endgroup$ – mtr Jun 18 '15 at 6:40

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