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I'm trying to determine the state transition matrix,$\Phi(t,t_0)$, of the following system: $$ \begin{bmatrix}x'_1\\x'_2\end{bmatrix}= \begin{bmatrix}-\sin(t)&t\\0&-\cos(t)\end{bmatrix} \times \begin{bmatrix}x_1\\x_2\end{bmatrix} $$ with initial condition $$x(t_0)=\begin{bmatrix}x_1(t_0)\\x'_2(t_0)\end{bmatrix}$$

The fundamental matrix I got from this system is: $$ \begin{bmatrix}(t^2/2)+\cos(t)&1-\sin(t)\\t^2&2-\sin(t)\end{bmatrix}$$

and from that, I got $$\Phi(t,t_0)= \begin{bmatrix}1&0\\0&1\end{bmatrix} $$ which I don't think is correct. Can someone tell me where I'm going wrong? Is it that my fundamental matrix is incorrect?

The general Fundamental Matrix that I got is: $$ \begin{bmatrix}(t^2-t^2_0/2)+\cos(t-t_0)&1-\sin(t-t_0)\\t^2-t_0^2&2-\sin(t-t_0)\end{bmatrix}$$

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  • $\begingroup$ The fundamental matrix is incorrect, because it does not satisfy $\Psi'(t)=A(t)\Psi(t)$. $\endgroup$
    – user239203
    Commented Dec 11, 2019 at 6:53
  • $\begingroup$ I thought that this equation only satisfied LTI systems. The system that I have is an LTV system. $\endgroup$ Commented Dec 11, 2019 at 8:21
  • $\begingroup$ Here is how I solved for X(t). Do you know where I went wrong? (imgur.com/a/ktEzphG) $\endgroup$ Commented Dec 11, 2019 at 8:21
  • $\begingroup$ If you write $f'(t)=a(t)f(t)$ and you want to integrate both sides, then you must write $f(t)-f(t_0)=\int_{t_0}^t a(\tau)f(\tau)\,d\tau$, not $\int_{t_0}^t a(\tau)f(t_0)\,d\tau$. You don't get to integrate just the things you like. $\endgroup$
    – user239203
    Commented Dec 11, 2019 at 8:47
  • $\begingroup$ $\Psi'(t)=A(t)\Psi(t)$ is just an algebraic consequence of the columns being solutions to an ODE in the form $x'(t)=A(t)x(t)$. $\endgroup$
    – user239203
    Commented Dec 11, 2019 at 8:53

1 Answer 1

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From $x_2'(t)=- x_2(t)\cos t$, you obtain (say, by separation of variables), $x_2(t)=x_2(t_0)e^{\sin t_0}e^{-\sin t}$.

Substitute in the first equation and you obtain $$x_1'(t)+x_1(t)\sin t=x_2(t_0)e^{\sin t_0}te^{-\sin t}$$

The fundamental function of (the homogenous part of) this last ODE is $e^{\cos t}$, and therefore \begin{align}&x_1(t)=x_1(t_0)e^{-\cos t_0}e^{\cos t}+e^{\cos t}\int_{t_0}^t x_2(t_0)e^{\sin t_0}ue^{-\cos u-\sin u}du=\\=&\left(x_1(t_0)e^{-\cos t_0}+x_2(t_0)e^{\sin t_0}\int_{t_0}^0 ue^{-\cos u-\sin u}du\right)e^{\cos t}+x_2(t_0)e^{\sin t_0}e^{\cos t}\int_{0}^t ue^{-\cos u-\sin u}du\end{align}

Which means that $$\begin{pmatrix}x_1(t)\\ x_2(t)\end{pmatrix}=\begin{pmatrix}A e^{\cos t}+Be^{\cos t}\int_0^t ue^{-\cos u-\sin u}\,du\\ Be^{-\sin t}\end{pmatrix}$$

With $A=x_1(t_0)e^{-\cos t_0}+x_2(t_0)e^{\sin t_0}\int_{t_0}^0 ue^{-\cos u-\sin u}\,du$ and $B=x_2(t_0)e^{\sin t_0}$. This means that a possible choice for $\Psi$ is $$\Psi(t)=\begin{pmatrix}e^{\cos t}&e^{\cos t}\int_0^tue^{-\cos u-\sin u}\,du\\ 0&e^{-\sin t}\end{pmatrix}$$

And now you can calculate easily $\Psi(\tau)(\Psi(t))^{-1}$.

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