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Solve the system $$x'= \left(\begin{matrix}1&1&0\\1&1&0\\0&0&3\end{matrix}\right)x +\left(\begin{matrix}e^t\\e^{2t}\\3e^{3t}\end{matrix}\right)$$ I got the eigenvalues $\lambda_1=0; \lambda_2=2; \lambda_3=3$ and the solution for homogeneous part is $$x=C_1\left(\begin{matrix}-1\\1\\0\end{matrix}\right)+C_2e^{2t}\left(\begin{matrix}1\\1\\0\end{matrix}\right)+C_3e^{3t}\left(\begin{matrix}1\\2\\1\end{matrix}\right)$$ I am stuck here. I don't know how to find the partial solution, please help

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  • $\begingroup$ @user73192: Did they provide initial conditions at say $t = 0), or even say $x(0) = x_0$? $\endgroup$ – Amzoti Aug 19 '13 at 0:08
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I will provide a guiding hint. Give it a go and respond back if lost.

Given:

$$x'= Ax + F(t) = \left[\begin{matrix}1&1&0\\1&1&0\\0&0&3\end{matrix}\right]x +\left[\begin{matrix}e^t\\e^{2t}\\3e^{3t}\end{matrix}\right]$$

So, $ A = \left[\begin{matrix}1&1&0\\1&1&0\\0&0&3\end{matrix}\right]$ and $F(t) = \left[\begin{matrix}e^t\\e^{2t}\\3e^{3t}\end{matrix}\right]$

  • We want to find the matrix exponential $e^{At}$
  • We then solve for $X(t) = e^{At}x(\tau) + \int_\tau^t e^{A(t-s)}F(s)ds$, where $\tau$ is the initial condition time, but you do not have that, so we need to leave it in a general form. In other words $x(\tau) = C$, some set of initial values for some given $t$.

So, the matrix exponential is given by (this is different than what you got):

$$e^{At} = \left[\begin{matrix} \dfrac{1}{2} (1 + e^{2 t}) & \dfrac{1}{2} (-1 + e^{2 t}) & 0 \\ \dfrac{1}{2} (-1 + e^{2 t}) & \dfrac{1}{2} (1 + e^{2 t}) & 0 \\ 0 & 0 & e^{3t} \end{matrix}\right]$$

  • Next, find $e^{A(t-s)}F(s)ds$, so we have (just multiply these two matrices):

  • $e^{A(t-s)}F(s) = \left[\begin{matrix} \dfrac{1}{2} (1 + e^{2 (t-s)}) & \dfrac{1}{2} (-1 + e^{2(t-s)}) & 0 \\ \dfrac{1}{2} (-1 + e^{2 (t-s)}) & \dfrac{1}{2} (1 + e^{2 (t-s)}) & 0 \\ 0 & 0 & e^{3(t-s)} \end{matrix}\right] \cdot \left[\begin{matrix}e^s\\e^{2s}\\3e^{3s}\end{matrix}\right]$

  • Lastly, find $X(t) = e^{At}x(\tau) + \int_\tau^t e^{A(t-s)}F(s)ds$

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  • $\begingroup$ Nice work again, as usual! +1 $\endgroup$ – amWhy Aug 18 '13 at 23:58

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