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I have the system:

$$x_1'=2x_1-x_2+e^{2t}\\ x_2'=4x_1+2x_2+4$$

So I searched for the homogeneous solutions and got:

$$X_H=c_1e^{2t} \left(\begin{matrix} \cos 2t \\ 2\sin 2t\\ \end{matrix}\right)+c_2e^{2t} \left(\begin{matrix} \sin 2t \\ -2\cos 2t\\ \end{matrix}\right) $$

Then I have to set: $$ e^{2t}\left(\begin{matrix} \cos 2t & \sin 2t\\ 2\sin 2t& -2\cos2\\ \end{matrix}\right) \left(\begin{matrix} c_1' \\ c_2'\\ \end{matrix}\right)= \left(\begin{matrix} e^{2t} \\ 4\\ \end{matrix}\right) $$

I spent like two hours solving the resulting integrals, and checked with wolfram (they seemed okay) and got:

$$X_P=\left(\frac {\sin 2t}{2}-\frac {\sin 2t+ \cos 2t}{2e^{2t}}\right)e^t\left(\begin{matrix} \cos2t\\ 2\sin2t\end{matrix}\right)+\left( \frac {-\cos 2t}{2} + \frac {\cos2t-\sin2t}{2e^{2t}}\right) e^{2t}\left(\begin{matrix} \sin 2t \\ -2\cos2t\end{matrix}\right)$$

But then I checked and this didn't solve the system... Did I miss any step?

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Your next equations are correct : $$X_H=c_1e^{2t} \left(\begin{matrix} \cos 2t \\ 2\sin 2t\\ \end{matrix}\right)+c_2e^{2t} \left(\begin{matrix} \sin 2t \\ -2\cos 2t\\ \end{matrix}\right) $$

$$ e^{2t}\left(\begin{matrix} \cos 2t & \sin 2t\\ 2\sin 2t& -2\cos 2t\\ \end{matrix}\right) \left(\begin{matrix} c_1' \\ c_2'\\ \end{matrix}\right)= \left(\begin{matrix} e^{2t} \\ 4\\ \end{matrix}\right) $$ Then, one cannot check your calculus because all the details are not written. The mistake is probably easy to find because there is a term $e^t$ instead of $e^{2t}$ in your final result. No $e^t$ should appear.

I obtained : $$x_1=c_1 e^{2t}\cos(2t)+c_2 e^{2t}\sin(2t)-\frac{1}{2}$$ $$x_2=2c_1 e^{2t}\sin(2t)-2c_2 e^{2t}\cos(2t)-1+e^{2t}$$

Hint :

$$ \left(\begin{matrix} c_1' \\ c_2'\\ \end{matrix}\right)= \left(\begin{matrix} \cos 2t & \sin 2t\\ 2\sin 2t & -2\cos 2t\\ \end{matrix}\right)^{-1} \left(\begin{matrix} 1 \\ 4 e^{-2t}\\ \end{matrix}\right) $$

$$ \left(\begin{matrix} c_1' \\ c_2'\\ \end{matrix}\right)= \left(\begin{matrix} \cos 2t +2e^{-2t}\sin 2t\\ \sin 2t -2e^{-2t}\cos 2t\\ \end{matrix}\right) $$ Then integration leads to $c_1$ and $c_2$

$$ \left(\begin{matrix} c_1 \\ c_2 \\ \end{matrix}\right)= \frac{1}{2}e^{-2t} \left(\begin{matrix} -\cos 2t -\sin 2t +e^{2t}\sin 2t\\ \cos 2t -\sin 2t -e^{2t}\cos 2t\\ \end{matrix}\right)+ \left(\begin{matrix} C_1\\ C_2\\ \end{matrix}\right) $$

Note that all this stuff isn't necessary to find a particular solution : just a bit of intuition and/or "trial and error" approach, on can find the terms $(-\frac{1}{2} \: ,\: -1+e^{2t})$ to be added to the homogeneous solution.

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  • $\begingroup$ Could you give me some insights into how you calculated $c_1(t), c_2(t)$? I keep failling... $\endgroup$ – YoTengoUnLCD Dec 1 '15 at 22:43
  • $\begingroup$ See the addition to my first answer. $\endgroup$ – JJacquelin Dec 2 '15 at 7:58

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