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\begin{array} { c } { \text { Solve the initial value problem } } \\ { \mathbf { x } ^ { \prime } = \left( \begin{array} { c c } { 1 } & { - 5 } \\ { 1 } & { - 3 } \end{array} \right) \mathbf { x } , \quad \mathbf { x } ( 0 ) = \left( \begin{array} { c } { 1 } \\ { 1 } \end{array} \right) } \\ { \text { and describe the behavior of the solution as } t \rightarrow \infty } \end{array}

My answer so far is: $$\vec{x}=c_1e^{(-1+i)t}\begin{pmatrix}2+i\\1\end{pmatrix}+c_2e^{(-1-i)t}\begin{pmatrix}2-i\\1\end{pmatrix}$$

The textbook has the answer: $$\mathbf{x}(t)=e^{-t}\begin{pmatrix}\cos(t)-3\sin(t)\\ \cos(t)-\sin(t)\end{pmatrix}$$

How can I simplify my solution?

UPDATE: I used the identity $e^{it} = \cos t+i\sin t$ $$c_1e^{-t}(\cos t+i\sin t)\begin{pmatrix}2+i\\1\end{pmatrix}+c_2e^{-t}(\cos t-i\sin t)\begin{pmatrix}2-i\\1\end{pmatrix}$$

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    $\begingroup$ $e^{(-1+i)t}=e^{-t}(\cos t+i\sin t)$ $\endgroup$ – Nosrati Dec 10 '18 at 4:56
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    $\begingroup$ $\sin$ and $\cos$ are bounded and $e^{-t}\to0$ as $t\to\infty$. $\endgroup$ – Nosrati Dec 10 '18 at 5:01
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    $\begingroup$ At first apply initial condition to renove constants $C_1$ and $C_2$. $\endgroup$ – Nosrati Dec 10 '18 at 5:21
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    $\begingroup$ Your solution shouldn't has (have) $i$ term! $\endgroup$ – Nosrati Dec 10 '18 at 5:29
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    $\begingroup$ My method if you interest: $x'=x-5y$ and $y'=x-3y$ then $y''+2y'+2y=0$ shows $y=e^{-t}\cos t$ and $y=e^{-t}\sin t$ $\endgroup$ – Nosrati Dec 10 '18 at 5:44
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Let $$X=e^{(-1+i)t}\begin{pmatrix}2+i\\1\end{pmatrix}$$ $$X_1=\Re(X)=e^{-t}\begin{pmatrix}2 \cos{(t)}-\sin{(t)}\\ \cos{(t)}\end{pmatrix}$$ $$X_2=\Im(X)= e^{-t}\begin{pmatrix}2 \sin{(t)}+\cos{(t)}\\ \sin{(t)}\end{pmatrix}$$ Then general real solution is $$X=c_1X_1+c_2X_2$$ From initial condition we get $$c_1\begin{pmatrix}2\\ 1\end{pmatrix}+c_2\begin{pmatrix}1\\ 0\end{pmatrix}=\begin{pmatrix}1\\ 1\end{pmatrix}$$ $c_1=1$, $c_2=-1$.

Answer: $$X=X_1-X_2=e^{-t}\begin{pmatrix}\cos{(t)}-3 \sin{(t)}\\ \cos{(t)}-\sin{(t)}\end{pmatrix}$$

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$x'=x-5y$ and $y'=x-3y$ then $y''+2y'+2y=0$. Hence by characteristic equation $\lambda^2+2\lambda+2=0$, $\lambda=-1\pm i$. This shows solutions $y=e^{-t}\cos t$ and $y=e^{-t}\sin t$.

Now replacing $x=y'+3y$ gives $x=2e^{-t}\cos t-e^{-t}\sin t$ and $x=e^{-t}\cos t+2e^{-t}\sin t$.

Therefore general solution is \begin{cases} x=C_1(2e^{-t}\cos t-e^{-t}\sin t)+C_2(e^{-t}\cos t+2e^{-t}\sin t),\\ y=C_1e^{-t}\cos t+C_2e^{-t}\sin t. \end{cases} with initial condition $C_1=1$ and $C_2=-1$, we have particular solution \begin{cases} x=e^{-t}\cos t-3e^{-t}\sin t,\\ y=e^{-t}\cos t-e^{-t}\sin t. \end{cases} the behaviour of solution is $|\mathrm{x}|\to0$, because \begin{cases} |x|=|e^{-t}\cos t-3e^{-t}\sin t|\leq\sqrt{10}e^{-t}\to0,\\ |y|=|e^{-t}\cos t-e^{-t}\sin t|\leq\sqrt{2}e^{-t}\to0. \end{cases} as $t\to\infty$.

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  • $\begingroup$ ,Your answer to behaviour of solutions as $t\rightarrow 0$ contains inequalities. Those inequalities contains$\sqrt{10}e^{-t}$ and $\sqrt{2}e^{-t}$.My question is why did you select those terms? $\endgroup$ – Dhamnekar Winod Dec 10 '18 at 15:21

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