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Find the general solution of $\vec{x^{'}}=\begin{pmatrix} 1&1\\4&1 \end{pmatrix}x+\begin{pmatrix}2\\-1\end{pmatrix}e^{t}$

I found the eigenvalues to be $\lambda_1=3 \;\; \lambda_2=-1$

Next I calculated the eigenvectors to be $e_1=\begin{pmatrix}1\\2 \end{pmatrix}$ and $e_2=\begin{pmatrix}1\\-2 \end{pmatrix}$ Next I form the fundamental matrix $$\phi(t)= \begin{pmatrix} e^{3t}&e^{-t}\\2e^{3t}&-2e^{-t} \end{pmatrix}$$

Finding the inverse yields $\phi^{-1}(t)=\dfrac{1}{4} \begin{pmatrix}2e^{-t}&e^{-t}\\2e^{3t}&-e^{3t} \end{pmatrix}$

I proceeded to multiply the matrix by $g(t)$ to obtain $$\begin{pmatrix} 4-1\\4e^{4t}+e^{4t} \end{pmatrix} \implies \begin{pmatrix} 3\\5e^{4t} \end{pmatrix}$$

I integrated that and multiplied by $\phi(t)$ to get $\dfrac{1}{4} \begin{pmatrix} 3te^{3t}+\frac{5}{4}e^{3t}\\6te^{3t}-\frac{5}{2}e^{3t} \end{pmatrix}$

I left the $\dfrac{1}{4}$ out from the inverse to avoid working with fractions. Now the book gets $x_h+\dfrac{1}{4}\begin{pmatrix}1\\-8 \end{pmatrix}e^{t}$ where $x_h$ is the solution to the homogenous. Is the book wrong or am I wrong somewhere?

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  • $\begingroup$ Yes that is the particular solution they got and I am lost how they got it. We both used the same eigenvalues and vectors so our solution to the homogenous matches $\endgroup$ – adam Apr 24 '14 at 15:24
  • $\begingroup$ Im sorry i forgot to include a $e^{t}$ in the question though it doesn't change the answer just a typo on my part $\endgroup$ – adam Apr 24 '14 at 15:29
  • $\begingroup$ Your right it isn't correct... I forgot to distribute a $e^{-2t}$ My gosh there isn't a problem I can do without making a mistake. $\endgroup$ – adam Apr 24 '14 at 15:31
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It looks like your $\phi^{-1}(t)$ went astray.

I got:

$$\phi^{-1}(t) = \begin{pmatrix} \dfrac{1}{2 e^{3 t}} & \dfrac{1}{4 e^{3 t}} \\ \dfrac{e^{t}}{2} & -\dfrac{e^{t}}{4} \end{pmatrix}$$

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