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 January 3rd, 2010, 08:32 PM #1 Newbie   Joined: Nov 2009 Posts: 10 Thanks: 0 Another system of differential equations $\begin{bmatrix}y'_{1}\\y'_{2}\\y'_{3}\ end{bmatrix}=\begin{bmatrix}2&0&0\\0&6&-4\\0&4&-2\end{bmatrix}\begin{bmatrix}y_{1}\\y_{2}\\y_{3}\e nd{bmatrix}+\begin{bmatrix}e^{2t}\cos{3t}\\-2\\-2\end{bmatrix}$ How do you solve the problem above? I tried solving it and I ended up with weird eigenvalues/vectors...
 January 4th, 2010, 12:37 AM #2 Global Moderator   Joined: Dec 2006 Posts: 21,114 Thanks: 2329 $y'_1\,=\,2y_1\,+\,e^{2t}\cos{3t}\text{ (which you can solve).}$ $\begin{bmatrix}y'_{2}\\y'_{3}\end{bmatrix} \,=\,\begin{bmatrix}6&-4\\\\\\\\\\4&-2\end{bmatrix}\begin{bmatrix}y_{2}\\y_{3}\end{bmat rix}\,+\,\begin{bmatrix}-2\\\\\\\\\\-2\end{bmatrix}\text{ (which is like the problem you posted earlier).}$ $\text{Substituting }\begin{bmatrix}y_2\\y_3\end{bmatrix}\,=\,e^{2t}\b egin{bmatrix}u_2\\u_3\end{bmatrix}\,+\,\begin{bmat rix}1\\\\\\\\1\end{bmatrix}\text{ gives }\begin{bmatrix}u'_{2}\\\\u'_{3}\end{bmatr ix}\,=\,\begin{bmatrix}4&-4\\\\\\\\\\4=&-4\end{bmatrix}\begin{bmatrix}u_{2}\\\\u_{3}\end{bm atrix}\text{ (so }u'_2\,=\,u#39;_3\,=\,\text{a\,constant).}$
 January 5th, 2010, 12:08 AM #3 Newbie   Joined: Nov 2009 Posts: 10 Thanks: 0 Re: Another system of differential equations Oh lol! I totally forgot about multiplying them and seperating them as equations! Thanks! (oh, and forgot to thank you for the other problem)

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