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[tex]X'\,=\,\left(\begin{array}{cc} 0 & 2 \\ -1 & 3 \end{array}\right)\,X\,+\left(\begin{array}{c} 2 \\ e^{-3t}\end{array}\right)[/tex]

Here is what I have:

[tex]r_1\,=\,2,\,\,\,\,\,\,r_2\,=\,1[/tex]

[tex]u_1'\,=\,-2\,e^{-2t}\,+2\,e^{-5t}[/tex]

[tex]u_2'\,=\,2\,e^{-t}\,-\,e^{-4t}[/tex]

[tex]\psi\,=\,\left(\begin{array}{cc} e^{2t} & 2\,e^t \\ e^{2\,t} & e^t \end{array}\right)[/tex]

After integrating the [itex]u_1[/itex] and [itex]u_2[/itex] terms and multiplying the answer by [itex]\psi[/itex], I get:

[tex]X\,=\,\left(\begin{array}{c} C_1\,e^{2t}\,+\,2\,C_2\,e^t\,+\,\frac{1}{10}\,e^{-3t}\,-\,3 \\ C_1\,e^{2t}\,+\,C_2\,e^t\,-\,\frac{3}{20}\,e^{-3t}\,-\,1 \end{array}\right)[/tex]

Then seperating and combining terms:

[tex]X\,=\,C_1\,e^{2t}\,\left(\begin{array}{c} 1 \\ 1 \end{array}\right)\,+\,C_2\,e^t\,\left(\begin{array}{c} 2 \\ 1 \end{array}\right)\,+\,\frac{1}{20}\,\left(\begin{array}{cc} 2\,e^{-3t}\,-\,60 \\ 3\,e^{-3t}\,-\,20 \end{array}\right)[/tex]

Does this look correct? Thanks.

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# Homework Help: Diff. EQ. Inhomogeneous Linear System (Double check please)

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