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bmxicle

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## Homework Statement

[tex] \mathbb{x'}= \begin{bmatrix} 1 & 1 \\ 4 & 1 \end{bmatrix} \mathbf{x} \ + \ \begin{bmatrix} 2e^{t} \\ -e^{t} \end{bmatrix} [/tex]

Find the general solution.

## Homework Equations

## The Attempt at a Solution

Well i found the eigenvalues of the matrix That i'll call A, to get the solution to the homogenous equation:

[tex] \mathbf{x_h} = c_1\begin{bmatrix} 1 \\ 2 \end{bmatrix} e^{3t} + c_2 \begin{bmatrix} -1 \\ 2 \end{bmatrix} e^{-t}[/tex]

Let F be the fundamental matrix such that

[tex] F = \begin{bmatrix} e^{3t} & -e^{-t} \\ 2e^{3t} & 2e^{-t}\end{bmatrix} [/tex]

[tex] \mathbf{x'} = A\mathbf{x} + \mathbf{g}[/tex]

If we assume a solution [tex] \mathbf{x_{p}} = F\mathbf{u}[/tex]

Then after a few calculations we get

[tex]\mathbf{u'} = F^{-1}\mathbf{g}[/tex]

[tex]\mathbf{u'} = \frac{1}{4e^{4t}} \begin{bmatrix} 2e^{-t} & e^{-t} \\ -2e^{3t} & e^{3t} \end{bmatrix}* \begin{bmatrix} 2e^t \\ -e^t \end{bmatrix}[/tex] [tex]

\mathbf{u'} = \begin{bmatrix} 16e^{-4t} -4e^{-4t} \\-16 - 4 \end{bmatrix} = \begin{bmatrix} 12e^{-4t} \\ -20 \end{bmatrix} [/tex]

[tex] u_{1} = \int 12e^{-4t} dt = -3e^-4t + c [/tex]

[tex] u_{2} = \int -20 dt = -20t + c [/tex]

[tex] \mathbf{x_{p}} = F\mathbf{u} = \begin{bmatrix} e^{3t} & -e^{-t} \\ 2e^{3t} & 2e^{-t}\end{bmatrix} * \begin{bmatrix} -3e^{-4t} \\ -20t \end{bmatrix} [/tex]

[tex] \mathbf{x_{p}} =\begin{bmatrix} -3e^{-t} + 20e^{-t} \\ - 6 e^{-t} -40te^{-t} \end{bmatrix}[/tex]

So then to get the final answer you just add the particular solution X_p to the homogeneous solution.

I've been working on a bunch of these different non-homogeneous equations, and i keep getting the wrong answer basically every time, so If someone sees where i might be going wrong it would be much appreciated.