# Non-Homogenous System answer not matching

Find the Genereal Solution of $\vec{x}^{'}=\begin{pmatrix}2&-1\\3&-2 \end{pmatrix}\vec{x}+\begin{pmatrix}1\\-1 \end{pmatrix}e^t$

I found the eigenvalues to be $\lambda=\pm 1$

Therefore the eigenvectors are $e_1=\begin{pmatrix}1\\1 \end{pmatrix}$ and $e_2=\begin{pmatrix} 1\\3 \end{pmatrix}$

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

Next I multiply by $g(t)$ $$\dfrac{1}{2}\begin{pmatrix} 3e^{-t}& -e^{-t}\\-e^{t}&e^{t} \end{pmatrix}\begin{pmatrix}e^t\\-e^t \end{pmatrix}=\begin{pmatrix} 3e^{-t}e^{t}+e^{-t}e^{t}\\-e^{t}e^{t}-e^{t}e^{t} \end{pmatrix}=\begin{pmatrix}4\\-2e^{2t} \end{pmatrix}$$

Taking the integral yields $$\begin{pmatrix} 4t\\-e^{2t} \end{pmatrix}$$

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

Now I write $x_h + 2\begin{pmatrix}1\\1 \end{pmatrix}te^{t}-\dfrac{1}{2}\begin{pmatrix} 1\\3 \end{pmatrix}e^{t}$

The book got $x_h + 2\begin{pmatrix}1\\1 \end{pmatrix}te^{t}+\begin{pmatrix} 1\\0 \end{pmatrix}e^{t}$

Where are they getting the $\begin{pmatrix}1\\0 \end{pmatrix}$ from?

Hint:

How about that $\dfrac{1}{2}$ term when you multiply by $g(t)$?

Update

• do you think it has something to do with $c_1 \begin{pmatrix}1\\1 \end{pmatrix}e^t$? I mean did they just omit certain terms to combine them as a constant to make the solution look "nice"? – adam Apr 24 '14 at 16:58
• wait so what is $x^{'}$ and $y^{'}$? – adam Apr 24 '14 at 17:31
• The top row of the matrix is $x(t)$? – adam Apr 24 '14 at 17:40