Checking solution for a system of ODEs

[SithsNGiggles]

Homework Statement

Find a particular solution to
$\begin{cases}x'=5x+4y+t\\ y'=x+8y-t\end{cases}$
using a variety of methods that are listed. I've been using undetermined coefficients on this problem thus far.

Homework Equations

The Attempt at a Solution

My answer is
$\begin{cases} x(t)= -4C_1e^{4t}+C_2e^{9t}-\frac{1}{3}t-\frac{5}{54}\\ y(t)= C_1e^{4t}+C_2e^{9t}+\frac{1}{6}t+\frac{7}{216} \end{cases}$

I'll show my work in another post. This problem just so happens to have its solution listed in the back of the book, which is
$\begin{cases}x(t)=\frac{1}{5}C_1\left(e^{9t}+e^{4t}\right) + \frac{4}{5}C_2\left(e^{9t}-e^{4t}\right) - \frac{18t+5}{54}\\ y(t) = \frac{1}{5}C_1\left(e^{9t}-e^{4t}\right) + \frac{1}{5}C_2\left(e^{9t}+e^{4t}\right) - \frac{t}{6}+\frac{7}{216}\end{cases}$

I've also checked with WolframAlpha, which gives me a similar result with powers of $e$ factored out, but without the rational coefficients:
http://www.wolframalpha.com/input/?i=x'=5x+4y+t,+y'=x+8y-t

I'm wondering why all these answers may be different. It could be that I'm missing something in my work. Any ideas?

I just don't see how the exponential terms in my answer can be written in the other ways.

 

[SithsNGiggles]

Here's my work. Rewriting the equations in matrix form, I get
$\begin{pmatrix}x\\y\end{pmatrix}' = \begin{pmatrix}5&4\\1&8\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} + t\begin{pmatrix}1\\-1\end{pmatrix}$

Using the usual eigenvalue method, I get a characteristic solution of
$\begin{pmatrix}x\\y\end{pmatrix}_c = C_1e^{4t}\begin{pmatrix}-4\\1\end{pmatrix} + C_2e^{9t}\begin{pmatrix}1\\1\end{pmatrix}$.

Then, using the undetermined coefficients method, as a guess I use
$\begin{pmatrix}x\\y\end{pmatrix}_p=t\vec{a}+\vec{b}$,

which gives me the particular solution
$\begin{pmatrix}x\\y\end{pmatrix}_p=t\begin{pmatrix}-\frac{1}{3}\\\frac{1}{6}\end{pmatrix} + \begin{pmatrix}-\frac{5}{54}\\\frac{7}{216}\end{pmatrix}$

So, my final solution, as before, is
$\begin{cases} x(t)= -4C_1e^{4t}+C_2e^{9t}-\frac{1}{3}t-\frac{5}{54}\\ y(t)= C_1e^{4t}+C_2e^{9t}+\frac{1}{6}t+\frac{7}{216} \end{cases}$