I have the system: \begin{align} 2\frac{dx}{dt} + \frac{dy}{dt} - 2x &= 1 \\ \frac{dx}{dt} + \frac{dy}{dt} - 7x-7y &=2 \end{align}
$y(0)=0, x(0)=0$
Which I am attempting to solve using Laplace Transform's. I change the system to
$2sx+sy-2x=1$
$sx-sy-7x-7y=2$
Solving I get $x=-{(s+7)\over(s^2-9s+14)}$ and $y={3(s+1)\over(s^2-9s+14)}$.
Plugging in the inverse laplace transforms for these I get $x=\frac{14}{5}e^{7t}+\frac{9}{5}e^{2t}$ and $y=\frac{24}{5e}^{7t} - \frac{9}{5}e^{2t}$. Which is evidently wrong.
Does anyone see what I did wrong here?
Using the standard Laplace transform then the equations become \begin{align} 2(s-1) \, \overline{x} + s \overline{y} &= \frac{1}{s} \\ (s-7) \overline{x} + (s-7) \overline{y} &= \frac{2}{s}. \end{align} Solving for $\overline{x}$ and $\overline{y}$ then \begin{align} \overline{x} &= - \frac{s+7}{s (s-2) (s-7)} \\ \overline{y} &= \frac{3 (s+1)}{s (s-2) (s-7)} \end{align} and lead to \begin{align} x(t) &= \frac{9}{5} \, e^{2 t} - \frac{2}{5} \, e^{7 t} - \frac{1}{2} \\ y(t) &= \frac{24}{35} \, e^{7 t} - \frac{9}{10} \, e^{2 t} + \frac{3}{14}. \end{align}
Note: The Laplace transform is defined by $$\mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{- s t} f(t) \, dt,$$ in shorthand notation $f(t) \doteqdot f(s)$, and \begin{align} \frac{dx}{dt} &\doteqdot s \overline{x} - x(0) \\ a &\doteqdot \frac{a}{s} \hspace{5mm} \text{$a$ is a constant} \\ e^{a t} &\doteqdot \frac{1}{s - a} \end{align}