I need to solve the following using Laplace transformations (and inverses):
$x' + 3x + y' = 1$
$x' - x + y' - y = e^t$
where:
$x(0)=y(0)=0$
Using Laplace transformation I got following equations:
$sX + 3X + sY = 1$
$sX - X + sY - Y = \frac{1}{(s-1)}$
where:
$X = L(x(t))(s)$
and:
$Y = L(y(t))(s)$
$\rightarrow (s+3)X + sY = 1$
$(s-1)X + (s-1)Y = \frac{1}{(s-1)}$
By subtracting I now have:
$(s+3)X + sY = 1$
$4X + Y = \frac{(s-2)}{(s-1)}$
Multiplying first equation by $(\frac{-1}{s})$ and adding it to second one I ended up with:
$((\frac{-s-3)}{s})+4)X = \frac{(s-2)}{(s-1)} - \frac{1}{s}$
Finally I have:
$X = \frac{1}{3} - \frac{1}{3(s-1)^2} - \frac{1}{3(s-1)}$
I now want to use inverse Laplace to $x(t)$ but we didn't define what is Laplace inverse of a constant (for $1/3$ in this case).
Once I have $x(t)$ I know how to finish a task, so my question is did I do something wrong so far in this task and if not what is Laplace inverse of $1/3$?
$$x' + 3x + y' = 1$$ Taking Laplace transform gives: $$sX(s)-x(0)+3X(s)+sY(s)-y(0)=\frac 1 s$$ $$sX(s)+3X(s)+sY(s)=\frac 1 s$$ $$x' - x + y' - y = e^t$$ Taking Laplace transform again gives: $$sX(s)-x(0)-X(s)+sY(s)-y(0)-Y(s)=\frac 1 {s-1}$$ $$sX(s)-X(s)+sY(s)-Y(s)=\frac 1 {s-1}$$ So we have the following system: $$ \begin{cases} (s+3)X(s)+sY(s)=\dfrac 1 s \\ X(s)+Y(s)=\dfrac 1 {(s-1)^2} \\ \end{cases} $$ $$ \begin{cases} X(s)=\frac 1 3 \left (\dfrac 1 s -\dfrac s {(s-1)^2} \right ) \\ Y(s)=\dfrac 1 {(s-1)^2} -X(s)\\ \end{cases} $$ Taking inverse Laplace Transform we have: $$ \begin{cases} x(t)=\dfrac 1 3 \left (1-e^t-te^t \right ) \\ y(t)=\dfrac 1 3(4te^t-1+e^t) \\ \end{cases} $$