$\displaystyle \frac{dx}{dt}+y=\sin t$
$\displaystyle \frac{dy}{dt}+x=\cos t$, given $\displaystyle x(0)=2, y(0)=0$
My Attempt:
Taking Laplace transforms on both sides $\displaystyle $ $\displaystyle [s\bar x-2]+\bar y=\frac{1}{s^2+1} $
$\displaystyle [s\bar y-0]+\bar x=\frac{s}{s^2+1} $
$\displaystyle \bar x=\begin{vmatrix} 2 &1 \\ \frac{s}{s^2+1} &s \end{vmatrix}\div \begin{vmatrix} s &1 \\ 1 &s \end{vmatrix}$ Taking inverse, $\displaystyle x=2\cos t$ and $\displaystyle x=-\sin t$
The given answer is: $\displaystyle x=e^t+e^{-t}$ and $\displaystyle x=e^{-t}-e^t+\sin t$
I can't find where I am going wrong. Please help.
We get the following system: $$ \left( \begin{array}{l l} s & 1\\ 1 & s \end{array} \right) \cdot \left( \begin{array}{l} \bar X\\ \bar Y \end{array} \right) = \left( \begin{array}{c} 2 + \frac{1}{s^2 + 1}\\ \frac{s}{s^2+1} \end{array} \right) $$ Then you will have to multiply the left hand side with $$ \left( \begin{array}{l l} s & 1\\ 1 & s \end{array} \right)^{-1} $$ from the left, then inverse transform the expression