If $x(t)$ and $y(t)$ are the general solution of the system of the differential equations: $$\frac{dx}{dt}+\frac{dy}{dt}+2y=\sin t $$ $$\frac{dx}{dt}+\frac{dy}{dt}-x-y=0$$ Then which of the following conditions holds for $x(t)$ and $y(t)$? ($a$ is any arbitrary constant)
(a) $x(t)+y(t)=ae^t$
(b) $x(t)+y(t)=a\sin t$
(c) $x(t)-y(t)=ae^{-t}$
(d) $x(t)-y(t)=ae^t+\sin t$
We can write the above system of equations as: $$Dx+(D+2)y=\sin t \label{eq:1}$$ and $$(D-1)x+(D-1)y=0$$
Now, how to solve for $x$ and $y$ ? The solution says solving above equation for $x$ and $y$ to get:
$(D-1)x=-\frac{\cos t - \sin t}{2}$ and $(D-1)y=\frac{\cos t - \sin t}{2}$
How to solve please let me know. Thanks.
Multiplying the $1$st equation by $(D-1)$ and the $2$nd one by $D$ and then subtracting, we get
\begin{align} \{(D-1)(D+2)-D(D-1)\}y&=\sin t\\ \implies (D-1)y&=\dfrac12\sin t\\ \end{align} So the auxiliary equation is
$m-1=0\implies m=1$
So C.F is $y_c=ce^t$
For P.I, \begin{align} y_p&=\dfrac{1}{D-1}\dfrac12 \sin t\\ &=-\dfrac14 \sin t\\ \end{align} So the general solution is $y=ce^t-\dfrac14 \sin t$
Now $Dy=ce^t-\dfrac14 \cos t$
So from the $2$nd equation, we get \begin{align} &(D-1)x+ce^t-\dfrac14 \cos t-ce^t+\dfrac14 \sin t=0\\ \implies &(D-1)x=\dfrac14(\cos t-\sin t)\\ \end{align} The auxiliary equation is
$m-1=0\implies m=1$
So C.F is $x_c=de^t$
For P.I, \begin{align} x_p&=\dfrac{1}{D-1}\dfrac14(\cos t-\sin t)\\ &=\dfrac18 (\sin t-\cos t)\\ \end{align} So the general solution is $x=de^t+\dfrac18 (\sin t-\cos t)$