Solve the given system of differential equations.[Dx+Dy+(D+1)z=0Dx+y=e^{t}Dx+y-2z=50sin(2t)

The given system of differential equations is,
$Dx+Dy+(D+1)z=0$
$Dx+y=e^t$
$Dx+y-2z=50\sin (2t)$
Putting the value of $Dx+y$ in the third equation, we get,
$Dx+y-2z=50\sin (2t)$
$\Rightarrow e^t-2z=50\sin (2t)$
$\Rightarrow z=\frac{1}{2}{e}^t-25\sin (2t)$
Now differentiating both sides with respect to t, we get,
$Dz=\frac{1}{2}{e}^t-50\cos (2t)$
$\Rightarrow (D+1)z=Dz+z=(\frac{1}{2}{e}^t-50\cos (2t))+\frac{1}{2}{e}^t-25\sin (2t)$
$\Rightarrow (D+1)z=e^t-25\sin (2t)-50\cos (2t)$
Putting $Dx=e^t-y$ and the value of $(D+1)z$ in first equation, we get,
$e^t-y+Dy+e^t-25\sin 2t-50\cos 2t=0$
$\Rightarrow Dy-y=25\sin 2t+50\cos 2t-2e^t$
So the integrating factor of this differential equation is,
$I.F=e^{\int (-1)dt}$
$=e^{-t}$
Hence the solution is,
$y{e}^{-t}=\int (25\sin 2t+50\cos 2t-2e^t)e^{-t}dt$
$=\int 25\sin 2t{e}^{-t}dt+\int 50\cos 2t{e}^{-t}dt-2\int e^t{e}^{-t}dt$
$=-10e^{-t}\cos 2t-5e^{-t}\sin 2t+20e^{-t}\sin 2t-10e^{-t}\cos 2t-2t+C$
$\Rightarrow y=-10\cos 2t-5\sin 2t+20\sin 2t-10\cos 2t-2t{e}^t+C{e}^t$
$\Rightarrow y=-20\cos 2t+15\sin 2t-2t{e}^t+C{e}^t$
Now putting this value of $y$ in the given second differential equation, we get:$Dx=e^t-y$
$=e^t-(-20\cos 2t+15\sin 2t-2t{e}^t+C{e}^t)$
$=e^t+20\cos 2t-15\sin 2t+2t{e}^t-C{e}^t$
Integrating both sides, we get,
$x=\int (e^t+20\cos 2t-15\sin 2t+2t{e}^t-C{e}^t)dt$
$=e^t+10\sin 2t+\frac{15}{2}\cos 2t+2t{e}^t-2e^t-C{e}^t+D$

Not exactly what you’re looking for?

Ask My Question

This is helpful 78