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

Tammy Todd 2021-02-21 Answered

Solve the given system of differential equations.
$Dx+Dy+\left(D+1\right)z=0$

$Dx+y={e}^{t}$

$Dx+y-2z=50\mathrm{sin}\left(2t\right)$

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## Expert Answer

krolaniaN
Answered 2021-02-22 Author has 86 answers
The given system of differential equations is,
$Dx+Dy+\left(D+1\right)z=0$
$Dx+y={e}^{t}$
$Dx+y-2z=50\mathrm{sin}\left(2t\right)$
Putting the value of $Dx+y$ in the third equation, we get,
$Dx+y-2z=50\mathrm{sin}\left(2t\right)$
$⇒{e}^{t}-2z=50\mathrm{sin}\left(2t\right)$
$⇒z=\frac{1}{2}{e}^{t}-25\mathrm{sin}\left(2t\right)$
Now differentiating both sides with respect to t, we get,
$Dz=\frac{1}{2}{e}^{t}-50\mathrm{cos}\left(2t\right)$
$⇒\left(D+1\right)z=Dz+z=\left(\frac{1}{2}{e}^{t}-50\mathrm{cos}\left(2t\right)\right)+\frac{1}{2}{e}^{t}-25\mathrm{sin}\left(2t\right)$
$⇒\left(D+1\right)z={e}^{t}-25\mathrm{sin}\left(2t\right)-50\mathrm{cos}\left(2t\right)$
Putting $Dx={e}^{t}-y$ and the value of $\left(D+1\right)z$ in first equation, we get,
${e}^{t}-y+Dy+{e}^{t}-25\mathrm{sin}2t-50\mathrm{cos}2t=0$
$⇒Dy-y=25\mathrm{sin}2t+50\mathrm{cos}2t-2{e}^{t}$
So the integrating factor of this differential equation is,
$I.F={e}^{\int \left(-1\right)dt}$
$={e}^{-t}$
Hence the solution is,
$y{e}^{-t}=\int \left(25\mathrm{sin}2t+50\mathrm{cos}2t-2{e}^{t}\right){e}^{-t}dt$
$=\int 25\mathrm{sin}2t{e}^{-t}dt+\int 50\mathrm{cos}2t{e}^{-t}dt-2\int {e}^{t}{e}^{-t}dt$
$=-10{e}^{-t}\mathrm{cos}2t-5{e}^{-t}\mathrm{sin}2t+20{e}^{-t}\mathrm{sin}2t-10{e}^{-t}\mathrm{cos}2t-2t+C$
$⇒y=-10\mathrm{cos}2t-5\mathrm{sin}2t+20\mathrm{sin}2t-10\mathrm{cos}2t-2t{e}^{t}+C{e}^{t}$
$⇒y=-20\mathrm{cos}2t+15\mathrm{sin}2t-2t{e}^{t}+C{e}^{t}$
Now putting this value of $y$ in the given second differential equation, we get: $Dx={e}^{t}-y$
$={e}^{t}-\left(-20\mathrm{cos}2t+15\mathrm{sin}2t-2t{e}^{t}+C{e}^{t}\right)$
$={e}^{t}+20\mathrm{cos}2t-15\mathrm{sin}2t+2t{e}^{t}-C{e}^{t}$
Integrating both sides, we get,
$x=\int \left({e}^{t}+20\mathrm{cos}2t-15\mathrm{sin}2t+2t{e}^{t}-C{e}^{t}\right)dt$
$={e}^{t}+10\mathrm{sin}2t+\frac{15}{2}\mathrm{cos}2t+2t{e}^{t}-2{e}^{t}-C{e}^{t}+D$
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