Solve the given system of differential equations.

Tammy Todd
2021-02-21
Answered

Solve the given system of differential equations.

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krolaniaN

Answered 2021-02-22
Author has **86** answers

The given system of differential equations is,

$Dx+Dy+(D+1)z=0$

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

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

Putting the value of$Dx+y$ in the third equation, we get,

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

$\Rightarrow {e}^{t}-2z=50\mathrm{sin}(2t)$

$\Rightarrow z=\frac{1}{2}{e}^{t}-25\mathrm{sin}(2t)$

Now differentiating both sides with respect to t, we get,

$Dz=\frac{1}{2}{e}^{t}-50\mathrm{cos}(2t)$

$\Rightarrow (D+1)z=Dz+z=(\frac{1}{2}{e}^{t}-50\mathrm{cos}(2t))+\frac{1}{2}{e}^{t}-25\mathrm{sin}(2t)$

$\Rightarrow (D+1)z={e}^{t}-25\mathrm{sin}(2t)-50\mathrm{cos}(2t)$

Putting$Dx={e}^{t}-y$ and the value of $(D+1)z$ in first equation, we get,

${e}^{t}-y+Dy+{e}^{t}-25\mathrm{sin}2t-50\mathrm{cos}2t=0$

$\Rightarrow Dy-y=25\mathrm{sin}2t+50\mathrm{cos}2t-2{e}^{t}$

So the integrating factor of this differential equation is,

$I.F={e}^{\int (-1)dt}$

$={e}^{-t}$

Hence the solution is,

$y{e}^{-t}=\int (25\mathrm{sin}2t+50\mathrm{cos}2t-2{e}^{t}){e}^{-t}dt$

$=\int 25\mathrm{sin}2t{e}^{-t}dt+\int 50\mathrm{cos}2t{e}^{-t}dt-2\int {e}^{t}{e}^{-t}dt$

$=-10{e}^{-t}\mathrm{cos}2t-5{e}^{-t}\mathrm{sin}2t+20{e}^{-t}\mathrm{sin}2t-10{e}^{-t}\mathrm{cos}2t-2t+C$

$\Rightarrow y=-10\mathrm{cos}2t-5\mathrm{sin}2t+20\mathrm{sin}2t-10\mathrm{cos}2t-2t{e}^{t}+C{e}^{t}$

$\Rightarrow y=-20\mathrm{cos}2t+15\mathrm{sin}2t-2t{e}^{t}+C{e}^{t}$

Now putting this value of$y$ in the given second differential equation, we get:$Dx={e}^{t}-y$

$={e}^{t}-(-20\mathrm{cos}2t+15\mathrm{sin}2t-2t{e}^{t}+C{e}^{t})$

$={e}^{t}+20\mathrm{cos}2t-15\mathrm{sin}2t+2t{e}^{t}-C{e}^{t}$

Integrating both sides, we get,

$x=\int ({e}^{t}+20\mathrm{cos}2t-15\mathrm{sin}2t+2t{e}^{t}-C{e}^{t})dt$

$={e}^{t}+10\mathrm{sin}2t+\frac{15}{2}\mathrm{cos}2t+2t{e}^{t}-2{e}^{t}-C{e}^{t}+D$###### Not exactly what you’re looking for?

Putting the value of

Now differentiating both sides with respect to t, we get,

Putting

So the integrating factor of this differential equation is,

Hence the solution is,

Now putting this value of

Integrating both sides, we get,

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