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+(D+1)z=0

Dx+y=et

Dx+y2z=50sin(2t)

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Expert Answer

krolaniaN
Answered 2021-02-22 Author has 86 answers
The given system of differential equations is,
Dx+Dy+(D+1)z=0
Dx+y=et
Dx+y2z=50sin(2t)
Putting the value of Dx+y in the third equation, we get,
Dx+y2z=50sin(2t)
et2z=50sin(2t)
z=12et25sin(2t)
Now differentiating both sides with respect to t, we get,
Dz=12et50cos(2t)
(D+1)z=Dz+z=(12et50cos(2t))+12et25sin(2t)
(D+1)z=et25sin(2t)50cos(2t)
Putting Dx=ety and the value of (D+1)z in first equation, we get,
ety+Dy+et25sin2t50cos2t=0
Dyy=25sin2t+50cos2t2et
So the integrating factor of this differential equation is,
I.F=e(1)dt
=et
Hence the solution is,
yet=(25sin2t+50cos2t2et)etdt
=25sin2tetdt+50cos2tetdt2etetdt
=10etcos2t5etsin2t+20etsin2t10etcos2t2t+C
y=10cos2t5sin2t+20sin2t10cos2t2tet+Cet
y=20cos2t+15sin2t2tet+Cet
Now putting this value of y in the given second differential equation, we get: Dx=ety
=et(20cos2t+15sin2t2tet+Cet)
=et+20cos2t15sin2t+2tetCet
Integrating both sides, we get,
x=(et+20cos2t15sin2t+2tetCet)dt
=et+10sin2t+152cos2t+2tet2etCet+D
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