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Question

Consider the differential equation d2x(t)dt2+3dx(t)dt+2x(t)=0
Given x(0)=20 and x(1)=10/e, where e=2.718, the value of x(2) is

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Solution

d2xdt2+3dxdt+2x=0,x(0)=20,x(1)=10C
A.E. is
D2+3D+2=0
D=1,2
x=C1et+C2e2t
x(0)=20
C1+C2=20 .... (1)
x(1)=10e
10e=C1e+C2e2
10=C1+C2e ....(2)
From (1) and (2)
C1=10e20e1,C2=10ee1
x(t)=(10e20e1)et+(10ee1)e2t
x(2)=(10e20e1)e2+(10ee1)e4
=0.8556

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