d2xdt2+3dxdt+2x=0,x(0)=20,x(1)=10C
A.E. is
D2+3D+2=0
D=−1,−2
x=C1e−t+C2e−2t
x(0)=20
⇒ C1+C2=20 .... (1)
x(1)=10e
⇒ 10e=C1e+C2e2
⇒ 10=C1+C2e ....(2)
From (1) and (2)
C1=10e−20e−1,C2=10ee−1
x(t)=(10e−20e−1)e−t+(10ee−1)e−2t
x(2)=(10e−20e−1)e−2+(10ee−1)e−4
=0.8556