1.3 Solutions 59
The complete solution is
y=U+V (4)
whereV=C 3 x+C 4 (5)
In order thatVbe a particular solution of (1), substitutey=C 3 x+C 4 ,in
(1) in order to determine C 3 and C 4.
− 5 C 3 +6(C 3 x+C 4 )=x
Equating the like coefficients
6 C 3 = 1 →C 3 = 1 / 6
6 C 4 − 5 C 3 = 0 →C 4 = 5 / 36
Hence the complete solution isy=C 1 e^2 x+C 2 e^3 x+x
6+
5
36
1.60
d^2 x
dt^2+ 2
dx
dt+ 5 x=0(1)Putx=eλt,ddxt=λeλt,d(^2) x
dt^2 =λ
(^2) eλtin(1)
λ^2 + 2 λ+ 5 =0(2)
its roots being,λ=− 1 ± 2 i
x=Ae−t(1−^2 i)+Be−t(1+^2 i)
x=e−t[Ccos 2t+Dsin 2t]
where C and D are constants to be determined from the initial conditions.
Att= 0 ,x=5. HenceC=5. Further
dx
dt
=−e−t[Ccos 2t+Dsin 2t+ 2 Csin 2t− 2 Dcos 2t]
Att=0, dx/dt=− 3
− 3 =−C+ 2 D=− 5 + 2 D
whenceD=1. Therefore the complete solution is
x=e−t(5 cos 2t+sin 2t)
1.61 (a) Let the mass 1 be displaced byx 1 and massm 2 byx 2. The force due to the
spring on the left acting on mass 1 is−kx 1 and that due to the coupling is
−k(x 1 −x 2 ).
The net force
F 1 =−kx 1 −k(x 1 −x 2 )=−k(2x 1 −x 2 )
The equation of motion for mass 1 is
mx ̈ 1 +k(2x 1 −x 2 )=0(1)
Similarly, for mass 2, the spring on the right exerts a force−kx 2 , and the
coupling spring exerts a force−k(x 2 −x 1 ). The net force
F 2 =−kx 2 −k(x 2 −x 1 )=−k(2x 2 −x 1 )