1.3 Solutions
59
The complete solution is
y = U + V
(4)
where V = C 3 x + C 4
(5)
In order that V be a particular solution of (1), substitute y = C 3 x + C 4 , in
(1) in order to determine C 3 and C 4 .
−5C 3 + 6(C 3 x + C 4 ) = x
Equating the like coefficients
6C 3 = 1 → C 3 = 1/6
6C 4 − 5C 3 = 0 → C 4 = 5/36
Hence the complete solution is
y = C 1 e
2x
+ C 2 e
3x
+
x
6
+
5
36
1.60
d
2 x
dt 2 + 2
dx
dt
+ 5x = 0
( 1 )
Put x = e
λt
,
dx
dt
= λe
λt
,
d
2 x
dt 2 = λ
2 e
λt in (1)
λ
2
+ 2λ + 5 = 0
( 2 )
its roots being, λ = −1 ± 2i
x = Ae
−t(1−2i)
+ Be
−t(1+2i)
x = e
−t [C cos 2t + D sin 2t]
where C and D are constants to be determined from the initial conditions.
At t = 0, x = 5. Hence C = 5. Further
dx
dt
= −e
−t [C cos 2t + D sin 2t + 2C sin 2t − 2D cos 2t]
At t = 0, dx/dt = −3
−3 = −C + 2D = −5 + 2D
whence D = 1. Therefore the complete solution is
x = e
−t (5 cos 2t + sin 2t)
1.61 (a) Let the mass 1 be displaced by x 1 and mass m 2 by x 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
m ¨
x 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 )