Solving Second Order Linear Differential Equations

1) a)
I d2 θ
dt2 =−n2 sin θ−2 μ ( dθ
dt − dφ
dt )
When θis small
tanθ >θ> sinθ
sin θ
cos θ >θ>sin θ Dividing both sides by sin θ yields
1
cos θ > θ
sinθ >1
When θ ⟶ 0 , cos θ ⟶ 1 i.e. lim
θ ⟶ 0
1
cos θ ⟶ 1.
Since1> θ
sin θ > 1, then it must be that sin θ ⟶ 1.
Thensin θ ≅θ,
I d2 θ
dt2 =−n2 θ−2 μ ( dθ
dt − dφ
dt )
Dividing by I,
d2 θ
dt2 =−n2
I θ−2 μ
I ( dθ
dt −dφ
dt )
d2 θ
dt2 + n2
I θ+2 μ
I
dθ
dt =2 μ
I ( dφ
dt )
d2 θ
dt2 +2 μ
I
dθ
dt + n2
I θ=2 μ
I ( dφ
dt )
d2 θ
dt2 +2 λ dθ
dt +k2 θ=2 λ ( dφ
dt )
Where λ= μ
I andk 2=n2
I ,
Where λand k are both constants and are expressed in terms of n, I and μ.
b) From Question μ= √ n2 I eq. (1)
λ= μ
I We have obtained from before eq. (2)

Putting the values of (1) in (2) we get,
λ= √n2 I
I eq. (3)
k 2=n2
I eq. (4)
n2 =k2 I
Therefore,
Putting eq. (4) in eq. (3) we get,
λ= √ k2 II
I
λ=k
c) Since λ=k
The equation becomes
d2 θ
dt2 +2 k dθ
dt + k2 θ=2 k ( dφ
dt )
This is a non-homogeneous equation. When φ=0 the related homogeneous
equation is
d2 θ
dt2 +2 k dθ
dt + k2 θ=0
Taking the LHS and finding the complimentary solution
m2 +2 km+ k2=0
¿
m=−k
This means that the roots of auxiliary equation are equal having value of
−k. Thus, the general solution is
θ= ( At +B ) e−kt eq. (5)
Where A and B are arbitrary constants.
Now as we know the given initial conditions from the question,
dθ(0)
dt =0∧θ ( 0 ) =θ0
Putting the conditions in the equation we get,

Now, consider the initial conditions θ=θ0 , dθ
dt =0 at t=0 to find the A and B.
When t=0 , θ=θ0
θ0 =e−k (0 ) (A (0)+ B)
θ0 =e0 ( B )
B=θ0
Differentiating eq. (5) gives
dθ
dt =−k e−kt ( At +B ) + A e−kt
When dθ
dt =0at t=0 and θ0 =B
0=−k e0 ( θ0 + A ( 0 ) ) + A e0
0=−k θ0 + A
A=k θ0
And A=k θ0
Putting this values in the equation,
θ= ( k θ0 t+θ0 ) e−kt
d)
We are given that φ ( t )= πt
2 T .So,
2 λ dφ
dt =2 λ d
dt ( πt
2 T )
¿ 2 λ × π
2T
d
dt ( t )
¿ 2 λ π
2 T

Noting thatλ=k,
2 k dφ
dt = 2 k π
2T
Let the particular solution be θp =β
d2 θ
dt2 =0
dθ
dt =0
Putting this in to the equation,
0+ 0+k2 ( β ) =2 k πt
2 T
β= π
Tk
Therefore,
θ= ( At +B ) e−kt + π
Tk
Putting the conditionsθ=0, t=0 and dθ
dt =0.
We get,
0=e−k ( 0 ) ( B+ A ( 0 ) ) + π
k T
0=e0 ( B ) + π
k T
0=B+ π
k T
B=−π
k T
Differentiating eq. (8) gives
dθ
dt =−k e−kt ( B + At ) + A e−kt + 0