SUG3026 Ordinary Differential Equations Complete Solution

This document provides a detailed solution to an Ordinary Differential Equations assignment, likely part of a coursework for the module SUG3026. The solution covers topics such as solving differential equations, applying partial derivatives, and interpreting mathematical results. The assignment includes problems related to the notation of a car compass using differential equations. The solution demonstrates analytical skills, problem-solving abilities, and proper referencing, including a bibliography. Numerical methods and graphical representations using Python are also incorporated to solve and visualize the differential equations. The solution includes finding complementary solutions, applying initial conditions, and analyzing the behavior of the system as time approaches infinity. Desklib offers a wide array of educational resources, including similar solved assignments and past papers, to aid students in their studies.

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
sinθ >1, then it must be that sin θ ⟶ 1.
Then sin 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 conditions θ=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

0=−k e−k (0 ) ( B+ A (0) ) + A e−k(0)
0=−k B+ A
A=kB
SinceB=−π
k T , then
A=k × (−π
k T )
A=−π
T
Therefore the total solution becomes,
θ=(−π
T t− π
Tk )e−kt + π
Tk
After rearranging,
θ ( t )= π
k T [1−e−kt (1+kt) ]
e)
dθ
dt = π
k T {0− [−k e−kt ( 1+ kt )+k e−kt ] }
dθ
dt = π
k T {0+k e−kt ( 1+kt )−k e−kt }
dθ
dt = π
k T k e−kt ( 1+kt )− π
k T k e−kt
dθ
dt = π
T e−kt [ ( 1+kt )−1 ] ≥ 0
Proved
Now when t tends to infinity then

We know thatθ ( t )= π
k T [1−e−kt (1+kt) ]. Taking its limit as t approaches infinity
yeilds
lim
t ⟶ ∞
θ ( t ) =
lim
t ⟶ ∞
π
k T [ 1−e−kt (1+kt ) ]
¿
lim
t ⟶ ∞
π
k T −
lim
t ⟶ ∞
π
k T e−kt −
lim
t ⟶ ∞
π
T t e−kt
¿ π
k T −0−0
lim
t ⟶ ∞
θ ( t )= π
k T
lim
t ⟶ ∞
θ ( t )= π
Tk
f) If kT >10 π
Then putting the value of kT >10 π in the equation and taking care of the inequality sign we
get,θ ( t )< π
10 π [ 1−e−kt (1+kt) ]
θ ( t )< 1
10 [ 1−e−kt (1+kt ) ]
So it is a good approximation.
2) a) Given, dz
dy =x ,and dy
dx =z2
.
Differentiating the first equation with y
dz
dx =x z2
Now we know that dy
dx =z2 so dx
dy = 1
z2 putting this value in the above equation,
xdx= dz
z2
∫ xdx=∫ dz
z2

x2
2 =−1
z +c
Putting the values (1, 1, 1) in the above equation and finding out the value of c we get
1
2 =−1+c
c=3 /2
x2
2 + 1
z = 3
2
The above diagram is for the equation dz
dy =x
1 1
1.5 1.333
2 1.666
2.5 1.999
3 2.333
3.5 2.666
4 3
4.5 3.333
5 3.666

The table and the graph for the second equation dy
dx =z2 are as follows:
1 1
1.5 1.5
2 2
2.5 2.5
3 3
3.5 3.5
4 4
4.5 4.5
5 5
Only this two graphs will be generated when the y value is modified as the third equation
dz
dx =x z2 won’t be impacted by the change in y value.
b) The graph of the dz
dx =x z2
Are as follows:

And the table is
1 1
1.5 1.5
2 3.0
2.5 10.5
3 120.75
3.5 17131.40625
4 391330571.54296875
Similarly the equation dz
dy =x won’t be valid but rest two graphs will be valid as the change
in x makes an effect on the z and as well as y.
The table and the graph for the second equation dy
dx =z2 are as follows:
1 1
1.5 1.5
2 2
2.5 2.5
3 3
3.5 3.5
4 4
4.5 4.5
5 5

Bibliography
Anynomous, n.d. Numerical Methods Using Python. [Online]
Available at: http://people.bu.edu/andasari/courses/numericalpython/python.html
[Accessed 16 Jan 2019].
equation, D., n.d. Second-Order Linear Differential Equations. [Online]
Available at: https://www.stewartcalculus.com/data/CALCULUS%20Concepts%20and%20Contexts/
upfiles/3c3-2ndOrderLinearEqns_Stu.pdf
[Accessed 16 Jan 2019].
Paul, n.d. Chapter 3 : Second Order Differential Equations. [Online]
Available at: http://tutorial.math.lamar.edu/Classes/DE/IntroSecondOrder.aspx
[Accessed 16 Jan 2019].