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ME5112 Math’s Assessment
Student Name
Institution
Date of Submission

1. Solving the differential equation
i.
y dy
dx − ( 1+ y2 ) x2=0
dy
dx = ( 1+ y2 ) x2
y
ii. dx
dt − x
t−1 =t2−1
dx
dt =t2−1+ x
t−1
When t >1
dx
dt > ( 2−1 )+ x
2−1
Hence dx
dt >0
2. Using the Maclaurin series
In this problem will apply the first four non-zero terms of the McLaurin’s series to
approximate f ( x ) =ln (1+2 x) and thereafter obtain the value of f ( 1.2 )to 4 decimal
places
The McLaurin’s series formula is given as;
f ( 0 )+ f '(0)
1 ! x + f ' '(0)
2! x2 + f ' ' ' (0)
3! x3+ f iv (0)
4 ! x4 … f n ( 0 )
n! xn
f ( x )=ln (1+2 x)
f ( 0 )=ln ( 1+2∗0 )=ln (1 )=0
f i ( x )= d
dx ln ( 1+2 x )= 2
1+ 2 x
f i ( 0 )= 2
1+ 2 x =2
f ii ( x ) = d
dx f i ( x )
f ii ( x ) = d
dx
2
1+2 x = −4
( 1+2 x ) 2

f ii ( 0 ) = 4
( 1+2∗0 )2 =−4
f iii ( x ) = d
dx
−4
( 1+2 x )2 = 16
( 1+2 x ) 2
f iii ( 0 )= 16
( 1+2∗0 )2 =16
f iv ( x ) = d
dx
16
( 1+2 x ) 2 = −96
( 1+2 x ) 4
f iv ( 0 )= −96
( 1+2∗0 )4 =−96
Now replacing the values in the series expansion formula we obtain
f ( 0 ) + f ' (0)
1 ! x + f ' '( 0)
2! x2 + f ' ' ' (0)
3! x3+ f iv( 0)
4 ! x4=0+ 2
1 ! x− 4
2 ! x2 + 16
3 ! x3− 96
4 x4
¿ 2 x−2 x2 + 8
3 x3−4 x4
this is theMcLaurin’s series expansion of
f ( x )=ln (1+2 x)
At f ( 1.2 ) we have 2 ( 1.2 ) −2 ( 1.2 ) 2 + 8
3 ( 1.2 ) 3−4 ( 1.2 ) 4 =−4.166
3. Taylor’s series
Using the first 5 terms of the Taylors series to estimate the value of
f ( 1.1 ) ¿ 2 decimal Places.
The formula is given by:
f ( x ) =f ( a)+ f ' ( a )
1! ( x−a)+ f ' ' (a)
2 ! (x−a)2 + f ' ' '(a)
3 ! (x−a)3 + f iv (a)
4 ! ( x−a)4 … f n ( a )
n ! ( x −a)n ¿
For the given problem the value of a is 1with this the above formula can be expressed as
f ( x )=f (1)+ f ' (1 )
1! (x−1)+ f ' '(1)
2 ! ( x−1)2 + f ' ' ' (1)
3 ! ( x−1)3+ f iv (1)
4 ! ( x −1)4 … f n ( 1 )
n ! ( x−1)n ¿
Now

f ( a )=f ( 1 )=1
1 =1
f ' ( a )=f ' ( 1 )= d
dx x−1=−1 x−2=−1
f ' ' ( a ) =f ' ' ¿
f '' ' ( a )=f ' '' (1 )= d
dx (2 x−3 ) =−6 x−4 =−6
f iv ( a ) =f iv ( 1 ) = d
dx (−6 x−4 ) =24 x−5=24
Replacing the obtained values in the previously stated formula gives
f ( x )=1− 1
1 ! ( x−1 ) + 2
2 ! ( x−1 )2− 6
3 ! ( x −1)3 + 24
4 ! ( x−1)4
which can Be simplified to
1− ( x−1 ) + ( x −1 ) 2− ( x−1 ) 3 + ( x−1 ) 4
This is the Taylor’s series expansion
Now obtaining
f ( 1.1 ) =1− ( 1.1−1 ) + ( 1.1−1 )2− ( 1.1−1 )3 + ( 1.1−1 ) 4
¿ 1−1.1+0.12 −0.13+ 0.14=0.9091
4. Integral
i.
∫
0
1
1
1+ x2 =¿∫
0
1
( 1+ x2 ) −1
¿
Let u=1+ x2
Then ∫
0
1
u−1 dx
du
dx =2 x
dx= du
2 x

¿
∫
0
1
( u )−1∗1
2 x du=∫
0
1 1
u ∗1
2 x du= 1
2 x lnu
¿ 1
2 ln ( 1+12 )= π
4
ii) Binomial series
f ( x )= 1
1+ x2
The formula is stated as
( 1+x ) k=1+kx + k ( k −1 )
2 ! x2+ k ( k −1 ) ( k−2 )
3 ! x3
k =2
5. Eigenvalues and eigenvectors
A=[6 2
2 3]
Eigenvlues
Given A to be a square matrix, v a vector and λ is a scalr tht satisfy the formula Av=λv .
λ Is defines as the eigenvalue with eigenvector v of A.
hence theEigenvalues of will be obtained from the roots of the equation det ( A−λI ) =0
Hence
det ( (6 2
2 3 )−λ (1 0
0 1) ): λ2−9 λ+14
Solving for λ2−9 λ+14=0
Using the quadratic equation we obtain the values of λ as 2nd 7
This are the eigenvalues of matrix A
Eigenvectors
Being that A is a square matrix, v vector and λa scalar satisfying the eqution Av=λv ,
therefore v is an eigenvector of A
The eigenvalues as obtained above are 2 and 7.

Therefore the eigenvectors of λ=2 ;
Solving ( A−2 I ¿=[6 2
2 3 ]−2 (1 0
0 1 )= (4 2
2 1)
Solving ( 4 2
2 1 )( x
y ) =( 0
0 )
By reducing (4 2
2 1 )we obtin (1 1
2
0 0 )
Hence ( A−2 I ¿ ( x
y )= ( 1 1
2
0 0 ) ( x
y ) =( 0
0 )
This reduces to x + 1
2 y=0 , if you let y be 1 weobtain
(
−1
2
1
)
And the eigenvectors for λ=7 :
Solving ( A−2 I ¿=[6 2
2 3 ]−7 (1 0
0 1)=(−1 2
2 −4 )
(−1 2
2 −4 )( x
y )=(0
0 )
reducing (−1 2
2 −4 ) gives ( 1 −2
0 0 )
gives theSystem ( A−7 I ¿ ( x
y )= (1 −2
0 0 )( x
y )=(0
0 )
Which gives the equation x−2 y=0. isolate x=2 y nd plug ∈( x
y )
If you let y=1 , you obtain (2
1 )

The eigenvector for the matric A is thereby given by ( −1
2
1 ) ,(2
1)
6. Newton’s law of cooling
Describe Euler’s method
Application
Euler’s method is a tool that is applied in numerical arithmetic to approximate values of
solutions involving differential equations.
It applies the formula yn= yn−1 +hF ( xn−1 , yn−1 ) , where y0 isthe initial value of y and h
the step size.
Formula
yn= yn−1 +hF (xn−1 , yn −1 )
dT
dt =−1
3 T +20−10 e−t
The initial starting point is 60 with a step size of 0.5.
We therefore go from x : 0 ,0.5 , 1,1.5,2,2.5,3
Now
y1= y0 +hF(x0 , y0)
¿ 60+0.5 ( −10 )
55
y2= y1 +hF (x1 , y1 )
¿ 55+0.5∗−4.3986=52.8
y3= y2 +hF (x2 , y2 )
¿ 52.8+0.5∗(−1.2788 )=52.16
y4 = y3 +hF ( x3 , y3 )
¿ 52.16+0.5∗( 0.3820 ) =52.35
y5= y 4 +hF ( x4 , y4 )
¿ 52.35+0.5∗( 1.1966 )=52.95

y6= y5+ hF(x5 , y5)
52.95+0.5∗( 1.5292 )=53.71
y7= y6+hF(x6 , y6)
¿ 53.71+0.5∗( 1.5988 ) =54.509
Hence the value of T (3) as approximated by the Euler’s method is 53.71℃
7. Graphs
a. Plotting the data in the table
V
(m^3) P(kPa)
1 520
2 350
3 237
4 140
5 98
6 73
0 1 2 3 4 5 6 7 8
0
100
200
300
400
500
600
A line graph of Pressure against Volume of a gas
Volume (m^3)
Pressure (kPa)

b. P=717 e−0.377 v
W is given by the formula∫
v 1
v 2
Pdv
Hence
∫
1
7
717 e−0.377 v dv
¿ 717∫
1
7
e−0.377 v dv
¿ 717 [ 1
−0.377 e−0.377 v
],¿ 1 ¿ 7
this gives 717 ¿]
¿ 717∗1.62993=1168.6598
W =1168.6598
c. Using Simpson’s rule
Formula
∫
a
b
f ( x ) dx ≈ Sn
Sn= ∆ x
3 [ f ( x0 ) +4 f ( x1 ) +2 f ( x2 ) +4 f ( x3 ) +2 f ( x4 ) +… f ( xn ) ]
∆ x= b−a
n = 7−1
6 =1
since n=6
Therefore
S6 =1
3 ¿]
¿ 1
3 [520+ 4 ( 350 ) +2 ( 237 ) +4 ( 140 ) +2 ( 98 ) +4 ( 73 ) +59]
¿ 1
3 [ 520+1400+474 +560+196+292+59 ] =1
3∗3501≈ 1167
Trapezium rule
Formula

∫
a
b
f ( x ) dx ≈ ∆ x
2 [f ( x0 ) +2 f ( x1 ) +2 f ( x2 )+2 f ( x3 ) + 2 f ( x4 ) +… f ( xn ) ]
∆ x= 7−1
6 =1
≈ 1
2 [ 520+2 ( 350 ) + 2 ( 237 ) + 2 ( 140 ) +2 ( 98 ) +2 ( 73 ) +59 ] =1
2∗2375
≈ 1187.5
d. Comparing the results
Using integration
W =1168.6598
Using Simpson’s rule
W =1167
Using Trapezium
W =1187.5
8. Torricelli’s
a. dv
dt =−a √ 2 gh
dv
dt =−a ( 2 gh )( 1
2 )
v=∫−a ( 2 gh )
( 1
2 ) dt
v=a ( 2 gh )
( 1
2 ) t+c
v=−a ( 2∗9.8∗h )
1
2 + c
a is the cross sectional area of a circle given by π r2= 22
2 ∗( 0.025 ) 2= 11
5600
v= −11
5600 ( 98
5 h )1
2 t+ c
Hence, making h the subject t of the formula gives

h=( v
−8.6963 )
2
dh
dt = v2 d
dt (−8.6963∗10−3 t ) 2
=−1.230∗10−5 √h
b . At v¿ 0 , t=c
But at time t=0 ,the tank if full hence v=c
Since is given by π r2 h= 22
7 ∗1.52∗4=¿28.2857,
Hence the tank will be empty after 28.29 seconds.

References
Autar, K., Egwu, K. & Duc, N., 2008. Numerical Methods with Applications. [Online]
Available at: http://nm.mathforcollege.com/topics/textbook_index.html
[Accessed 31 December 2018].
Dani, S., 2012. Ancient Indian Mathematics –. A Conspectus". Resonance. , 17(3), p. 236–246.