Solved ENM2600 Engineering Mathematics Assignment 1 with Explanations

This document presents a detailed solution to an Engineering Mathematics assignment, covering topics such as finding roots of complex equations and representing them on the Argand diagram, determining distances between roots, calculating moduli and arguments, and expressing roots in polar form. It further explores Cartesian equations of loci, including circles, ellipses, straight lines, and rays, derived from complex number relationships. The solution also addresses separable and inseparable differential equations, integrating factors, and particular solutions. The final question involves modeling battery discharge using differential equations and determining the discharge time. Desklib offers a wide array of solved assignments and past papers for students seeking academic assistance.

Running head: ENGINEERING MATHEMATICS 1
Engineering Mathematics
Name
Institution

ENGINEERING MATHEMATICS 2
Question 1
Given that , z3 +6 z2+37 z+58=0
Using the rational root theorem, a0=58 , an=1
The dividers of 58 are :1,2,29,58
Checking out the dividers 1,2,29,58 we get that 2 is a root and we factor out (z +2) and divide
the equation z3 +6 z2+37 z +58 by (z +2) to obtain:
z3 +6 z2 +37 z+58
z +2 =z2+4 z +2 9
z3 +6 z2+37 z +58= ( z+2 ) ( z2 +4 z +29 )=0
( z +2 )=0 , z=−2
Also, z2+ 4 z +2 9=0. We find z using the quadratic formula
z=−b ± √b2−4 ac
2 a =−4 ± √42−4 (29)
2 =−4 ± √−100
2 =¿
−4 ± i √100
2 =−4 ±10 i
2 =−2 ±5 i
z=−2+5 i , z =−2−5 i
Thus, z=2 , z=−2+5 i, z=−2−5 i
Figure 1 shows the roots plotted on the argand diagram

ENGINEERING MATHEMATICS 3
Figure 1: Argand Diagram
Part b
z=2∧z=−2+5 i,
Distance between the roots= √ (5−0)2 +(−2−2)2= √ 41
z=2∧z=−2−5 i ,
Distance between the roots= √ (−5−0)2+(−2−2)2= √ 41
z=−2+5 i∧z=−2−5 i,
Distance between the roots= √(−5−5)2 +(−2−(−2))2=10
Part c

ENGINEERING MATHEMATICS 4
Modulus of z=2
¿ √ (2)2=2
Argument ( z=2 ) =0
Modulus of of z=−2+5 i
¿ √ (−2)2 +(5)2= √ 29
Argument of ( z=−2+5 i ) =180−tan−1
( 5
2 )=111.8014 °
Modulus of z=−2−5 i
¿ √ (−2)2 +(−5)2= √ 29
Argument of ( z=−2−5 i )=180+tan−1
(5
2 )=248.1986 °
Part d
Polar form of z=2 ¿ , z=r ∠θ=2 ∠ 0
Polar form of z=−2+5 i ¿ , z =r ∠ θ= √29 ∠111.8014 °
Polar form of z=−2−5 i¿ , z=r ∠θ= √ 29 ∠248.1986 °
Question 2
Part a
2−4 j+ ( z−2 j ) ( z −2 j )¿
¿ 4 j+3∨¿=7−(−4 j+2)¿ ¿

ENGINEERING MATHEMATICS 5
2−4 j+ ( x + yj−2 j ) ( x+ yj−2 j ) ¿
√ 42 +32 =7−(−4 j+2)¿
2−4 j+ ( x +( y−2) j ) ( x +( y−2) j )
5 =7−(4 j+2)
2−4 j+ x2− ( y−2 ) 2
5 =5−4 j
2+ x2 − ( y−2 ) 2
5 =5
10+x2− ( y −2 )2=25
x2− ( y−2 ) 2=15. Hence, it is a hyperbola
Part b
arg ( z−4+ j ) =arg(− j √ 3+3)
arg ( x+ yj−4+ j ) =arg(− j √ 3+3)
arg ( x−4+ ( y +1 ) j ) =arg(− j √ 3+3)
tan−1
( y +1
x−4 )=π −tan−1 (¿¿ √ 3/3)¿ ¿
tan−1
( y +1
x−4 )=π − π
6 = 5
6 π
y +1
x−4 =tan ( 5
6 π )
y +1
x−4 =−1
√3

ENGINEERING MATHEMATICS 6
y +1=−1
√ 3 ( x−4 )
The equation is that of a straight line which passes via ( 4 ,−1 )with the gradient ¿− 1
√ 3
Part c
|z−4−7 j|+| z−4 +3 j|=16
|x + yj−4−7 j|+|x + yj−4 +3 j|=1 6
|( x−4 ) +( y−7) j|+|( x−4 ) +( y +3) j|=16
√ ( x−4)2 +( y−7)2+ √ ( x−4)2+( y +3)2=1 6
√ (x−4)2 +( y−7)2=16− √ (x−4)2+( y +3)2
(x−4)2+( y −7)2= (16− √ ( x−4 )2 + ( y +3 )2 )2
(x−4)2+( y −7)2=216−32 √ ( x−4 ) 2 + ( y +3 ) 2 + ( x−4 ) 2 + ( y +3 ) 2
( y−7)2 − ( y+ 3 ) 2=216−32 √ ( x−4 )
2 + ( y +3 ) 2
y2−14 y+ 49− y2−6 y−9=216−32 √ ( x−4 ) 2 + ( y +3 ) 2
−20 y +40−256=−32 √ ( x−4 )
2 + ( y +3 )
2
20 y +216=32 √ ( x−4 ) 2 + ( y +3 ) 2
( 20 y+ 216
32 )
2
= ( x −4 ) 2+ ( y+3 ) 2

ENGINEERING MATHEMATICS 7
( 20 y2+8640 y +46656
1024 )=x2−8 x+16 + y2+ 6 y+ 9
If we rearrange and simplify the equation we obtain,
256
251 ( x−4 )2 + y2− 624
251 y−35.187=0
( x−4 )
0.981
2
+ y2− 624
502 y +1.545−36.732=0
( x−4 )
36.016
2
+ ( y−1.24 )
36.732
2
=1
Thus, the equation defines an ellipse
f = √36.732−36.016=0.84 6
Coordinates of foci are ( 0.85 , 4 ) ∧(0.85 ,1.24 )
Question 3
Part a
y' =ln (x− y ) is inseparable since we cannot move the variables x and y to opposite sides of the
equation.
Part b
y' = ln ( x )
ln ( y)
The equation is separable as follows.

ENGINEERING MATHEMATICS 8
y' = dy
dx = ln ( x )
ln ( y )
ln ( y )dy =ln ( x ) dx
∫ ln ( y ) dy =∫ ln ( x ) dx
y ln ( y )− y =xln ( x )−x +C
Part c
( x2 y − y ) y'=xy2+ x .
The equation is separable as shown below
( x2 y − y ) y'= ( x2 y− y ) dy
dx =xy2 + x
y ( x2−1 ) dy
dx =x ( y ¿¿ 2+1)¿
y
( y ¿¿ 2+1)dy= x
(x ¿¿ 2−1)dx ¿ ¿
∫ y
( y¿ ¿2+1) dy =∫ x
( x¿ ¿ 2−1)dx ¿ ¿
We let u=( y ¿¿ 2+1)¿ such that, dy = 1
2 y du
So that,
∫ y
( y¿ ¿2+1) dy =∫ y
u
1
2 y du=∫ 1
u
1
2 du ¿

ENGINEERING MATHEMATICS 9
¿ 1
2∫ 1
u du=1
2 ln ( u ) +C=1
2 ln( y ¿¿ 2+1)+ C ¿
Then, we let (x ¿¿ 2−1)=u ¿ so that, dx= 1
2 x du
∫ x
( x ¿¿ 2−1)dx =∫ x
u
1
2 x du=∫ 1
u
1
2 du=¿ ¿
1
2∫ 1
u du=1
2 ln ( u ) +C=1
2 ln( x ¿¿ 2−1)+C ¿
1
2 ln( y¿ ¿2+1)+C= 1
2 ln (x ¿¿ 2−1)+C ¿ ¿
ln ( y ¿¿ 2+ 1)=ln (x ¿¿ 2−1)+C ¿ ¿
Question 4
e− y (1+ y' )=1
( 1+ dy
dx )= 1
e− y =ey
dy
dx =e y−1
dy
e y−1 =dx
Integrating both sides
∫ dy
e y−1 =∫dx=x +C
Let e y=u , du=e y dy

ENGINEERING MATHEMATICS 10
∫ dy
e y−1 =∫ du
u(u−1) =∫ −1 du
u + ¿∫ 1 du
( u−1 ) ¿
¿−ln ( u ) +ln ( u−1 ) =ln ( u−1
u )
ln (u−1
u )=x +C
ln ( ey −1
e y )=x +C
ex +C= e y−1
ey
Substituting y ( 0 ) =−ln (4)
e0+C = ey −1
e y =e− ln (4)−1
e−ln (4 )
eC=1−eln ( 4 ) =1−4=−3
lneC=ln (−3)
C=ln ( −3 )
ex
3 = e y
ey +1
ex= 3 e y
e y +1
x ( y )=ln ( 3 e y )−ln (e y +1)

ENGINEERING MATHEMATICS 11
x ( y )=ln ( 3 ) + ln ( ey ) −ln (e y+1)
The particular solution is, x ( y )=ln ( 3 ) + y−ln (ey +1)
Question 5
( x +1 ) y'−2 y=ex (x+1)3
The equation is linear, non-homogeneous, and non-separable
We rewrite the equation in the form
y' ( x ) + p ( x ) y=q( x )
p ( x )= −2
x +1 ,q ( x ) =ex (x+ 1)2
We find the Integrating Factor (IF)
d ( IF )
dx =IF ( −2
x+1 )
d ( IF )
dx
IF = ( −2
x+1 )
d
dx ln ( IF )= ( −2
x +1 )
ln ( IF )=∫ ( −2
x+1 )dx=−2 ln ( x +1 ) +C=ln ( eC
( x +1 )2 )
( IF ) = eC
( x+1 ) 2 = 1
( x +1 ) 2

ENGINEERING MATHEMATICS 12
Putting the equation in the form, ( IF . y )' =IF .q (x) we obtain
d
dx ( 1
( x +1 ) 2 y ) =q ( x ) = 1
( x+ 1 ) 2 ex ( x+1)2=ex
1
( x+1 )2 y =∫ ex dx=ex +C
y ( x ) =ex ( x +1 ) 2 +C ( x +1 ) 2
Applying the initial condition y ( 0 ) =3
y ( 0 ) =e0 ( 0+1 ) 2 +C ( 0+ 1 ) 2=3
1+C=3 , C=2
Substituting C=2 we obtain:
y ( x ) =ex ( x +1 ) 2 +2 ( x+1 )2 =(e x+ x ) ( x +1 ) 2
y ( x ) =(ex+ x ) ( x +1 ) 2
Plotting y ( x ) we obtain