Engineering Mathematics 2: End of Term Worksheet Solutions

This document presents the solutions to an Engineering Mathematics 2 assignment. The first section solves a first-order differential equation using Euler's numerical method, providing approximations for y(0.2), y(0.4), and y(0.6). The second part derives the general solution to the one-dimensional heat equation using separation of variables, resulting in a solution involving trigonometric and exponential functions. The subsequent sections address probability problems, including calculating the probability of a defective item from three production lines, determining probabilities using the binomial distribution, and calculating the probability of defective pixels in an image. This assignment covers a range of topics within Engineering Mathematics, offering detailed solutions to aid in understanding and problem-solving.

Q1
a. ,
Assuming an interval of 0≤ x ≤ 1 and h=0.2
dy
dx =x− y +1
For y(0.2)
f 0=f ( 0 ,0.2 ) = ( 0.2 ) − ( 0 ) +1
= 1.2
y(0.2)= y0 +h f 0
= 1 + 0.2(1.2)
= 1.24
For y(0.4)
f 0=f ( 0,0. 4 ) = ( 0. 4 )− ( 0 ) +1
= 1.4
y(0.2)= y0 +h f 0
= 1 + 0.2(1.4)
= 1.28
For y(0.6)
f 0=f ( 0,0.6 ) = ( 0. 6 ) − ( 0 ) +1

= 1.6
y(0.2)= y0 +h f 0
= 1 + 0.2(1.2)
= 1.32
b. ,
The solution would be given as
U = Ae λt+ Ae− λt
Inserting the bounderly conditions
X (0) = A + B = 0 , therefore A=-B
X(1) = A eλ +B e−λ = 0 , and A(e2 λ−1) =0
Since |λ | > 0 thereby having A=B=u =0 this is the case with λ<0
Now trying the same for λ> 0.
X(t) = A cos (√ λ x) + B sin ((√ λ x)
When applying boundary conditions.
0 = X(0) = A and B sin √ λ =0 since B≠ 0
Sin √ λ=0
√ λ=nπ where n = 1, 2, 3, . . .
Now solving for U
U' ( t ) = −n2 π2 T ( t )
The solution is
T N =cn e−n2 π2 t and n= 1,2,3,4…

Putting things together
un ( x , t ) =Bn sin ( nπx ) e−n2 π 2 t and Bn = cn bn
Q 2
(i) Defective item = Adefective or Bdefective or Cdefective
= (0.6 * 0.1) + (0.06 *0.3) + (0.1 * 0.15)
= 0.0930
(ii) Adefective =0.6∗0.1
= 0.06
(iii) Cdefective= 3
10∗0.15
= 0.045
Q3
- n p k = n!
( n−k ) !
So since there are 3 items
= n!
( n−kapple ) ! + n!
( n−klemon ) ! + n!
( n−korange ) !
= 12 !
( 12−3 ) ! + 12!
( 12−4 ) ! + 12 !
( 12−5 ) !
= 108240

Q 4
(a)
Probability of the defective pixels
 Number of defective pixels = 512 x 1%
= 5.12 pixel
Probability of a defective pixel = 5.12/512
= 1%
(b) 50% = (512 - 512
2 ) r
r=19.53
(c) Pr (x=k) = λk e−1
k !
= 0.5124 e−4
4 !
= 0.52 X 10−4