Engineering Mathematics 1

Engineering Mathematics 1
ENGINEERING MATHEMATICS
Name
Course
Professor
University
City/state
Date

Engineering Mathematics 2
Engineering Mathematics
Question 1
The function used is dy
dx =2 x
y ²
The first step is to separate the variables as follows (Dawkins, 2018); (Dobson & Slomson,
(n.d.)):
y2 dy = 2x dx
The second step is to integrate both sides of the equation as follows (Khan Academy, 2017):
y ³
3 = 2 x ²
2 +C → y ³
3 =x2 +C
y3 = 3(x2 + C) → y= 3
√3( x2 +C), where is just an arbitrary constant
The third step is to find the value of C (Pierce, 2017), using the initial condition y(0) = 1
y= 3
√ 3( x2 +C); 1= 3
√ 3(02+ C); 1=3
√ 3C →13=3 C ;C= 1
3
Therefore y= 3
√3 ( x2 + 1
3 ); y= 3
√3 x ²+1
The exact solution is y= 3
√3 x ²+1 or y3=3 x2 +1
Question 2
Euler’s method uses the first two terms of the Taylor’s series of the differential equation as
follows (Bourne, 2018):

Engineering Mathematics 3
y(x + h) ≈ y(x) + hy'(x) or y ( x+ h ) = y ( x ) + h dy
dx
The initial value (xo, yo) = (0, 1) i.e. when xo= 0, yo = 1
At the initial point, dy
dx =2 x
y ² =2 (0)
1² =0
y1 = y(0.1) = yo + h(dy/dx)o = 1 + 0.1(0) = 1
y2 = y(0.2) = y1 + h dyi
dx 1 ¿=1+0.1 ( 2 x 0.1
12 x 12 )=1.2
The remaining approximate values of y from x = 0 to x = 5 at h = 0.1 have been done in Excel
spreadsheet.
Question 3
The exact values of y are obtained using the equation The exact solution is y= ( 3 x2 +1 )
1
3 . These
values are also provided in the spreadsheet.
Question 4
The error has been found by determining the absolute difference between the approximate and
exact values of y. The error is provided in the spreadsheet.
Question 5
The approximate and exact values of y when h = 0.05, h = 0.2 and h =0.5 have been provided in
the spreadsheet. The table of values is as shown in Table 1 below
Table 1: Approximate and exact values of y with different values of h

Engineering Mathematics 4
x y (approximate)
y
(exact)