This document provides an overview of engineering mathematics II with a focus on first order ordinary differential equations (ODEs). It explains what first order ODEs are, how to solve separable and reducible first order ODEs, and provides examples of applying first order ODEs to model real-world scenarios like population growth, decay, and radioactive decay. The objectives are to explain first order ODEs, separable equations, and apply the concepts to real life applications.

1. KNF1023
Engineering
Mathematics II
First Order ODEs
Prepared By
Annie ak Joseph
Prepared By
Annie ak Joseph Session 2008/2009

2. Learning Objectives
Explain the first order of ODEs
Explain the separable equations
Apply first order in real life application

3. First Order differential equations
ODEs in y=y(x) are 1st order if y’(x) is the
highest order derivative of y present in the
equations.
First order ODEs in y=y(x) may be written in
the form
dy
= G ( x, y )
dx
For cases where G(x,y) assumes certain
specific forms, methods of solving the ODEs
are available.

4. 1st Order ODEs in Separable Form
A 1st order ODE is said to be in separable
form if it can be written as
dy
= f ( x) • f ( y )
dx
Which is simplified to
dy
∫ f ( y)
= ∫ f ( x)dx
This provides us with a method for solving the
ODE if we can work out the integrals on both
sides. Notice the integration on the left hand
side is with respect y to and that on the right
hand side is with respect to x.

5. Example 1
Solve the ODE y , ( x) = ( y + 1) 2 (3x 2 + 2) subject to
y(0)=1
The ODE is 1st order and in separable form
Rewrite it as
y , ( x)
∫ ( y + 1) 2 = (3 x 2 + 2)
1
∫ ( y + 1) 2 ( )
dy = ∫ 3 x 2 + 2 dx

6. Continue…
−1
− ( y + 1) = x 3 + 2 x + C
− (1 + 1) −1 = 0 3 + 2(0) + C
1
Then C=−
2
−1 3 1
− ( y + 1) = x + 2x −
2
For the explicit form, we can write it as
2x3 + 4x + 1
y=
− 2x3 − 4x + 1

7. Example 2
, 2x 1
Solve y ( x) = = ( 2x)
y +1 y +1
The ODE is 1st order and separable form
Rewrite it as
dy 2x
=
dx y + 1
( y + 1)dy = 2 xdx
Integrate on Both sides
∫ ( y + 1)dy = ∫ 2 xdx

8. Continue…
∫ ydy + ∫ dy = 2∫ xdx
y2 2x 2
+y= +C
2 2
2 2
y + 2 y = 2 x + 2C

9. Example 3
x2 + 7 x + 3
Solve the initial value problem y' =
y2
with y(0)=3
y 2 y ' = x2 + 7 x + 3
dy
2
y = x2 + 7x + 3
dx
2 2
y dy = ( x + 7 x + 3)dx
∫ ( )
y 2 dy = ∫ x 2 + 7 x + 3 dx

10. Continue…
y 3 x3 x2
= + 7 + 3x + C
3 3 2
3 21 2 3
y = x + x + 9 x + 3C
2
21 2 3
y = x + x + 9 x + 3C
3
2
From the initial condition y(0)=3.
3 = y (0) = 3 0 + 0 + 0 + 3C = 3 3C
27
C= =9
3

11. Continue…
Thus C=9 and substitute it into equation we
get
21 2
3
y = x + x + 9 x + 3(9)
3
2
21 2
3
y = x + x + 9 x + 27
3
2

12. 1ST ORDER ODEs REDUCIBLE TO
SEPARATE FORM

13. 1ST ORDER ODEs REDUCIBLE TO
SEPARATE FORM

14. 1ST ORDER ODEs REDUCIBLE TO
SEPARATE FORM

15. Example 4
3
 y  x
Solve the ODE y' =  +  subject to
 x  y
y(1)=1
Rewrite the equation as
3
dy  y   x 
= + 
dx  x   y 
 
Use
y
y = x.u ( x) or u=
x

16. Continue…
Substitute it in the equation
3
dy  y   x 
= + 
dx  x   y 
 
3
dy 1
=u+ 
dx u
Differential the y = x.u ( x ) , thus we get
dy
= x.u , ( x) + u
dx

17. Continue
3
1
x ⋅ u '( x) + u = u +  
u
3
du 1
x =u +  −u
dx u
3
du  1 
x = 
dx  u 
This is a separable ODE and can be written
as u 3 du = dx
x
Integrate both sides, we get ∫ u 3 du = ∫ dx
x

18. Continue…
u4
+ C1 = ln( x) + C 2
4
u4
= ln( x) + (C 2 − C1 )
4
u4
= ln( x) + C
4
Replacing u back by y/x we obtain
4
1 y
  = ln( x) + C
4 x
or
y 4 = 4 x 4 ln( x) + 4Cx 4

19. Continue…
as the general solution of the ODE (in
implicit form)
Now y(1)=1 gives
14 = 4(1) 4 ln(1) + 4C (1) 4 or 1
C=
4
Thus, the required particular solution is
y 4 = 4 x 4 ln( x) + ( x) 4

20. Example 5
Solve the first order differential
equation.
dy 2 y
x = y + x cos  
dx x

21. Application in Growth and Decay
First order ODEs in separable form can be
found in population dynamics which is
concerned with the growth and decay of a
population. The population may be a lump
of decaying radioactive substance, a
colony of bacteria thriving on an agar
culture or a group of people living in a
community.

22. Theory
The rate at which the “Something” grows
or decay is directly proportional to the
current population (until such time as
resources become scarce or overcrowding
becomes a limiting factor).
If we let y(t) represent the number of
“something” at time t, then the rate of
change of the “something” with respect to
time is y’(t). Thus, since y’(t) is
,
proportional to y(t), we have y (t ) = ky(t )

23. Theory
For k > 0, equation is called an
exponential growth law and for k < 0 it is
an exponential decay law.

24. Example 1: Exponential Growth of a
Bacterial Colony
A freshly inoculated bacterial culture
contains 100 cells. When the culture is
checked 100 minutes later, it is
determined that there are 450 cells
present. Assuming exponential growth,
determine the number of cells present
at any time t (measured in minutes).

25. Solution:
Exponential growth means that
,
y (t ) = ky (t )
dy
= ky (t )
dt
dy
= kdt
y (t )
dy
∫ y (t )
= ∫ kdt
ln y (t ) = kt + C
y (t ) = e kt +C

26. Continue…
y (t ) = e kt .e C
Then the equation become
y (t ) = Ae kt
Where A and k are constants to be
determined. Notice that if we set the
starting time be t=0, we have y(0) = 100

27. Continue…
Setting t = 0, we now have
100 = y(0) = Ae0=A
And hence,
y(t) = 100 ekt
We can use the second observation to
determine the value of the growth constant,
k. We have
450 = y(100) = 100e100k

28. Continue…
Dividing both sides by 100, we hav
4.5 = e100k
ln 4.5 = 100k
ln 4.5
k= ≈ 0.01504
100
We now have a formula representing the number
of cells present at any time t:
kt  ln 4.5 
y (t ) = 100e = 100 exp  t
 100 

29. Example 2: Radioactive Decay
If you have 50 grams of 14 C today, how much
will be left in 100 years?
(half-life for 14 C is approximately 5730 years)

30. Solution:
Let y(t) be the number of grams of 14 C present
at time t. Then, we have
y , (t ) = ky(t )
dy
= ky (t )
dt
dy
= kdt
y (t )
dy
∫ y (t )
= ∫ kdt

31. Continue...
ln y (t ) = kt + C
kt + C
y (t ) = e
kt C
y (t ) = e .e
Assume C
e =A
kt
y (t ) = Ae

32. Continue...
Our initial condition is y(0) = 50 and so
k ( 0)
y (0) = 50 = Ae
Thus,
A = 50
Substitute A=50 and we get
kt
y (t ) = 50e
In order to determine the value of the decay
constant k, we use the half-life:
5730 k
25 = y (5730) = 50e

33. Continue...
25
= e 5730k
50
1
ln = 5730k
2
The value of K can be calculated using the scientific calculator
1
ln
k= 2 ≈ −1.20968 ×10 − 4
5730
To see how much will be left in 100 years, we substitute the
value that we calculated
and into the equation − 1 . 20968 × 10 − 4 × 100
y (100 ) = 50 e ≈ 49.3988 grams

34. Prepared By
Annie ak Joseph
Prepared By
Annie ak Joseph Session 2008/2009