Differential equations can be used to model real-world phenomena like disease spread. An influenza outbreak in a village of 500 people is modeled using a differential equation where the rate of spread is proportional to the number of infected and uninfected individuals. The equation is solved, giving the number of infected people over time. After 3 months there were 30 infected, and the model predicts 205 infected after 5 months. Differential equations allow quantification of how factors like disease transmission rates impact outcomes over time.

1. Application
Of
Differential Equation in Our Real Life
Introduction:
A differential equation is a mathematical equation for an unknown function of
one or several variables that relates the values of the function itself and its derivatives
of various orders. Differential equations play a prominent role in engineering, physics,
economics, and other disciplines. One thing that will never change is the fact that the
world is constantly changing. Mathematically, rates of change are described by
derivatives. If we try and use math’s to describe the world around us like the growth
of plant, the growth of population, the fluctuations of the stock market, the spread of
diseases, or physical forces acting on an object. Most real life differential equation
needs to be solved numerically and many methods have been developed over the last
century and half and the goal has been to find methods that work for large classes of
differential equations. There seems to have been very little work published that
examines methods specialized to a single Differential Equation. By using Differential
Equation we can easily solve our day to day life problems.
Problem:
Influenza virus is one of the main problems in Bangladesh. Many people in our
country are affected by this virus. The person carrying an influenza virus returns to an
isolated village of 500 peoples. It is assumed that the rate at which the virus spreads is
proportional not only to the number of infected peoples but also to the people not
infected. Find the number of infected people after 5 months when it is further
observed that after 3 months. Number of infected peoples in 3 months is 30.
Mathematical Formulation:
Let Ni denote the number of infected people at any time t, N is the total number of
people and Nt is the time period.

2. Assuming that no one leaves the village throughout the duration of this disease, now
we can solve the initial value problem.
).........().........( iNNkN
dt
dN
ii
i

The initial condition is, N (0) = 1
Conditions:
1. When t = 0 then Ni = 1
2. When t = 3 then Ni = 30
Solution:
Equation (i) is separable. Separating variables, we have
)..(....................
)(
iikdt
NNN
dN
ii
i


Integrating equation (ii),
 

dtk
NNN
dN
ii
i
)(
 






 dtkdN
NNNN
i
ii
111
 






 dtkNdN
NNN
i
ii
11
ckNtNNN ii  )ln(ln
ckNt
NN
N
i
i


 ln
kNt
i
i
Ae
NN
N 



kNti
Ae
N
N 
 1
1 kNti
Ae
N
N
).......(....................
1
iii
Ae
N
N kNti

 

3. When t=0 then Ni=1
From equation (iii), now we can write,
1
500
1


A
5001  A
499 A
When t = 3 then Ni = 30
By using this condition now we can determine k from eq.(iii),
3500
4991
500
30 

 k
e
500)4991(30 1500
  k
e
03.01500
  k
e
00234.0 k
Then the equation (iii) becomes,
).........(....................
1499
500
17.1
iv
e
N ti

 
When t = 5 then the equation (iv) becomes,
517.1
4991
500



e
Ni
or, 85.5
4991
500



e
Ni
=205 peoples
Number of infected peoples during a time interval:
Period of time, Nt (Months) Number of infected people, Ni
3 30
4 88
5 205
6 345
7 439
8 479
9 493
10 497

4. Interpretation of Result:
The graph shows that there is a gradual increase in the number of infected
peoples. Day by day, most of the people of this village are infected this disease.
NumberofInfectedPeople
Time Period