Differential Equations


   
 
Applications
As we have mentioned earlier that almost every branch of study deals with problems involving differential equation, in this section we solve few examples to exhibit the application of differential equation in various branches of knowledge.
 
Example 1:
 
A wet porous substance in the open air loses its moisture at the rate proportional to the moisture content. If a sheet hung in the wind loses half of its moisture during the first hour, when will it have lost?
 
(i) 95% moisture, weather condition remaining constant
 
(ii) 90% moisture, weather condition remaining the same
 
Solution:
 
Let m0 be the moisture content initially and let m be the moisture content after t hours.
 
According to problem:
 
 
 
 
 
when t = 0, m= m0
 
 
 
 
log m = kt + log m0
 
 
 
Substituting in (2), we have
 
 
\ Equation (2) becomes
 
 
(i) Again when the sheet losses 95% of the moisture, m =
 
Equation (3) becomes
 
 
 
 
(ii) when the sheet loses 90% of moisture, m =
 
Equation (3) becomes
 
 
 
 
Example 2:
 
The decay rate of radium at any point t is proportional to its mass at that time. The mass is M0 at time t = 0. Form the differential equation and find the time when the mass will be halved.
 
Solution:
 
The differential equation for the decay is given by
 
 
where M is the mass of the radioactive substance after t hours.
 
 
 
 
= log M = - kt + log C
 
where log C is the constant of integration
 
 
 
 
 
Given at t = 0, M = M0
 
(Substituting m = M0 t = to in (1))
 
The differential equation is
 
M = M0 e-kt ….(2)
 
When the mass is halved,
 
 
From (2), we have
 
 
 
 
 
 
Example 3:
 
Suppose the growth of a population is proportional to the population itself. If the population of a colony doubles in 50 days, in how, many days will the population become triple.
 
Solution:
 
Let the initial population be P0 and P be the population of the colony at any instant t.
 
Then according to the problem
 
 
 
 
when t = 0, P= P0
 
 
Equation (1) becomes
 
log P = kt + log P0 ….(2)
 
when P0 is doubled,
 
P = 2P0 where t = 50 days
 
From equation (2)
 
log (2 P0) = 50 k + log P0
 
 
 
When P is tripled, the equation (2) becomes,
 
 
 
 
 
Example 4:
 
Find the equation of the curve that passes through the point (3,- 4) such that the slope of the tangent at the point (x, y) on it equals .
 
Solution:
 
According to the problem
 
 
Integrating, we have
 
 
 
 
y = x2c
 
This is the family of equations with parameter.
 
The family of curves is represented by y = x2c, since one of these curves passes through (3,- 4), we have
 
 
 
 
Þ 4x2 + 9y = 0 is the required equation.
 
  
     
   
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