DIFFERENTIAL EQUATION

Definitions:

A differential equation is an equation which contains at least one differential coefficient or derivative. Thus,

                       

are examples of differential equations.

Order of a differential equation is the order of the highest derivative involved in the equation

                       

Degree of a differential equation is highest power to which derivative of the highest order is raised, after the equation has been rationalized and cleared of fractions with respect to all derivative

 

The solution of a differential equation is an equation relating the variable involved but containing no differential coefficient or derivative. The general solution contains an arbitrary constant. A particular may contain if given the value x and the corresponding of y. These values are called boundary conditions or initial conditions.

 

FORMATION OF DIFFERENTIAL EQUATION

In forming differential equation, all the arbitrary constants are eliminated. We illustrate this with the following examples.

 

EXAMPLE 1

Form a differential equation from the following functions

Note: The number of arbitrary constants will give the order of the differential equation.

A function with 1 arbitrary constant, gives 1^st order equation.

A function with 2 arbitrary constants, gives 2^nd order equation.

A function with n arbitrary constants, gives n^th order equation.

 

 

 

 

SOLUTION OF DIFFERENTIAL EQUATION

To solve a differential equation, we have to provide a function for which differential equation is true. That is, find we find a way to eliminate the differential coefficient, so that it leaves us with a function expressed in terms of x, y and a constant. This chapter will be devoted to first order differential equation

 

Method 1: By direct integration

If the equation is of the form , we integrate directly so that

EXAMPLE 2

Method 2: By separating the variables (variable separable)

A differential equation that can be written in the form

                                   

is said to separable. The general solution can be obtained can be obtained by separating the variables and integrating both sides, that is

                                   

EXAMPLE 3

EXAMPLE 4

EXAMPLE 5

EXAMPLE 6

If (x² – 1)  + 2y = 0 find the value of y in terms of x given that y =3 when x = 2

SOLUTION

 

Method 3: Homogeneous equations

A differential equation [  ] is said to be homogenous when M and N are homogenous of the same degree in x and y. Such homogeneous function can be solved using the substitution

y = vx.

where v is a function of x

In this case the equation  can be written in the form

                       

Since  is a homogeneous function of degree zero in x and y. Differentiating y = vx  gives with respect to x

Differentiating of  with respect to x gives

                                        

Also  this makes

which when related to  gives,

                         

Integrating

                       

The integral will be in terms of  v and x. Replace v = y/x to give the required solution.

 

EXAMPLE 7

SOLUTION

EXAMPLE 8

Solve the equation

SOLUTION

Method 4: First order Linear Differential Equation

A first – order linear differential equation is one with general form

                       

Where p and q are function of x alone, or are constants

 

Solution of a First – Order Linear Differential Equation

The first – order linear differential equation  has the general solution

                       

Where C is an arbitrary constant and I(x) is called the integrating factor

                       

EXAMPLE 9

Find the general solution of the differential

                       

SOLUTION

This is a first – order linear differential with p(x) = 3/x and q(x) = x. The integrating factor is

                       

And the general solution of the first – order linear equation is

                       

EXAMPLE 10

Find the general solution of the given first – order linear differential equation

                       

SOLUTION

 

EXAMPLE 11

Find the general solution of the given first – order linear differential equation

           

SOLUTION

EXAMPLE 12

Determine the general solution of the equation

                         

SOLUTION

QUESTION 13

Find the particular solution of the given differential equation that satisfies the condition

                       

SOLUTION

 

Method 5: First – Order Equation with Exact Differential

Any differential equation of the form

                       

such that the variables are not separable, is an exact differential equation if it arranged in such way that its left – hand side is an exact differential of some function u(x, y) that is, , we ca say  is an exact differential equation, and its solution is u(x, y) = C , C is an arbitrary constant. The total differential of a function of a function u(x, y) is defined by

           

that

           

Since

           

The condition  is necessary condition for  to be an exact differential equation

 

EXAMPLE 14

Verify that each of the following differential equation is exact and solve

 

APPLICATION OF DIFFERENTIAL EQUATION

GROWTH AND DECAY PROBLEM

Micro – organism such as Bacteria and Viruses a found to reproduce at an alarming rate. The reason for this is that they reproduce by binary fission (i.e. each cell reproduces by dividing into two cells.

            In the case of Bacteria culture, it was discovered that the rate of growth is directly proportional to the current population (until such time as resources becomes scarce or overcrowding becomes a limiting factor).

If we let y(t) represent the number of bacteria in a culture at time t, then the rate of change of population  with respect to time is y¹(t). Thus, since y¹(t) is proportional to y(t), we have

                       

Where k is the proportionality constant (the growth constant)

The equation  is a differential equation, which we have to solve.

                       

Integrate both sides of equation (2) with respect to t, we obtain

                       

Evaluating these integral, we obtain

                       

Since C is an arbitrary constant

           

For k > 0, equation (4) is called exponential growth law and for k < 0, it is exponential decay law.

 

 

EXAMPLE 15: GROWTH PROBLEM

Suppose a bacterial culture doubles in population every 4 hour. If the population is initially 100, find an equation for the population at any time. Determine when the population will reach 6000. Assuming the growth is exponential

 

SOLUTION

Exponential growth means that

and from equation (4)

                       

Where A and k are constant to be determine. Let the starting time be t = 0, we have

                       

And hence

                       

To determine k, let time t = 4 and y(t) = 2A = 200

                       

Taking the natural logarithms of both sides

                       

We now have a formula representing the number of bacteria in the culture at any time as

                       

One more thing to do, to determine the time when the population will be 6000 (i.e. y(t) = 6000)     

                       

Take the natural logarithm of both sides

                       

 

The population will be 6000 in approximately 23.64 hours

 

EXAMPLE 16

A virus infection is thought to spread through a population in way after t weeks; a fraction of x of the population has been infected, where

                                   

If 25% of the population is infected when t = 0, Calculate the time taken that elapses before 50% of the population is infected. Estimated the time for which everybody is infected.

SOLUTION

 

 

EXAMPLE 17

The population of a certain bacteria is increasing at 20% per year.  If the present population is

500,000, estimate the population in 20 years time.

SOLUTION

The differential equation is given as  = kp where k is the growth rate

 

RADIOACTIVE DECAY

Experiment have shown that the rate at which a radioactive elements decay is directly proportional

to amount present. Let y(t) be the amount present (mass) of a radioactive element present at time t.

Then we have the rate of change (rate of decay) of  y(t) satisfies

                       

The equation will be

                       

Where A and k (the decay constant) are constants.

 

EXAMPLE 18

The radioactive isotope plutonium – 241 decays so as to satisfy the differential equation

                       

Where Q is measured in milligram and t in years

(1)                Determine the half year τ of plutonium – 241

(2)                If 50 milligram of plutonium of plutonium – 241 are present today, how much will remain in ten years.

 

 

SOLUTION

(1)        To determine the half life of plutonium (that is, when Q_o/2 and Q_o is the initial amount of plutonium – 241 present)

                       

(2)