What is Differential Equation?

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differential equations.

The weight and height of a normal child will increase as it keeps growing old. How will you say this mathematically? The rate in which the weight of the child increases is dependent to the child’s height. The rate in which something changes is written as a derivative. One quantity is dependent on the other. The relationship between the dependent variable, independent variable and their derivatives is called a differential equation.

"differential-equation"

Differential Equations with One Independent Variable (ODE)      

The differential equation where the dependent variable depends on only one independent variable is called as an ordinary differential equation (ODE). The notations used for the derivatives are dy dx , d 2 y d x 2 , y , y , etc.

Differential Equation with Two or More Independent Variables (PDE)

The differential equation where, there are two or more than two independent variable is called as a partial differential equation (PDE). The notations used for partial derivatives are x y , 2 x y 2 , etc.

Order of a Differential Equation        

It is the order of the highest order derivative present in the equation.    

For example-

2 x 2 d 2 y d x 2 +4x dy dx +y=0

Here, order of highest order derivative is 2. So, the order is 2.

Degree of a Differential Equation       

It is the power of the highest order derivative in the differential equation. 

For example-

dy dx 2 + dy dx =4

Here, power of the derivative is 2. So, the degree is 2.     

The degree should be a positive integer.  

How is a Differential Equation Formed?

It can be formed from its general solution. The order depends on the arbitrary constants in the general solution.  

For example,

Consider, y=A x 2 +C where, A and C are arbitrary constants.

As there are 2 arbitrary constants, the differential equation has order 2. Differentiate the equation 2 times in x terms.

So,

y=A x 2 +C dy dx =2Ax

Again differentiate in x terms.

d 2 y d x 2 =2A

Multiply the equation by x

x d 2 y d x 2 =2Ax

Substitute dy dx =2Ax

x d 2 y d x 2 = dy dx

Hence the differential equation is x d 2 y d x 2 = dy dx

Solution of the Ordinary Differential Equations

The solution is the equation which does not contain derivative of variables which are dependent or independent. It satisfies the differential equation.   

The two types of solution for a differential equation are as follows.      

1) General solution       

2) Particular solution      

In general solution, the number of arbitrary constants and the order of differential equation are equal.      

The particular solution is obtained by putting values of arbitrary constant in the general solution.

Let us take a differential equation dy dx =cosx.

Here, the solution is y=sinx+c and c is an arbitrary constant. So the differential equation has infinitely many solutions.

y=sinx+cis the general solution and by replacing value of c=3 in general solution y=sinx+3 it becomes a particular solution.

Variable Separation Method to solve differential equation

Variable separation is to rearrange the terms of equation as, all one variable dependent term in a side and another variable dependent term on other side and to solve it.

Suppose the first order and the first degree differential equation is written in the form of Adx=Bdy

Where, A is function of x and B is function of y then this is called as variable separable form.

The solution can be found by integrating both sides.

Adx+ Bdx=c

Here, c is an arbitrary constant.

Homogenous function

If f λx,λy = λ n f x,y where, λis a constant then the function f x,y is called as a homogeneous function.

Homogenous differential equation

If f x,y and g x,y are homogenous functions of same degree then f x,y dx+g x,y dy=0is called as homogenous differential equation.

Method of solving is given below.

f x,y dx+g x,y dy=0 is a homogenous differential equation.

First, express the equation in form:

dy dx = f x,y g x,y

Substitute y=ux:

dy dx =u+x du dx

Now, the solution is in two variables x and u.

Substitute back u= y x

For example,

x 2 ydx x 3 + y 3 dy=0 …… (1)

Simplify the equation:

x 2 ydx= x 3 + y 3 dy dy dx = x 2 y x 3 + y 3

Put y=ux.

Differentiate y in x terms:

dy dx =u+x du dx …… (2)

Substitute y=ux and equation (2) in equation (1):

u+x du dx = x 2 ux x 3 + u 3 x 3 = x 3 u x 3 1+ u 3 = u 1+ u 3

x du dx = u 1+ u 3 u x du dx = u 4 1+ u 3

Separate the variable:

1+ u 3 u 4 du= dx x

Integrate on both sides:

1+ u 3 u 4 du = dx x 1 u 4 + 1 u du =log x +c 1 3 u 3 +log u =log x +c

Substitute back u= y x :

1 3 y x 3 +log y x =log x +c x 3 3 y 3 +log y log x =log x +c log y x 3 3 y 3 =c

Linear Differential equation

If the variable which is dependent and its differential coefficient are of degree 1, then it is a linear differential equation.

dy dx +Ay=B

Here, A and B are functions of x.

Method to solve first order linear equation

First, find integrating factor (IF) e Adx .

Write solution in form:

y e Adx = B× e Adx dx +c

For example:

dy dx +y= e x

Compare the equation with dy dx +Ay=B:

So A=1and B= e x .

Find IF:

IF= e Adx = e dx = e x

Write the solution:

y e x = e x e x dx +c y e x =x+c y= x+c e x

Applications of Differential Equation

Differential equation has wide range of applications such as rate change problems, physical problems, and electrical problems.

Population growth and growth of bacteria

The population of a country increase with time.

Suppose P is population of the country at time t. Then the rate of increase in population of the country is proportional to original population.

dP dt P dP dt =kP

Where, k is proportionality constant.

By integrating it gives population at time t.

Radioactive decay

The rate of disintegration of radioactive element is always proportional to the amount present at time t.

If z is the amount of radioactive material present at time t, then, dz dt z dz dt =kz

Where, k is proportionality constant. Negative sign represents that x decreases as the time increases.

After solving differential equation it gives the amount of radioactive substance at time t.

Half-life period - Time taken for half of substance to disintegrate is called as half-life period.

Example:

The rate of the disintegration of the radioactive element at time t is proportional to its mass. Find the time during which the original mass of 1.5gmwill disintegrate into its mass of 0.5gm.

Let x be the mass of the radioactive element.

dx dt =kx

Separate the variable:

dx x =kdt

Integrate on both sides:

dx x = kdt logx=kt+c …… (1)

Substitute t=0,x=1.5

log1.5=k 0 +c log1.5=c

Substitute x=0.5and log1.5=cin equation (1)

log 0.5 =kt+log 1.5 kt=log 1.5 log 0.5 kt=log 1.5 0.5 t= log3 k

The original mass of 1.5gmwill disintegrate into its mass of 0.5gmin log3 k units of time.

Newton’s law of cooling

The rate of change of temperature of heated body at any time is proportional to the difference between temperature of the body and surrounding medium.

Suppose θ is the temperature of the body at time t and φbe the temperature of the surrounding.

dθ dt θφ dθ dt =k θφ

Where, k is proportionality constant. Negative sign is there as the temperature decreases.

After solving differential equation it gives temperature of the body at time t.

Example:

If a body cools from 80 Cto 50 C at room temperature of 25 Cin 30 minutes, Find temperature of the body after 1 hour.

Let θ be the temperature of the body at time t.

So,

dθ dt =k θ25

Separate the variables:

dθ θ25 =kdt

Integrate on both sides:

dθ θ25 = kdt log θ25 =kt+c …… (1)

Substitute t=0,θ= 80 :

log θ25 =k 0 +c log 8025 =c log55=c

Substitute t=30,θ=50and log55=cin equation (1):

log 5025 =30k+log 55 log 25 log 55 =30k log 25 55 =30k k= 1 30 log 5 11

Substitute t=60, then,

log θ25 =60k+log 55

Substitute k= 1 30 log 5 11

log θ25 =60 1 30 log 5 11 +log 55 =2log 5 11 +log 55 =log 125 11

Find θ:

θ25= 125 11 θ= 36.36 C

Temperature of the body after 1 hour is 36.36 C.

Formulas

  • Power rule for differentiation: d dx x m =m x m1 , where m is a real number.
  • Product rule for differentiation: d dx uv =v d dx u +u d dx v , where u and v are differentiable functions in x.
  • Power rule for integration: x m dx= x m+1 m+1 ,  if  m1 log x ,    if  m=1 , where m is a real number.

Practice Problems

1. Find the order of the differential equation:

4x d 2 y d x 2 + dy dx +y=0

Solution:

The order of highest order derivative is 2. So, the order is 2.

2. Find the degree of the differential equation:

4x d 2 y d x 2 3 + dy dx +y=0

Solution:

The power of the highest order derivative is 3. So, the degree is 3.

3. Find the differential equation for y=P e 3x +Q e 3x .

Solution:

The given equation has 2 arbitrary constants P and Q. Differentiate the equation 2 times with respect to x.

So,

dy dx =3P e 3x 3Q e 3x =3(P e 3x Q e 3x )

Again differentiate with respect to x.

d 2 y d x 2 =3(3P e 3x +3Q e 3x ) =9(P e 3x +Q e 3x )

Substitute y for P e 3x +Q e 3x in the second derivative.

d 2 y d x 2 =9y

Hence, the differential equation is d 2 y d x 2 =9y.

4. Find the solution of d 2 y d x 2 +x=0.

Solution:

Simplify the given equation:

d 2 y d x 2 =x

Integrate on both sides of the equation:

d 2 y d x 2 = x dy dx = x 2 2 + c 1

Again integrate on both sides:

dy dx = x 2 2 + c 1 y= x 3 6 + c 1 x+ c 2

where, c 1 and c 2 are arbitrary constants.

5. Solve y 2 dx+ xy+ x 2 dy=0.

Solution:

Simplify the equation:

y 2 dx=(xy+ x 2 )dy dy dx = y 2 xy+ x 2 ..….(1)

Put y=ux, then differentiate with respect to x. dy dx =u+x du dx .…..(2)

Substitute y=uxand equation 2 in equation 1, then simplify.

u+x du dx = u 2 x 2 xux+ x 2 = u 2 x 2 x 2 (u+1) = u 2 u+1 x du dx = u 2 u+1 u x du dx = 2 u 2 u u+1

Separate the variables:

u+1 2 u 2 u du= dx x

Integrate on both sides:

u+1 2 u 2 u du = dx x 2( 1 8 ln 16 u 2 +8u + 3 8 (ln 4u+2 ln(4 u )))=ln x +C

Substitute back u= y x

2 1 8 ln 16y x 2 2 + 8y x + 3 8 ln 4y x +2 ln 4 y x =ln x +C 1 4 ln 16y x 2 2 + 8y x + 3 4 ln 4y x +2 ln 8 y x =ln x +C

6. Solve the differential equation x+y dy dx =1.

Solution:

Obtain the given equation in the form dx dy +Ax=B.

dy dx = 1 x+y dx dy =x+y dx dy x=y

Compare the simplified equation with the equation dx dy +Ax=B.

A=1,B=y

Find the integrating factor, IF:

IF= e Ady = e (1)dy = e y

Write the solution:

x e y = y e y dy +C x e y =y e y e y +C x=y1+C e y x+y+1=C e y

Context and Applications

This topic is significant in the professional exams for both undergraduate and graduate courses, especially for

  • Bachelor of Science in Mathematics
  • Master of Science in Mathematics

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