Unit – 5

First Order Ordinary Differential Equations

5.1 Exact differential Equations

Introduction-

In many engineering problems, differential equations have very important and significant role.

In this unit we will consider differential equations of first order and their solutions.

We study the solutions of differential equations which are: variables separable, homogeneous, non-homogeneous, exact, non-exact using integrating factors, linear, Bernoulli, higher degree, Clairaut’s, Lagrange’s. We consider geometrical applications and physical problems of law of natural growth, natural decay, Newton’s law of cooling, velocity of escape from earth and simple electrical circuits.

A “differential equation” (D.E.) is an equation involves an unknown function y of one or more independent variables x, t , . . . And its derivatives.

Differential equations are classified into two categories “ordinary and partial” depending on the number of independent variables appearing in the equation.

An ordinary differential equation is a D.E. In which the dependent variable y depends only on one independent variable say x (so that the derivatives of y are ordinary derivatives).

Definition-

An exact differential equation is formed by differentiating its solution directly without any other process,

Is called an exact differential equation if it satisfies the following condition-

Here
is the differential co-efficient of M with respect to y keeping x constant and
is the differential co-efficient of N with respect to x keeping y constant.

Step by step method to solve an exact differential equation-

1. Integrate M w.r.t. x keeping y constant.

2. Integrate with respect to y, those terms of N which do not contain x.

3. Add the above two results as below-

Example-1: Solve

Sol.

Here M = and N =

Then the equation is exact and its solution is-

Example-2: Solve-

Sol. We can write the equation as below-

Here M = and N =

So that-

The equation is exact and its solution will be-

Or

Example-3: Determine whether the differential function ydx –xdy = 0 is exact or not.

Solution.  Here the equation is the form of  M(x , y)dx + N(x , y)dy = 0

But, we will check for exactness,

These are not equal results, so we can say that the given diff. Eq. Is not exact.

Example: Solve (5 + 3 – 2x) dx + (2y – 3 – 5) dy = 0

Solution: Here, M = 5 + 3 – 2x, N = 2y – 3 – 5

Since,
the given equation is exact.

Now (terms is not containing ) (y constant)

Ans

Example: Solve:

Solution: Here we have

   (1)

   (2)

Comparing (1) and (2), we get

Here,

So the given differential equation is exact differential equation.

Hence solution is (terms is not containing )

Put so that

ans

Example: Solve

Solution: We

)

Given equation is exact.

Its solution is  (terms not containing )

Ans

Example: Solve:

Solution:  

Which is in the form

and

and

Hence the given differential equation is exact.

Solution is (terms not containing x)

... (1)

Now,

Equation (1) becomes

Equation reducible to exact form-

1. If M dx + N dy = 0 be an homogenous equation in x and y, then 1/ (Mx + Ny) is an integrating factor.

Example: Solve-

Sol.

We can write the given equation as-

Here,

M =

Multiply equation (1) by we get-

This is an exact differential equation-

2. I.F. For an equation of the type

IF the equation Mdx + Ndy = 0 be this form, then 1/(Mx – Ny) is an integrating factor.

Example: Solve-

Sol.

Here we have-
 

Now divide by xy, we get-

Multiply (1) by , we get-

Which is an exact differential equation-

3. In the equation M dx + N dy = 0,

(i) If be a function of x only = f(x), then is an integrating factor.

(ii) If be a function of y only = F(x), then is an integrating factor.

Example: Solve-

Sol.

Here given,

M = 2y and N = 2x log x - xy

Then-

Here,

Then,

Now multiplying equation (1) by 1/x, we get-

4. For the following type of equation-

An I.F. Is

Where-

Example: Solve-

Sol.

We can write the equation as below-

Now comparing with-

We get-

a = b = 1, m = n = 1, a’ = b’ = 2, m’ = 2, n’ = -1

I.F. =

Where-

On solving we get-

h = k = -3

Multiply the equation by , we get-

It is an exact equation.

So that the solution is-

5.2 Linear and Bernoulli’s Equations

Linear Equations

A differential equation is said to be linear if the dependent variable and its differential coefficients occur only in the first degree and not multiplied together.

Thus the standard form of the linear equation of the first order, commonly known as L              eibnitz’s linear equation is

To solve the equation, multiply both sides bi

i.e.

Integrating both sides we get, as the required solutuion.

Obs. The factor on multiplying by which the left hand side of (1) becomes the differential coefficient of the single function, is called the integrating factor (I.F.) of the linear equation (1)

It is important to remember that (I.F.)=

And the solution is

Example. Solve

Solution. Dividing throughout by (x+1),given equation becomes

which is Leibnitz’s equation

Thus the solution of (1) is

Example. Solve

Solution. Given solution can be written as

Thus solution of (1) is

Or,

Example. Solve

Solution. Putting the given equation becomes

Which is Leibnitz’s equation in z.

Thus the solution of (1) is

Hence the solution of the given equation is

Example. Solve

We have , which is a Leibnitz’sEquation in x

Thus the solution of (1) is

Example. Solve

Solution. This equation contains and is therefore not a linear in y, but since only z occurs, it can be written as

Which is a Leibnitz’s equation in z.

Thus the solution is

Bernoulli’s equation-

The equation

Is reducible to the Leibnitz’s linear equation and is usually called Bernoulli’s equation.

Working procedure to solve the Bernoulli’s linear equation-

Divide both sides of the equation -
 

By, so that

Put so that

Then equation (1) becomes-

)

Here we see that it is a Leibnitz’s linear equations which can be solved easily.

Example: Solve

Sol.

We can write the equation as-

On dividing by , we get-

Put so that

Equation (1) becomes,

Here,

Therefore the solution is-

Or

Now put

Integrate by parts-

Or
 

Example. Solve

Solution. We have,

The given equation reduces to a linear differential equation in z.

Hence the solution is

Example. Solve

Solution.

Equation (1) becomes

Solution is

Example: Solve

Sol. Here given,

Now let z = sec y, so that dz/dx = sec y tan y dy/dx

Then the equation becomes-

Here,

Then the solution will be-

Example: Solve-

Sol. Here given-

We can re-write this as-

Which is a linear differential equation-

The solution will be-

Put

5.3 Euler’s Equations

The general first order differential equation

With the initial condition

In this  method  the solution is in the form  of a  tabulated  values.

Integrating both side of the equation (i) we get

Assuming that  in  this gives Euler’s formula

In general  formula

, n=0,1,2,…..

Error estimate for the Euler’s method

Example1: Use Euler’s method to  find y(0.4) from the differential equation

    with h=0.1

Given  equation  

Here

We  break  the  interval in four steps.

So that

By Euler’s  formula

, n=0,1,2,3   ……(i)

For n=0 in equation (i) we  get

For n=1 in equation (i) we  get

.01

For n=2 in equation (i) we  get

For n=3 in equation (i) we  get

Hence y(0.4)  =1.061106.

Example2: Using  Euler’s method  solve the differential equation  for  y at x=1 in five steps

Given equation  

Here 

No. Of steps  n=5 and  so that

So that

Also

By Euler’s  formula

, n=0,1,2,3,4   ……(i)

For n=0 in equation (i) we  get

For n=1  in equation (i) we  get

For n=2  in equation (i) we  get

For n=3  in equation (i) we  get

For n=4  in equation (i) we  get

Hence 

Example3: Given  with  the initial condition  y=1 at   x=0.Find  y for x=0.1 by  Euler’s   method(five  steps).

Given equation is  

Here   

No. Of steps  n=5 and so that

So that

Also

By Euler’s  formula

, n=0,1,2,3,4   ……(i)

For n=0 in equation (i) we  get

For n=1 in equation (i) we  get

For n=2 in equation (i) we  get

For n=3 in equation (i) we  get

For n=4 in equation (i) we  get

Hence  

Modified Euler’s Method:

Instead of approximating as in Euler’s method.  In the  modified Euler’s  method we  have the iteration formula

Where is the nth approximation to .The iteration started with the Euler’s  formula

Example1: Use modified  Euler’s method  to compute y  for  x=0.05.  Given that

Result correct  to three decimal places.

Given  equation  

Here

Take  h  =  = 0.05

By modified Euler’s  formula the  initial iteration is

)

The iteration  formula by  modified Euler’s method  is

-----(i)

For n=0  in equation  (i)  we get

Where   and  as  above

For n=1  in equation  (i)  we get

For n=3  in equation  (i)  we get

Since  third and  fourth approximation are equal .

Hence y=1.0526  at x =  0.05  correct  to  three decimal places.

Example2: Using modified   Euler’s method,  obtain a solution of  the equation

Given  equation 

Here

By modified Euler’s  formula the  initial iteration is

The iteration  formula by  modified Euler’s method  is

-----(i)

For n=0  in equation  (i)  we get

Where   and  as  above

For n=1  in equation  (i)  we get

For n=2 in equation  (i)  we get

For n=3  in equation  (i)  we get

Since third and fourth approximation are equal.

Hence y=0.0952 at  x=0.1

To calculate  the value  of  at x=0.2

By modified Euler’s  formula the  initial iteration is

The iteration  formula by  modified Euler’s method  is

-----(ii)

For n=0  in equation  (ii)  we get

1814

For n=1  in equation  (ii)  we get

1814

Since first and second approximation   are  equal .

Hence  y  = 0.1814  at  x=0.2

To  calculate the value  of  at x=0.3

By modified Euler’s  formula the  initial iteration is

The iteration  formula by  modified Euler’s method  is

-----(iii)

For n=0  in equation  (iii)  we get

For n=1  in equation  (iii)  we get

For n=2 in equation  (iii)  we get

For n=3 in equation  (iii)  we get

Since third and fourth approximation are same.

Hence y  = 0.25936 at x = 0.3

5.4 Equations Not of First Degree: Equations Solvable For P

The differential equation will involve in higher degree and will not denoted by p. The differential equation will be of the form

Case 1. Equations solvable for p

Example. Solve

Solution.

Which gives on integration

Example. Solve

Solution. Given equation is where

Factorizing,

Thus we have

From (1)

From (2)

Integrating, Thus x y=c or , constitute the required solution.

Otherwise combining these into one, the required solution can be written as

Example. Solve

Solution. We have

From (1)

Integrating,

or,

From (2)

Integrating

Thus combining (3) and (4), the required general solution is

5.5 Equations Solvable For Y

Case II. Equation solvable by y

(i)                Differentiate the given equation with respect to x.

(ii)              Eliminate p from the given equation and the equation obtained as above.

(iii)           The eliminant is the required solution.

Example. Solve.

Solution.

Differentiating (1) with respect to x we obtain

On integration we get p=c

Putting the value of p in (1) we get

Example. Solve

Solution.

Differentiating it with respect to x we get

This is Leibnitz’s linear equation in x and p. Here

Therefore the solution of (ii) is

Substituting this value of x in (i) we get

The equations (iii) and (iv) taken together with parameter p constitute the general solution (i).

5.6 Equations Solvable for X and Clairaut’s Type

Case III. Equations solvable by x.

(i)                Differentiate the given equation with respect to y.

(ii)              Solve the equation as obtained in (1) for p.

(iii)           Eliminate p by putting the value of p in the given equation.

(iv)            The eliminant is the required solution.

Example. Solve

Solution.

Differentiating (2) with respect to x we get

Putting the value of p in (1)

Example. Solve

Solution. Given equation on solving for x takes the form x

Differentiating with respect to y,

Integrating,

Thus eliminating from the given equation and (i) we get which is required solution.

Class IV. Clairaut’s equation

The equation is known as Clairaut equation

Differentiating (1) with respect to x

Putting the value of p in (1) we have

y=ax+f(a)

Which is the required solution.

Method. In the Clairaut equation, on replacing p by a constant, we get the solution of the problem.

Example. Solve.

Solution. or

Which is Clairaut equation.

Here its solution is

Example. Solve

Solution. Put, so that

Then the given equation becomes

which is Clairaut form and its solution is

References:

1. G.B. Thomas And R.L. Finney, Calculus And Analytic Geometry, 9th Edition, Pearson, Reprint, 2002.
2. Erwin Kreyszig, Advanced Engineering Mathematics, 9th Edition, John Wiley & Sons, 2006.
3. W. E. Boyce And R. C. Diprima, Elementary Differential Equations And Boundary
4. Value Problems, 9th Edition, Wiley India, 2009.
5. S. L. Ross, Differential Equations, 3rd Ed., Wiley India, 1984.
6. E. A. Coddington, An Introduction To Ordinary Differential Equations, Prentice Hall India, 1995.
7. E. L. Ince, Ordinary Differential Equations, Dover Publications, 1958.

1        J. W. Brown And R. V. Churchill, Complex Variables And Applications, 7th Ed., Mc- Graw Hill, 2004.

2        N.P. Bali And Manish Goyal, A Text Book Of Engineering Mathematics, Laxmi Publications, Reprint, 2008.

3        B.S. Grewal, Higher Engineering Mathematics, Khanna Publishers, 36th Edition, 2010.

4        Hk dass