Solution of second order differential equations with variable coefficients

Objective:

To know second order differential equations with variable coefficients.

To study the methods of solving second order differential equations with variable coefficients.

Modules

Module I- Transformation of the equation by changing the dependent variable.

Module II- Equations of the form , in which is explicitly absent.

Module III- Equations of the form , in which is explicitly absent.

Module IV- Equations of the form, which are homogeneous in and (but not in).

Module V- Equations of the form, which are exact.

Introduction:

The general form of a linear differential equation of second order with variable coefficients is

,

where the coefficient of is unity and andare functions of .

In this section we shall discuss the methods of solving such equations by the method of reduction of order.

Module I- Transformation of the equation by changing the dependent variable

Let the equation be

.

Let us assume that one solution of the corresponding homogeneous equation, namely

is known. Let it be.

We then assume that is a solution.

Then

and.

Substituting these values of in the given equation, we get

i.e.,, since is a solution of the homogeneous equation

.

Dividing throughout by, we get

,

i.e.,, where and .

Putting, we obtain

,

which is a linear differential equation of the first order in and can be solvable. Thus by changing the dependent variableto, we have reduced the order of the equation by one.

Note: To find one solution of the equation , the following hints may be useful:

(i)If, then is a solution of the equation.

(ii)If, then is a solution of the equation.

(iii)If, then is a solution of the equation.

Example: Solve the equation, by finding one solution of the corresponding homogeneous equation by inspection and reducing the order of the equation.

Solution:

The homogeneous equation corresponding to the given equation is

,

which is of the form .

Here,and.

The condition is satisfied.

is a solution of the homogeneous equation.

Letbe a solution of the given equation.

Then

Using these values of in the given equation, it becomes

i.e.,

i.e.,, where .

This is a linear equation of the first order.

Integrating factor= .

Solution of this equation is

.

i.e.,.

Solving this equation, we get

.

Therefore, the solution of the given equation is

.

Note: If i.e., the given equation is homogeneous, then from the equation, we have

or

i.e.,.

Integrating both sides with respect, we get

Therefore,.

Solving this equation for and hence we will get the required solution of the given equation.

Example: Solve the equation, given that is a solution.

Solution:

Note that if is a solution of the equation, then will also be a solution of, where.

The given equation can be rewritten as

.

Here and.

Sinceis a solution, we have is also a solution of the equation, where

.

Hence the general solution of the given equation is

or.

Now, we shall discuss the method of reduction of order for some special types of second order differential equations.

Module II- Equations of the form , in which is explicitly absent

In this type of equations puttingso that.

Then theequation gets transformed as, which is only a first order equation.

Solving this transformed equation, we get

Again, solving this equation, we get

.

Extension:

Equations of the form can be solved by putting and reducing the order successively.

Example: Solve the equation, given that and .

Solution:

The given equation does not contain explicitly.

Putting and treating as a function of , we have

and the equation becomes

i.e.,, which is a linear equation of first order in .

Integrating factor.

And the solution is

i.e.,

or

Solving for, we get

i.e., the solution of the given equation is

Using the condition, we get .

Using the condition, we get is arbitrary.

Taking, the required solution is

ModuleIII- Equations of the form , in which is explicitly absent

Putting and treat as a function of. Then

The equation get transformed as , which is a first order equation.

Solving this transformed equation, we get

Solving this equation further, we get

Extension:

Equations of the form can be solved by

the above technique, by putting and treating as a function of.

Example: Solve the equation.

Solution:

The given equation is

,

which does not contain explicitly.

Putting and treating as a function of , we have .

Substituting these values of and in the given equation, we get

i.e.,, which is in the separable form.

Integrating both sides, we get

i.e.,.

Solving for, we get

.

i.e.,.

Therefore, the required solution is

Module IV- Equations of the form, which are homogeneous in and (but not in)

By putting, the order of the equation can be reduced by one and hence solved. When, and. Thus the order of the transformed equation in the dependent variable will be one less than that of the given equation.

Example: Solve the equation.

Solution:

The given equation is ,

which is homogeneous of degree 2 in .

Putting , and, we get

i.e.,.

Solving for z, we get

.

Therefore, .

or.

Module V- Equations of the form, which are exact

In this type the equation can be expressed as .

The first integral of this equation is, which is a first order equation, solving which we get the solution of the given second order equation.

Note: The equation is exact if and only if.

Proof:

Let.

Comparing like terms, we get

and

Differentiating both sides of , we get

Therefore,.

Conversely, when, we have

=, since

=

=

=

=

Thus,is an exact equation, when .

Example: Show that the equation is exact and hence solve it.

Solution:

The given equation is

.

Comparing with the standard form , we have

.

Now,.

Thus, the condition for exactness is satisfied.

Now, the given equation can be rewritten as

.

i.e.,.

or.

Integrating, we get

.

or.

i.e.,,

which is a linear differential equation of the first order in y.

The integrating factor =.

And the required solution is

.

i.e.,.

Assignment questions

  1. Solve the equation by reduction of order after finding one solution of the corresponding homogeneous equation.
  2. If is a solution of, solve it by the method of reduction of order.

Reduce the order of the following equations by suitable substitutions and hence solve them

  1. .
  2. .

Show that the following equations are exact and hence solve them:

  1. .
  2. .

Quiz questions

  1. A solution of the equation , ifis

(a)(b) (c)

  1. What is the substitution to be made to convert the equation which is homogeneous in , into a first order differential equation

(a)(b) (c)

  1. The differential equation is exact if

(a) (b)(c)

Answers

(1)(b)(2) (a)(3) (c)

Glossary

Linear differential equation: A differential equation is said to be linear if the dependent variable and its differential coefficients occur only in the first degree and are not multiplied together. The standard form is, where and are constants or functions of alone and the solution is

Exact equations: A differential equation is said to be exact if it can be derived from its primitive directly by differentiation, without any subsequent multiplication, elimination or reduction etc.

Integral: The solution of a differential equation is also called the integral.

Summary

The general form of a linear differential equation of second order with variable coefficients is , where the coefficient of is unity and andare functions of .

If is a solution of the equation, then will also be a solution of, where.

Equations of the form can be solved by putting and reducing the order successively.

Equations of the form can be solved by putting and treating as a function of.

For the equations of the form, which are homogeneous in and (but not in), putto reduce the order of the equation by one and hence we can solve it.

Equations of the form, which are exact, can be expressed as so that the first integral of this equation is, which is a first order equation and then solving to get the solution of the given second order equation.

The equation is exact if and only if.

Frequently asked questions (FAQs)

  1. Solve the equation, by the method of reduction of order.

Solution:

Here

Note that.

Therefore,is a solution.

Let be a solution.

Then .

Substituting, we get

i.e.,.

i.e.,,where.

This is a linear equation.

I.F. =.

The solution is

i.e.,

.

Integrating, we get

.

Therefore, the required solution of the given equation is

.

  1. Solve the equation, given that is a solution.

Solution:

Since is a solution, is also a solution of the given equation, where, since

.

Integrating, we get

or .

Therefore, the general solution is

.

i.e.,.

  1. Solve the equation; .

Solution:

The given equation is can be considered as one not containing explicitly.

Putting and treating as a function of x, we have.

Then the equation becomes

Integrating,

i.e.,.

Now, .

Thus,

Therefore, .

Solving this, we get

.

Now, .

Therefore, the required solution is

.

(Note that this problem can be considered as one not containing explicitly.

By putting and treating as a function of y, in the equation we may proceed to obtain the same solution as obtained in the above method).

  1. Solve ,

Solution:

Here, .

Now .

Hence the given equation is exact. It can be written as

i.e.,.

Therefore, the first integral is

.

i.e.,,

which is linear.

I.F.= .

The solution is

i.e.,

Books

1)Ahsan, Zafar, Differential Equations and Their Applications, 2nd ed.,Prentice-Hall, New Delhi (2006).

2)Simmons, F., Differential Equations, Tata McGraw-Hill, New Delhi (1996).

3)Ross, L., Differential Equations, 3rd ed., Wiley, New Delhi (2009).

4)Braun, M., Differential Equations and Their Applications, Springer- Verlag, Berlin (1975).

5)Kelly, L.M., Elementary Differential Equations, 6th ed., McGraw Hill, NewYork (1965).

6)Raisinghania, M.D., Ordinary and partial Differential Equations, S.Chand, New Delhi(2010).