Unit 3: Linear differential equation-II - Subjective Questions

Definition: A linear differential equation with constant coefficients is an equation of the form

where are constants and is a known function of .

Using the differential operator , the equation can be written as

where

The complete solution is

where is the complementary function obtained from , and is the particular integral satisfying .

For a homogeneous equation

we form the auxiliary equation by replacing with :

The roots of this polynomial determine the complementary function.

Rules for roots:

• If is a real root, the corresponding term is .

• If is a repeated root of multiplicity , the corresponding terms are

• If the roots are , the corresponding real terms are

• If are repeated times, multiply the corresponding expression successively by .

Thus, the complementary function is obtained by combining the terms associated with all roots of the auxiliary equation.

The equation is

1. Complementary function:

The auxiliary equation is

which gives

Therefore, the complementary function is

2. Particular integral:

For the right-hand side ,

Using the rule , we observe that . Hence, resonance occurs. Write

Let . Then

Thus,

A suitable particular solution is , since . Therefore,

Hence, the complete solution is

Let the equation be

The particular integral is written as

Important rules are:

• For an exponential function,

  provided that .

• For trigonometric functions, replace by and use the real or imaginary part:

• If for , or for trigonometric functions, the expression is undefined. In that case, multiply the trial particular integral by a suitable power of until it becomes valid.

These rules convert differentiation into algebraic substitution and simplify the calculation of the particular integral.

The equation is

The auxiliary equation of the associated homogeneous equation is

with roots . Since the forcing function is , its frequency corresponds to the roots . Therefore, resonance occurs.

Normally, the trial form would be

but this duplicates terms in the complementary function. Hence, multiply by :

Using the operator shift rule, let . A direct substitution gives

Therefore, the particular integral is

Resonance: Resonance occurs when the forcing term has the same exponential or trigonometric form as a term in the complementary function. The trial particular integral must then be multiplied by , or by a higher power of if the corresponding root is repeated.

The method of undetermined coefficients is used to determine a particular integral when the forcing function belongs to a suitable class of functions whose derivatives have the same general form.

Suppose

Choose a trial form for containing unknown constants, substitute it into the differential equation, and compare coefficients to determine those constants.

Common trial forms:

• If is a polynomial of degree , choose

• If , choose

• If or , choose

• If , choose

  where is an unknown polynomial of degree .

• If , use

If the trial form duplicates a complementary-function term, multiply it by the lowest required power of .

The differential equation is

1. Complementary function:

The auxiliary equation is

so

Thus,

2. Particular integral:

Since the right-hand side is a quadratic polynomial, assume

Then

Substitution gives

On comparing coefficients:

Therefore,

Hence,

The complete solution is

Operator method:

• Represents differentiation by .
• Expresses the equation as .
• Uses inverse-operator rules and shift rules.
• Is especially convenient for exponential and trigonometric forcing functions.

Method of undetermined coefficients:

• Assumes a trial form for the particular integral.
• Contains unknown constants or polynomial coefficients.
• Determines these constants by substitution and coefficient comparison.
• Is convenient for polynomial, exponential, trigonometric, and their combinations.

Similarities:

• Both methods first determine the complementary function from the auxiliary equation.

• Both produce the complete solution in the form

• Both require modification of the trial form in the case of resonance.

Difference: The operator method is primarily algebraic and uses inverse operators, whereas the undetermined-coefficient method directly constructs and verifies a suitable form for .

We have

1. Complementary function:

The auxiliary equation is

whose roots are

Therefore,

2. Particular integral:

Using the shift rule,

where . Thus,

Hence,

Since annihilates and , resonance occurs. Applying the repeated-root rule gives

Therefore, the complete solution is

Variation of parameters is a method in which the constants in the complementary function are replaced by functions of .

Let and be two linearly independent solutions of the homogeneous equation

The complementary function is

For the non-homogeneous equation, assume

Impose the auxiliary condition

Then

Solving the two simultaneous equations gives

where the Wronskian is

Therefore,

The particular solution is

Hence, the complete solution is .

The Wronskian of two functions and is defined by

In variation of parameters, the functions and are obtained from

The determinant of this system is the Wronskian. Thus,

For the method to work, and must be linearly independent, so that

on the interval under consideration. If the Wronskian is zero identically, the two functions are linearly dependent and cannot form a fundamental set of solutions.

The equation is

The complementary solutions are

Thus,

The Wronskian is

For variation of parameters,

Integrating,

Also,

Since ,

The term produces a complementary-function term and may be omitted. Therefore,

Hence,

Method of undetermined coefficients:

• Assumes a particular solution with unknown constants.
• Works well when the forcing function is a polynomial, exponential, sine, cosine, or a finite combination of these.
• Advantage: Computation is usually short and algebraic.
• Limitation: It is not generally applicable to functions such as , , or arbitrary functions.

Method of variation of parameters:

• Replaces the constants in the complementary function by functions of .
• Uses the fundamental solutions and their Wronskian.
• Advantage: It applies to a much wider class of forcing functions.
• Limitation: The integrations involved may be lengthy or difficult.

Both methods yield a particular solution, which is then added to the complementary function to obtain the complete solution.

The equation is

The complementary function is

The Wronskian is

Let

Using variation of parameters,

Integrating, one possible choice is

Therefore,

After simplifying and omitting terms belonging to the complementary function,

Thus, the complete solution is

An Euler-Cauchy equation is a linear differential equation in which the coefficient of each derivative is a corresponding power of . Its general second-order form is

Use the substitution

Then

and

Therefore,

The equation becomes

which is a linear differential equation with constant coefficients in . It can then be solved using the operator method.

Assume a solution of the form

Then

Substituting into the equation gives

After simplifying,

Since ,

or

Thus,

so is a repeated root.

For a repeated root , the complementary function is

Therefore,

For the homogeneous Euler-Cauchy equation

assume . The auxiliary equation is

or equivalently,

The forms of the complementary function are:

• Distinct real roots and :

• Repeated real root :

• Complex roots :

The logarithmic terms appear because the independent variable transformation is .

The equation is

1. Complementary function:

Assume . Then

so

which gives . Hence,

2. Particular integral:

Since the right-hand side is , assume

Then

Substitution gives

so

Therefore, and

Thus, the complete solution is

Let

Using the transformation formulas,

Substitute these into the equation:

This gives

Therefore, the transformed equation is

In operator notation, it is

This is a linear differential equation with constant coefficients and can be solved using the methods for constant-coefficient equations.

Put

Then

The equation becomes

or

In operator form,

Complementary function:

The repeated root is , so

Particular integral:

Let . Then

Thus,

so

Integrating twice and ignoring complementary-function terms,

Therefore,

Since and ,