CHAPTER 6 FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS 51

Author: Harold Jan R. Terano, ECE, MET

FIRST-ORDER LINEAR

DIFFERENTIAL EQUATIONS

Chapter 6 Outline: 6.1 First-Order Linear Differential Equations 6.2 Solutions to First-Order Linear Differential Equations

Overview: Another type of differential equation is the first-order linear differential equation. With the aid of integrating factor, the solution of this differential equation is obtained.

Objectives:

Upon completion of this chapter, the students will be able to:

1. Define first-order linear differential equation. 2. Determine first-order linear differential equations. 3. Solve first-order linear differential equations.

Augustin-Louis Cauchy (August 21, 1789 May 23, 1857) was a French mathematician who was an early pioneer of analysis. He started the project of formulating and proving the theorems of infinitesimal calculus in a rigorous manner, rejecting the heuristic principle of the generality of algebra exploited by earlier authors. He defined continuity in terms of infinitesimals and gave several important theorems in complex analysis and initiated the study of permutation groups in abstract algebra. A profound mathematician, Cauchy exercised a great influence over his contemporaries and successors. His writings cover the entire range of mathematics and mathematical physics. Cauchy verified the existence of recurrent elliptic functions, gave the first impetus to the general theory of functions and laid the foundation for the modern treatment of the convergence of infinite series. He perfected the method of integration of linear differential equations and invented the calculus of residues.

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52 FIRST-ORDER LINEAR CHAPTER 6 DIFFERENTIAL EQUATIONS

Author: Harold Jan R. Terano, ECE, MET

6.1 First-Order Linear Differential Equations

A first-order linear differential equation can be written in the form + ( ) = ( ), where and are all function of alone.

To solve for the general solution of the differential equation, we write the equation in the form ( , ) + ( , ) = 0, thus, from + ( ) = ( ), we have, + ( ) = ( )

[ ( ) ( )] + = 0

Now, set = ( ) ( ) and = 1, thus, = ( ) and = 0. The differential equation is not exact, therefore, we can apply the concept of integrating factors. From,

= ( ) = ( )

we obtained ( ) as the integrating factor.

From the original differential equation,

[ ( ) ( )] + = 0

Multiply both sides of the equation by the integrating factor ( ) ,

thus, we have,

{[ ( ) ( )] + = 0} ( )

( ) ( ) ( ) ( ) + ( ) = 0

Since,

( ) ( ) + ( ) = ( )

Then,

( ) ( ) ( ) = 0

Integrating both sides of the equation, the general solution is,

( ) ( ) ( ) =0

( ) ( ) ( ) =

( ) = ( ) ( ) + or

( ) = ( ) ( ) +

where = ( ) .

CHAPTER 6 FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS 53

Author: Harold Jan R. Terano, ECE, MET

6.2 Solutions to First-Order Linear Differential Equations

A solution to a first-order linear differential equation can be obtained by the formula ( ) = ( ) ( ) + , which is the obtained solution to a first-order linear differential equation of the form + ( ) = ( ).

Example(a). Obtain the general solution of ( + ) = 0. Solution:

Write the equation in the form + ( ) = ( ), therefore,

=

Find ( ) and ( ), thus,

( ) = and ( ) =

For integrating factor,

= ( ) =

= =

= =

By ( ) = ( ) ( ) + , the general solution is,

= +

= +

= + or = +

Example(b). Obtain the general solution of + = 4 . Solution:

Write the equation in the form + ( ) = ( ), therefore,

+ = 4

Find ( ) and ( ), thus,

( ) = and ( ) = 4

For integrating factor,

= ( ) = =

By ( ) = ( ) ( ) + , the general solution is,

= 4 +

= 4 + = 4 +

54 FIRST-ORDER LINEAR CHAPTER 6 DIFFERENTIAL EQUATIONS

Author: Harold Jan R. Terano, ECE, MET

Example(c).

Obtain the general solution of + = 0. Solution:

Write the equation in the form + ( ) = ( ), therefore,

+ =

Find ( ) and ( ), thus,

( ) = and ( ) =

For integrating factor,

= ( ) = = ( )

= + 1

By ( ) = ( ) ( ) + , the general solution is,

( + 1) = ( + 1) +

( + 1) = ( + ) +

For , use integration by parts, therefore,

=

Substituting to the formula, the general solution is,

( + 1) = + +

( + 1) = + or =

Example(d). Obtain the general solution of sec + ( csc 1) = 0. Solution:

Write the equation in the form + ( ) = ( ), therefore,

+ cot = cos

Find ( ) and ( ), thus,

( ) = cot and ( ) = cos

For integrating factor,

= ( ) = = ( )

= sin

By ( ) = ( ) ( ) + , the general solution is,

sin = cos sin +

sin = sin +

= sin + csc

CHAPTER 6 FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS 55

Author: Harold Jan R. Terano, ECE, MET

Example(e). Obtain the general solution of = ln . Solution:

For this example, we can also use the form + ( ) = ( ) and the formula will be ( ) = ( ) ( ) + , therefore, = ln

Find ( ) and ( ), thus,

( ) = and ( ) = ln

For integrating factor,

= ( ) = = =

= =

By ( ) = ( ) ( ) + , the general solution is,

= ln +

= ln + or = ln +

Example(f). Obtain the particular solution of = ( + sin ) , that satisfies the condition, = 0.

Solution:

Write the equation in the form + ( ) = ( ), therefore,

= sin

Find ( ) and ( ), thus,

( ) = and ( ) = sin

For integrating factor,

= ( ) = =

= =

By ( ) = ( ) ( ) + , the general solution is,

= sin +

= cos +

= cos +

To find the particular solution, set, = and = 0, then solve for .

56 FIRST-ORDER LINEAR CHAPTER 6 DIFFERENTIAL EQUATIONS

Author: Harold Jan R. Terano, ECE, MET

When = and = 0, = 0.

Then, the particular solution is,

= cos

Example(g). Obtain the particular solution of + ( + 3 ) = 0, that satisfies the condition (0) = 1. Solution: Write the equation in the form + ( ) = ( ), therefore,

+ 3 =

Find ( ) and ( ), thus,p

( ) = 3 and ( ) =

For integrating factor,

= ( ) =

=

By ( ) = ( ) ( ) + , the general solution is,

= ( ) +

= +

= +

= +

To find the particular solution, set = 0 and = 1, then solve for

.

When = 0 and = 1, = .

Then, the particular solution is,

= +

Supplementary Problems

I. Obtain the general solution of the following differential equations.

1. + = Ans: = +

2. + 2 = Ans: = +

3. + = Ans: = +

4. sin + cos = 1 Ans: sin = +

CHAPTER 6 FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS 57

Author: Harold Jan R. Terano, ECE, MET

5. + 6 = 2 Ans: = +

6. (1 2 ) = 0 Ans: = +

7. + 3 = 1 Ans: = +

8. = + 3 Ans: + =

9. = ( 2 ) Ans: 4 = 2 1 +

10. = 2 cot 2 Ans: 4 = 1 2 cot 2 + csc 2

II. Obtain the particular solutions satisfying the indicated conditions.

1. + = 6 , (1) = 2 Ans: ( ) = 1

2. + 2 = 1, (0) = 2 Ans: = 3

3. ( 2 ) + = 0, (1) = 3 Ans: ( ) = 4

4. = ( 4 ) , (0) = Ans: 4 = 1 2

5. ( tan sin 2 ) + = 0, (0) = 1 Ans: = cos (1 2 cos )

58 FIRST-ORDER LINEAR CHAPTER 6 DIFFERENTIAL EQUATIONS

Author: Harold Jan R. Terano, ECE, MET