Chapter 7

CHAPTER 7 BERNOULLIS DIFFERENTIAL EQUATIONS 59

Author: Harold Jan R. Terano, ECE, MET

Chapter 7 Outline: 7.1 Bernoullis Differential Equations 7.2 Solutions to Bernoullis Differential Equations

Overview: This chapter deals with the solution to first order non-linear differential equations. These equations are called as the Bernoullis differential equations. This equation is non-linear and is reducible to a linear form by a method of substitution and thus, solution to a first-order linear differential equation is applicable.

Objectives:

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

1. Define Bernoullis differential equation. 2. Determine Bernoullis differential equations. 3. Reduce Bernoullis differential equation into a first-order linear

differential equation. 4. Solve Bernoullis differential equations.

BERNOULLIS DIFFERENTIAL

EQUATIONS

Jacob Bernoulli (also known as James or Jacques) (December 27, 1654/ January 6, 1655 August 16, 1705) was one of the many prominent mathematicians in the Bernoulli family. He was an early proponent of Leibnizian calculus and had sided with Leibniz during the LeibnizNewton calculus controversy. He is known for his numerous contributions to calculus, and along with his brother Johann, was one of the founders of the calculus of variations. However, his most important contribution was in the field of probability, where he derived the first version of the law of large numbers in his work Ars Conjectandi. Jacob Bernoulli's paper of 1690 is important for the history of calculus, since the term integral appears for the first time with its integration meaning. In 1696, he solved the equation, now called the Bernoullis differential equation.

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60 BERNOULLIS CHAPTER 7 DIFFERENTIAL EQUATIONS


7.1 Bernoullis Differential Equations

A differential equation of the form + ( ) = ( ) is a Bernoullis differential equation, where and are all function of alone. To solve for the general solution of the differential equation, divide both sides of the equation by thus, we have,

+ ( ) = ( )

By transformation, we can set = and = ( + 1) or ( + 1) = . Substituting to the above equation we have,

( + 1) + ( ) = ( )

+ ( + 1) ( ) = ( + 1) ( )

Therefore, the new equation formed was now a linear differential equation. By the solution of linear differential equation, set ( ) = ( + 1) ( ) and ( ) = ( )( + 1), thus, the integrating factor is,

( ) = ( ) ( )

= ( ) ( )

Then, the general solution is,

( ) ( ) = ( + 1) ( ) ( ) ( ) +

where, = .

7.2 Solutions to Bernoullis Differential Equations

A solution to a Bernoullis differential equation can be obtained by the formula ( ) ( ) = ( + 1) ( ) ( ) ( ) + , where = , as shown in the previous section.

Example(a). Obtain the general solution of + = . Solution: Divide both sides of the equation by , therefore, + =

Then, set = and = 3 , and substitute to the above equation, thus, + =

3 = 3

The equation is now a linear differential equation.



By the solution of linear differential equation, set ( ) = 3 and ( ) = 3 , and the integrating factor is,

( ) =

= Then, = 3 +

= 3 +

For 3 , use integration by parts, therefore,

3 = + + Substituting to the above equation, we have,

= + + Since = , thus the general solution is,

= + +

= + +

Example(b). Obtain the general solution of [ + (1 + ln )] = 0. Solution:

Writing the equation in the form + ( ) = ( ), we have,

= (1 + ln )

Divide both sides of the equation by , therefore,

= 1 + ln

Then, set = and = , and substitute to the above equation, thus, + = 1 + ln


By the solution of linear differential equation, set ( ) = and ( ) = 1 + ln , and the integrating factor is, ( ) =

=

= Then, = (1 + ln ) +

= ( + ln ) +

= + ln +



For ln , use integration by parts, therefore,

ln = ln +

Substituting to the above equation, we have,

= + ln +

= + ln +

Since = , thus the general solution is,

= + ln +

4 = 2 ln +

(1 + ln ) = 4

Example(c). Obtain the general solution of + = cos . Solution: Divide both sides of the equation by , therefore,

+ = cos

Then, set = and = , and substitute to the above equation, thus, + = cos

= cos


By the solution of linear differential equation, set ( ) = 1 and ( ) = cos , and the integrating factor is,

( ) = = Then, = cos +

For cos , use integration by parts, therefore,

cos = (sin cos )


= (sin cos ) +

= (sin cos ) +




= (sin cos ) +

(sin cos ) = 2 Example(d). Obtain the general solution of + tan = sec . Solution: + tan = sec

+ tan = sec


+ = sec

Then, set = and = , and substitute to the above equation, thus, + tan = sec

tan = sec


By the solution of linear differential equation, set ( ) = tan and ( ) = sec , and the integrating factor is,

( ) = =

= cos Then, cos = sec cos

cos = +

cos = +


cos = +

( ) = cos

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

+ =


+ =



Then, set = and = 4 , and substitute to the above equation, thus, + =

4 = 4



( ) =

= Then, = 4 +

= +

= 1 +


= 1 +

1 + = 1

Example(f). Obtain the particular solution of + = , that satisfies the condition (1) = 0. Solution: + =

+ =


+ =

Then, set = and = 2 and substitute to the above equation, thus, + =

2 = 2



( ) =

= Then,



= 2 +

For 2 , use integration by parts, therefore,

2 = + +


= + +

= + +


= + +

(2 + 1 + ) = 2

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

.

When = 1 and = 0, = 1.

Then, the particular solution is,

(2 + 1 + ) = 2

Example(g). Obtain the particular solution of cos = cos , that satisfies the condition ( ) = 2. Solution: Divide both sides of the equation by , therefore,

= cos

Then, set = and = , and substitute to the above

equation, thus,

+ cos = cos


By the solution of linear differential equation, set ( ) = cos and ( ) = cos , and the integrating factor is,

( ) =

= Then, = cos +

= +

= 1 +




= 1

1 = 1

To find the particular solution, set = and = 2, then solve for . When = and = 2, = .

Then, the particular solution is,

1 = 1

3 2 = 2

Supplementary Problems

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

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

2. ( )

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

3. + = Ans: =

4. + = Ans: (4 3 ) =

5. + = Ans: = 1 +

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

7. + = Ans: (2 + ) = 1

8. + = ( 1) Ans: ( ln 1) = 1

9. + + ( + 2) = 0 Ans: 3 + 20 + 60 + = 225

10. + ( tan ) = 0 Ans: ( 2 sin 2 ) = 2cos

II. Obtain the particular solutions satisfying the indicated conditions.

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

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



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

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

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

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