CHAPTER 7
BERNOULLI’S DIFFERENTIAL EQUATIONS
Chapter 7
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BERNOULLI’S DIFFERENTIAL EQUATIONS
Outline: 7.1 7.1 7.2 7.2
Bern oull i’s Dif fer ent ial Equations Solut ion s to Bernoul li ’s Dif ferenti al Equations
acob Bernoulli Bernoulli (also known as ames or acques) acques) (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 Leibniz–Newton 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, However, his most important contribution contribution was in the field of probability, probability, where he derived the first version of the law of large numbers in his work Ars Conjectandi. Jacob Bernoulli's Bernoulli's paper paper of 1690 is is important for the history history of calculus, calculus, since the term integral appears for the first time with its integration meaning. In 1696, he solved the equation, now called the Bernoulli’s differential equation. www.wikipedia.com
Overview: This chapter deals with the solution to first order non-linear differential equations. These equations are called as the Bernoulli’s differential equations. This equation is non-linear an d is r educible to a linear f orm by a m ethod of substitution substitution an d thu s, solution solution t o a fir st-or t-or der der linear diff er ential equation i s applicable.
Objectives: Upon comp letion of th is chapt er, the stu stu dents wi ll be able to: 1 . Define Be Ber noulli’s diff er enti al equation . 2 . Deter Deter m ine Be Ber noulli’s diff er enti al equation s. 3 . Reduce Reduce Ber noulli’s differ enti al equat equat ion into a f ir st-order lin ear diff erential equation. 4 . Solve Ber Ber nou lli’s diff erenti al equat ions.
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BERNOULLI’S DIFFERENTIAL EQUATIONS
CHAPTER 7
7 .1
Ber n o u l l i ’ s D i f f e r e n t i a l Eq u a t i o n s ( ) = ( ) is a Bernoulli’s A differential equation of the form +
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,
+ ( ) = ( ) ( − = + 1) By transformation, we can set = and ( − + 1) = . Substituting to the above equation we have, ( − + 1) + ( ) = ( ) + (− + 1)( ) = (− + 1)( )
or
Therefore, the new equation formed was now a linear differential equation. ( ) = ( + 1) ( ) By the solution of linear differential equation, set + 1), thus, the integrating factor is, and ( ) = ( )(
− ∫ = ∫ = ∫ ( )
(
(
−
) ( )
)
( )
Then, the general solution is,
∫ = (− + 1) ∫ () ∫ + (
where,
7 .2
)
( )
(
)
( )
= .
So l u t i o n s t o Ber n o u l l i ’ s D i f f e r en t i a l Eq u a t i o n s
A solution to a Bernoulli’s differential equation can be obtained by the () ∫ ( ) ( ) ( ) ∫ () = ( + 1) + , where = , as formula shown in the previous section.
−
∫
Example(a).
Obtain the general soluti on of Solution:
equation, thus,
. + =
Divide both sides of the equation by
+ = Then, set = and
, therefore,
= −3 , and substitute to the above
− + = − 3 = −3
The equation is now a linear differential equation.
CHAPTER 7
BERNOULLI’S DIFFERENTIAL EQUATIONS
By the solution of linear differential equation, set ( ) = 3 , and the integrating factor is,
−
() = −3
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and
∫ = ∫ = 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 Solution:
− [ + (1 + l n )] = 0.
Writing the equation in the form
( ) = ( ), we have, +
− = (1+ln ) Divide both sides of the equation by
− = 1 + ln Then, set = − and
equation, thus,
, therefore,
= , and substitute to the above
+ = 1 + ln
The equation is now a linear differential equation. By the solution of linear differential equation, set
() = 1 + ln , and the integrating factor is, ∫ = ∫ = = Then, = ∫(1+ln ) + = ∫( + ln ) + = + ∫ ln + ( )
( ) =
and
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BERNOULLI’S DIFFERENTIAL EQUATIONS
For
CHAPTER 7
∫ 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 + l n ) = − 4 Example(c).
Obtain the general soluti on of Solution:
+ = cos .
Divide both sides of the equation by
+ = cos Then, set = and
equation, thus,
, therefore,
= − , and substitute to the above
− + = cos − − = cos
The equation is now a linear differential equation. By the solution of linear differential equation, set ( ) = cos , and the integrating factor is,
−
() = −1
∫ = ∫ = Then, = − ∫ cos + For − ∫ cos , use integration by parts, therefore, − ∫ cos = − (sin − cos ) ( )
Substituting to the above equation, we have,
Since
= − (sin − cos ) + = − (sin − cos ) + = , thus the general solution is,
and
CHAPTER 7
BERNOULLI’S DIFFERENTIAL EQUATIONS
63
= − ( sin − cos ) + (sin − cos ) = − 2
Example(d).
+ tan = sec . Solution: + tan = sec + tan = sec Divide both sides of the equation by , therefore, + = sec − Then, set = and = , and substitute Obtain the general solution of
to the above
equation, thus,
− + tan = sec − − tan = sec
The equation is now a linear differential equation. and
By the solution of linear differential equation, set ( ) = sec , and the integrating factor is,
− ∫ = ∫ = = cos Then, cos = ∫− sec cos cos = − ∫ + cos = − + Since = , thus the general solution is, cos = − + ( − ) = cos ( )
Example(e).
Obtain the general solution of Solution:
+ ( − ) = 0.
+ ( − ) = 0 + =
Divide both sides of the equation by
+ =
, therefore,
() = − tan
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BERNOULLI’S DIFFERENTIAL EQUATIONS
Then, set
=
CHAPTER 7
and
equation, thus,
= −4 , and substitute to the above
− + = − − 4 = 4
The equation is now a linear differential equation. By the solution of linear differential equation, set ( ) = 4 , and the integrating factor is,
−
( ) = −4 and
∫ = ∫ = ( )
Then,
= − ∫ 4 + = + = 1 + Since = , thus the general solution is, = 1 + 1 + = 1
Example(f).
Obtain the particular solution of condition ( 1) = 0.
Solution:
+ = ,
that satisfies the
+ = + =
Divide both sides of the equation by
+ = Then, set = and equation, thus,
, therefore,
− = 2 and substitute to the above
− + = − − 2 = 2
The equation is now a linear differential equation. By the solution of linear differential equation, set ( ) = 2 , and the integrating factor is,
−
∫ = ∫ = ( )
Then,
() = −2
and
CHAPTER 7
BERNOULLI’S DIFFERENTIAL EQUATIONS
For
65
= −2 ∫ + −2 ∫ , use integration by parts, therefore, −2 ∫ = + +
Substituting to the above equation, we have,
= + + = + + Since = , thus the general solution is, = + + (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 condition
() = 2.
− cos = cos , that satisfies the
Solution:
Divide both sides of the equation by
− = cos Then, set = − and
, therefore,
= , and substitute to the above
equation, thus,
+ cos = cos The equation is now a linear differential equation. By the solution of linear differential equation, set ( ) = cos , and the integrating factor is,
∫ = ∫ = Then, = ∫ cos + = + = 1 + ( )
( ) = cos and
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BERNOULLI’S DIFFERENTIAL EQUATIONS
CHAPTER 7
= − , thus the general solution is, − = 1 − − 1 = 1 To find the particular solution, set = and = 2, then solve for Since
.
When
= and = 2, = .
Then, the particular solution is,
− 1 = 1 3 − 2 = 2 Su p p l e m en t a r y Pr o b l em s I . O b t a i n t h e g en e r a l so l u t i o n o f t h e f o l l o w i n g d i f f e r en t i a l eq u a t i o n s .
1. 2. 3.
+ 2 = − = ( ) + = (
)
+ = 5. + = 6. + = 4.
7. 8.
+ = + = ( − 1)
9. + + ( + 2) = 0 10. + ( tan − ) = 0
Ans: Ans: Ans:
(1 + ) = 2 (2 + + ) = −4( + 1) =
Ans:
(4 − 3) =
Ans:
= 1 +
Ans:
(2 + ) = 1 (2 + ) = 1 ( − ln − 1) = 1
Ans:
3 + 20 + 60 +
Ans:
( − 2 − sin2 ) = 2cos
Ans: Ans:
= 225
I I . O b t a i n t h e p a r t i c u l a r so l u t i o n s sa t i s f y i n g t h e i n d i c a t e d co n d i t i o n s .
+ 2 = , (0) = 1 2. − = , (1) = −1 1.
Ans: Ans:
(1 − 2) = (3 − 2) =
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+ 3 = , (0) = −3 4. (2 − ) + = 0, (0) = 1 5. + cot − sin = 0, = 1 3.
BERNOULLI’S DIFFERENTIAL EQUATIONS
Ans: Ans: Ans:
1 − 2 = 3 1 + = 2 (2 + 2 − ) sin = 2
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BERNOULLI’S DIFFERENTIAL EQUATIONS
CHAPTER 7
