Differential Equations BERNOULLI EQUATIONS. Graham S McDonald. A Tutorial Module for learning how to solve Bernoulli differential equations

Transcription

1 Differential Equations BERNOULLI EQUATIONS Graham S McDonald A Tutorial Module for learning how to solve Bernoulli differential equations Table of contents Begin Tutorial c 2004 [email protected]

2 Table of contents. Theory 2. Exercises 3. Answers 4. Integrating factor method 5. Standard integrals 6. Tips on using solutions Full worked solutions

3 Section : Theory 3. Theory A Bernoulli differential equation can be written in the following standard form: + P (x)y = Q(x)yn, where n (the equation is thus nonlinear). To find the solution, change the dependent variable from y to z, where z = y n. This gives a differential equation in x and z that is linear, and can be solved using the integrating factor method. Note: Dividing the above standard form by y n gives: ( where we have used y n + P (x)y n = Q(x) + P (x)z = Q(x) ( n) = ( n)y n ).

4 Section 2: Exercises 4 2. Exercises Click on Exercise links for full worked solutions (there are 9 exercises in total) Exercise. The general form of a Bernoulli equation is + P (x)y = Q(x) yn, where P and Q are functions of x, and n is a constant. Show that the transformation to a new dependent variable z = y n reduces the equation to one that is linear in z (and hence solvable using the integrating factor method). Solve the following Bernoulli differential equations: Exercise 2. x y = xy2 Theory Answers IF method Integrals Tips

5 Section 2: Exercises 5 Exercise 3. + y x = y2 Exercise y = ex y 4 Exercise 5. x + y = xy3 Exercise x y = x2 cos x y 2 Theory Answers IF method Integrals Tips

6 Section 2: Exercises 6 Exercise 7. 2 (4x + 5)2 + tan x y = cos x y3 Exercise 8. x + y = y2 x 2 ln x Exercise 9. = y cot x + y3 cosecx Theory Answers IF method Integrals Tips

7 Section 3: Answers 7 3. Answers. HINT: Firstly, divide each term by y n. Then, differentiate z with respect to x to show that 2. y = x2 3 + C x, 3. y = x(c ln x), 4. y 3 = e x (C 3x), 5. y 2 = 2x+Cx 2, 6. y = x2 (sin x + C), 7. y 2 = 2 cos x (4x + 5)3 + C cos x, 8. xy = C + x( ln x), ( n) = y n,

8 Section 3: Answers 8 9. y 2 = sin2 x 2 cos x+c.

9 Section 4: Integrating factor method 9 4. Integrating factor method Consider an ordinary differential equation (o.d.e.) that we wish to solve to find out how the variable z depends on the variable x. If the equation is first order then the highest derivative involved is a first derivative. If it is also a linear equation then this means that each term can involve z either as the derivative OR through a single factor of z. Any such linear first order o.d.e. can be re-arranged to give the following standard form: + P (x)z = Q (x) where P (x) and Q (x) are functions of x, and in some cases may be constants.

10 Section 4: Integrating factor method 0 A linear first order o.d.e. can be solved using the integrating factor method. After writing the equation in standard form, P (x) can be identified. One then multiplies the equation by the following integrating factor : IF= e P (x) This factor is defined so that the equation becomes equivalent to: d (IF z) = IF Q (x), whereby integrating both sides with respect to x, gives: IF z = IF Q (x) Finally, division by the integrating factor (IF) gives z explicitly in terms of x, gives the solution to the equation.

11 Section 5: Standard integrals 5. Standard integrals f (x) x n x f(x) f (x) f(x) xn+ n+ (n ) [g (x)] n g (x) ln x g (x) g(x) [g(x)] n+ n+ (n ) ln g (x) e x e x a x ax ln a (a > 0) sin x cos x sinh x cosh x cos x sin x cosh x sinh x tan x ln cos x tanh x ln cosh x cosec x ln tan x 2 cosech x ln tanh x 2 sec x ln sec x + tan x sech x 2 tan e x sec 2 x tan x sech 2 x tanh x cot x ln sin x coth x ln sinh x sin 2 x cos 2 x x 2 x 2 sin 2x 4 sinh 2 x sinh 2x 4 x 2 + sin 2x 4 cosh 2 x sinh 2x 4 + x 2

12 Section 5: Standard integrals 2 f (x) f (x) f (x) f (x) a 2 +x 2 a tan x a a 2 x 2 2a ln a+x a x (0< x <a) (a > 0) x 2 a 2 2a ln x a x+a ( x > a>0) a 2 x 2 sin x a a 2 +x 2 ( a < x < a) x2 a 2 ln ln x+ a 2 +x 2 a x+ x 2 a 2 a (a > 0) (x>a>0) a2 x 2 a 2 2 [ sin ( ) x a a2 +x 2 a 2 2 ] x2 + x a 2 x 2 a a 2 a [ [ sinh ( x a cosh ( x a ) + x a 2 +x 2 a 2 ] ) + x ] x 2 a 2 a 2

13 Section 6: Tips on using solutions 3 6. Tips on using solutions When looking at the THEORY, ANSWERS, IF METHOD, INTEGRALS or TIPS pages, use the Back button (at the bottom of the page) to return to the exercises. Use the solutions intelligently. For example, they can help you get started on an exercise, or they can allow you to check whether your intermediate results are correct. Try to make less use of the full solutions as you work your way through the Tutorial.

14 Solutions to exercises 4 Full worked solutions Exercise. + P (x)y = Q(x)yn DIVIDE by y n : y n + P (x)y n = Q(x) SET z = y n : = ( n)y( n ) SUBSTITUTE ( n) = y n ( n) + P (x)z = Q(x) + P (x)z = Q (x) linear in z where P (x) = ( n)p (x) Q (x) = ( n)q(x). Return to Exercise

15 Solutions to exercises 5 Exercise 2. This is of the form + P (x)y = Q(x)yn where DIVIDE by y n : SET z = y n = y : where P (x) = x Q(x) = x and n = 2 y 2 x y = x = y 2 = y 2 x z = x + x z = x

16 Solutions to exercises 6 Integrating factor, IF = e x = e ln x = x x + z = x2 d [x z] = x2 xz = x 2 Use z = y : xz = x3 3 + C x y = x3 3 + C y = x2 3 + C x. Return to Exercise 2

17 Solutions to exercises 7 Exercise 3. This is of the form where P (x) = x, + P (x)y = Q(x)yn Q(x) =, and n = 2 DIVIDE by y n : y 2 + x y = SET z = y n = y : = y 2 = y 2 + x z = x z =

18 Solutions to exercises 8 Integrating factor, IF = e x = e ln x = e ln x = x x x 2 z = x [ d x z] = x x z = x z x = ln x + C Use z = y : yx = C ln x y = x(c ln x). Return to Exercise 3

19 Solutions to exercises 9 Exercise 4. This of the form + P (x)y = Q(x)yn DIVIDE by y n : SET z = y n = y 3 : where P (x) = 3 Q(x) = e x and n = 4 y y 3 = e x = 3y 4 = 3 y z = ex z = 3ex

20 Solutions to exercises 20 Integrating factor, IF = e = e x x e e x z = 3e x e x d [e x z] = 3 e x z = 3 e x z = 3x + C Use z = y : e x 3 y = 3x + C 3 y 3 = ex (C 3x). Return to Exercise 4

21 Solutions to exercises 2 Exercise 5. Bernoulli equation: + y x = y3 with P (x) =, Q(x) =, n = 3 x DIVIDE by y n y 3 : y 3 + x y 2 = SET z = y n z = y 2 : = 2y 3 2 = y x z = 2 x z = 2

22 Solutions to exercises 22 Integrating factor, IF = e 2 x = e 2 ln x = e ln x 2 = x 2 x 2 2 x 3 z = 2 x 2 [ d x z ] = 2 2 x 2 x 2 z = ( 2) ( ) x + C z = 2x + Cx 2 Use z = y 2 : y 2 = 2x+Cx 2. Return to Exercise 5

23 Solutions to exercises 23 Exercise 6. This is of the form + P (x)y = Q(x)yn where where P (x) = 2 x Q(x) = x 2 cos x and n = 2 DIVIDE by y n : y x y = x 2 cos x SET z = y n = y : = y 2 = y x z = x2 cos x 2 x z = x2 cos x

24 Solutions to exercises 24 Integrating factor, IF =e 2 x =e 2 x =e 2 ln x =e ln x 2 = x 2 x 2 2 x 3 z = x2 x 2 cos x [ ] d x 2 z = cos x x 2 z = cos x x 2 z = sin x + C Use z = y : x 2 y = sin x + C y = x2 (sin x + C). Return to Exercise 6

25 Solutions to exercises 25 Exercise 7. Divide by 2 to get standard form: + (4x + 5)2 tan x y = 2 2 cos x y3 This is of the form + P (x)y = Q(x)yn where P (x) = 2 tan x Q(x) = (4x + 5)2 2 cos x and n = 3

26 Solutions to exercises 26 DIVIDE by y n : y tan x (4x + y 2 5)2 = 2 cos x SET z = y n = y 2 : = 2y 3 = 2 y tan x z = (4x + 5)2 2 cos x (4x + 5)2 tan x z = cos x

27 Solutions to exercises 27 ] Integrating factor, IF = e tan x = e sin x cos x [ f (x) e f(x) Use z = y 2 : = e ln cos x = cos x cos x cos x tan x z =cos x(4x+5)2 cos x cos x sin x z = (4x + 5)2 d [cos x z] = (4x + 5)2 cos x z = (4x + 5) 2 ( ) cos x z = 4 3 (4x + 5)3 + C cos x y 2 = 2 (4x + 5)3 + C y 2 = 2 cos x (4x + 5)3 + C cos x. Return to Exercise 7

28 Solutions to exercises 28 Exercise 8. Standard form: + ( ) x y = (x ln x)y 2 P (x) = x, Q(x) = x ln x, n = 2 DIVIDE by y 2 : y 2 + ( x) y = x ln x SET z = y : = y 2 = y 2 + ( x) z = x ln x x z = x ln x

29 Solutions to exercises 29 Integrating factor: IF = e x = e ln x = e ln x = x x x 2 z = ln x d [ x z] = ln x x z = ln x + C [ Use integration by parts: u dv = uv v du, with u = ln x, dv = ] x z = [ x ln x x x ] + C Use z = y : xy = x( ln x) + C. Return to Exercise 8

30 Solutions to exercises 30 Exercise 9. Standard form: (cot x) y = (cosec x) y3 DIVIDE by y 3 : y 3 (cot x) y 2 = cosec x SET z = y 2 : = 2y 3 = 2 y 3 2 cot x z = cosec x + 2 cot x z = 2 cosec x

31 Solutions to exercises 3 Integrating factor: IF = e 2 cos x sin x e 2 f (x) f(x) = e 2 ln(sin x) = sin 2 x. sin 2 x + 2 sin x cos x z = 2 sin x [ d sin 2 x z ] = 2 sin x z sin 2 x = ( 2) ( cos x) + C Use z = y 2 : sin 2 x y 2 = 2 cos x + C y 2 = sin2 x 2 cos x+c. Return to Exercise 9