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Diﬀerential Equations BERNOULLI EQUATIONS Graham S McDonald A Tutorial Module for learning how to solve Bernoulli diﬀerential equations q Table of contents q Begin Tutorial c 2004 g.s.mcdonald@salford.ac.uk Table of contents 1. 2. 3. 4. 5. 6. Theory Exercises Answers Integrating factor method Standard integrals Tips on using solutions Full worked solutions Section 1: Theory 3 1. Theory A Bernoulli diﬀerential equation can be written in the following standard form: dy + P (x)y = Q(x)y n , dx where n = 1 (the equation is thus nonlinear). To ﬁnd the solution, change the dependent variable from y to z, where z = y 1−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: 1 dy + P (x)y 1−n y n dx dz 1 i.e. + P (x)z (1 − n) dx where we have used Toc dz dx = Q(x) = Q(x) dy = (1 − n)y −n dx . Back Section 2: Exercises 4 2. Exercises Click on Exercise links for full worked solutions (there are 9 exercises in total) Exercise 1. The general form of a Bernoulli equation is dy + P (x)y = Q(x) y n , dx where P and Q are functions of x, and n is a constant. Show that the transformation to a new dependent variable z = y 1−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. dy 1 − y = xy 2 dx x q Theory q Answers q IF method q Integrals q Tips Toc Back Section 2: Exercises 5 Exercise 3. dy y + = y2 dx x Exercise 4. dy 1 + y = ex y 4 dx 3 Exercise 5. dy x + y = xy 3 dx Exercise 6. dy 2 + y = −x2 cos x · y 2 dx x q Theory q Answers q IF method q Integrals q Tips Toc Back Section 2: Exercises 6 Exercise 7. dy (4x + 5)2 3 + tan x · y = y 2 dx cos x Exercise 8. x dy + y = y 2 x2 ln x dx Exercise 9. dy = y cot x + y 3 cosecx dx q Theory q Answers q IF method q Integrals q Tips Toc Back Section 3: Answers 7 3. Answers 1. HINT: Firstly, divide each term by y n . Then, diﬀerentiate z dy dz 1 with respect to x to show that (1−n) dx = y1 dx , n 2. 3. 4. 1 y 1 y 1 y3 = −x + 3 2 C x , = x(C − ln x) , = ex (C − 3x) , 1 2x+Cx2 5. y 2 = 6. 7. 8. 1 y 1 y2 1 xy , = x2 (sin x + C) , = 1 12 cos x (4x + 5)3 + C cos x , = C + x(1 − ln x) , Toc Back Section 3: Answers 8 9. y 2 = sin x 2 cos x+C 2 . Toc Back 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 dz involve z either as the derivative dx 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: dz + P1 (x)z = Q1 (x) dx where P1 (x) and Q1 (x) are functions of x, and in some cases may be constants. Toc Back Section 4: Integrating factor method 10 A linear ﬁrst order o.d.e. can be solved using the integrating factor method. After writing the equation in standard form, P1 (x) can be identiﬁed. One then multiplies the equation by the following “integrating factor”: IF= e P1 (x)dx This factor is deﬁned so that the equation becomes equivalent to: d dx (IF z) = IF Q1 (x), whereby integrating both sides with respect to x, gives: IF z = IF Q1 (x) dx Finally, division by the integrating factor (IF) gives z explicitly in terms of x, i.e. gives the solution to the equation. Toc Back Section 5: Standard integrals 11 5. Standard integrals f (x) x n 1 x x f (x)dx xn+1 n+1 f (x) [g (x)] g (x) a sinh x cosh x tanh x cosech x sech x sech2 x coth x sinh2 x cosh2 x g (x) g(x) x n f (x)dx [g(x)]n+1 n+1 (n = −1) (n = −1) e sin x cos x tan x cosec x sec x sec2 x cot x sin2 x cos2 x ln |x| ex − cos x sin x − ln |cos x| ln tan x 2 ln |sec x + tan x| tan x ln |sin x| x sin 2x 2 − 4 x sin 2x 2 + 4 ln |g (x)| ax (a > 0) ln a cosh x sinh x ln cosh x ln tanh x 2 2 tan−1 ex tanh x ln |sinh x| sinh 2x −x 4 2 sinh 2x +x 4 2 Toc Back Section 5: Standard integrals 12 f (x) 1 a2 +x2 1 a f (x) dx tan−1 x a f (x) 1 a2 −x2 1 x2 −a2 f (x) dx 1 2a 1 2a ln ln a+x a−x x−a x+a (0 < |x| < a) (|x| > a > 0) (a > 0) √ 1 a2 −x2 sin−1 x a √ 1 a2 +x2 √ 1 x2 −a2 ln ln √ x+ a2 +x2 a √ x+ x2 −a2 a (a > 0) (x > a > 0) √ x a2 +x2 a2 √ 2 2 + x x2−a a (−a < x < a) √ a2 − x2 a2 2 sin−1 +x √ x a √ √ a2 +x2 a2 2 a2 2 sinh−1 − cosh−1 x a x a + a2 −x2 a2 x2 −a2 Toc Back Section 6: Tips on using solutions 13 6. Tips on using solutions q 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. q 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. q Try to make less use of the full solutions as you work your way through the Tutorial. Toc Back Solutions to exercises 14 Full worked solutions Exercise 1. dy + P (x)y = Q(x)y n dx 1 dy y n dx DIVIDE by y n : SET z = y 1−n : i.e. i.e. SUBSTITUTE i.e. where + P (x)y 1−n = Q(x) dz dx dy = (1 − n)y (1−n−1) dx 1 dz (1−n) dx 1 dz (1−n) dx dz dx = 1 dy y n dx + P (x)z = Q(x) linear in z + P1 (x)z = Q1 (x) P1 (x) = (1 − n)P (x) Q1 (x) = (1 − n)Q(x) . Return to Exercise 1 Back Toc Solutions to exercises 15 Exercise 2. This is of the form dy + P (x)y = Q(x)y n where dx 1 where P (x) = − x Q(x) = x and n = 2 1 dy 1 i.e. − y −1 = x DIVIDE by y n : y 2 dx x dz dy 1 dy SET z = y 1−n = y −1 : i.e. = −y −2 =− 2 dx dx y dx 1 dz ∴ − − z=x dx x 1 dz i.e. + z = −x dx x Toc Back Solutions to exercises 16 Integrating factor, ∴ i.e. i.e. i.e. 1 Use z = y : IF = e x 1 x dx = eln x = x dz + z = −x2 dx d [x · z] = −x2 dx x2 dx x3 +C 3 3 xz = − xz = − x y = −x + C 3 i.e. 1 x2 C =− + . y 3 x Return to Exercise 2 Toc Back Solutions to exercises 17 Exercise 3. This is of the form 1 where P (x) = x , dy + P (x)y = Q(x)y n dx Q(x) = 1, and n=2 i.e. i.e. ∴ i.e. 1 dy y 2 dx dz dx 1 + x y −1 = 1 DIVIDE by y n : SET z = y 1−n = y −1 : dy dy = −1 · y −2 dx = − y12 dx dz 1 − dx + x z = 1 dz dx 1 − x z = −1 Toc Back Solutions to exercises 18 dx x Integrating factor, IF = e− = e− ln x = eln x −1 = 1 x ∴ i.e. i.e. i.e. 1 Use z = y : 1 dz x dx d dx 1 x z x − 1 x 1 x2 z 1 = −x 1 · z = −x dx x ·z =− = − ln x + C = C − ln x 1 yx i.e. 1 y = x(C − ln x) . Return to Exercise 3 Toc Back Solutions to exercises 19 Exercise 4. This of the form dy + P (x)y = Q(x)y n dx where and DIVIDE by y n : SET z = y 1−n = y −3 : i.e. i.e. ∴ i.e. 1 3 Q(x) = ex n = 4 1 dy 1 + y −3 = ex y 4 dx 3 P (x) = dz dy 3 dy = −3y −4 =− 4 dx dx y dx − 1 dz 1 + z = ex 3 dx 3 dz − z = −3ex dx Back Toc Solutions to exercises 20 Integrating factor, ∴ i.e. i.e. i.e. Use z = 1 y3 : IF = e− e−x dx = e−x dz − e−x z = −3e−x · ex dx d −x [e · z] = −3 dx e−x · z = −3 dx e−x · z = −3x + C e−x · 1 y3 = −3x + C i.e. 1 = ex (C − 3x) . y3 Return to Exercise 4 Toc Back Solutions to exercises 21 Exercise 5. dy y 1 Bernoulli equation: + = y 3 with P (x) = , Q(x) = 1, n = 3 dx x x DIVIDE by y n i.e. y 3 : SET z = y 1−n i.e. z = y −2 : i.e. ∴ i.e. 1 dy y 3 dx 1 + x y −2 = 1 dz dx dy = −2y −3 dx 1 dy y 3 dx dz − 1 dx = 2 dz 1 − 1 dx + x z = 1 2 dz dx 2 − x z = −2 Toc Back Solutions to exercises 22 Integrating factor, ∴ i.e. i.e. i.e. Use z = 1 y2 : 1 dz x2 dx d dx 1 x2 z IF = e−2 − 2 x3 z dx x = e−2 ln x = eln x −2 = 1 x2 2 = − x2 1 x2 z 2 = − x2 1 = (−2) · (−1) x + C z = 2x + Cx2 y2 = 1 2x+Cx2 . Return to Exercise 5 Toc Back Solutions to exercises 23 Exercise 6. This is of the form dy + P (x)y = Q(x)y n where dx 2 where P (x) = x Q(x) = −x2 cos x and n = 2 1 dy 2 DIVIDE by y n : i.e. + y −1 = −x2 cos x 2 dx y x dz 1 dy dy SET z = y 1−n = y −1 : i.e. = −1 · y −2 =− 2 dx dx y dx 2 dz ∴ − + z = −x2 cos x dx x dz 2 i.e. − z = x2 cos x dx x Toc Back Solutions to exercises 24 Integrating factor, IF = e ∴ i.e. i.e. i.e. 2 −x dx = e−2 dx x = e−2 ln x = eln x −2 = 1 x2 1 dz 2 x2 − 3 z = 2 cos x 2 dx x x x d 1 · z = cos x dx x2 1 · z = cos x dx x2 1 · z = sin x + C x2 1 x2 y 1 Use z = y : = sin x + C i.e. 1 = x2 (sin x + C) . y Return to Exercise 6 Toc Back Solutions to exercises 25 Exercise 7. Divide by 2 to get standard form: dy 1 (4x + 5)2 3 + tan x · y = y dx 2 2 cos x dy This is of the form + P (x)y = Q(x)y n dx where P (x) = 1 tan x 2 (4x + 5)2 2 cos x 3 Q(x) and n = = Toc Back Solutions to exercises 26 DIVIDE by y n : i.e. 1 dy 1 (4x + 5)2 + tan x · y −2 = 3 dx y 2 2 cos x SET z = y 1−n = y −2 : i.e. dz dy 2 dy = −2y −3 =− 3 dx dx y dx ∴ − 1 dz 1 (4x + 5)2 + tan x · z = 2 dx 2 2 cos x i.e. (4x + 5)2 dz − tan x · z = dx cos x Toc Back Solutions to exercises 27 Integrating factor, IF = e − tan x·dx =e sin − cos x dx x ≡e f (x) f (x) dx = eln cos x = cos x ∴ i.e. i.e. i.e. i.e. Use z = 1 y2 : cos x dz (4x+5)2 −cos x tan x · z = cos x dx cos x dz cos x − sin x · z = (4x + 5)2 dx d [cos x · z] = (4x + 5)2 dx cos x · z = cos x · z = cos x y2 (4x + 5)2 dx 1 4 1 · (4x + 5)3 + C 3 = 1 12 (4x + 5)3 + C i.e. Toc 1 C 1 = (4x + 5)3 + . 2 y 12 cos x cos x Return to Exercise 7 Back Solutions to exercises 28 Exercise 8. Standard form: i.e. DIVIDE by y 2 : SET z = y −1 : ∴ i.e. dy dx + 1 x y = (x ln x)y 2 1 P (x) = x , Q(x) = x ln x , n = 2 1 dy y 2 dx 1 x + y −1 = x ln x dz dx dy dy = −y −2 dx = − y12 dx 1 x dz − dx + dz dx z = x ln x − 1 x · z = −x ln x Toc Back Solutions to exercises 29 Integrating factor: IF = e− ∴ i.e. i.e. dx x = e− ln x = eln x − 1 xz 1 x2 z −1 = 1 x 1 dz x dx d dx 1 xz = − ln x = − ln x = − ln x dx + C v du dx, dx =1] 1 x · x dx + C [ Use integration by parts: dv u dx dx = uv − with u = ln x , i.e. 1 Use z = y : 1 xz 1 xy dv dx = − x ln x − = x(1 − ln x) + C . Return to Exercise 8 Toc Back Solutions to exercises 30 Exercise 9. Standard form: DIVIDE by y 3 : SET z = y −2 : ∴ i.e. dy dx − (cot x) · y = (cosec x) y 3 − (cot x) · y −2 = cosec x 1 dy y 3 dx 1 dy y 3 dx dz dx dy = −2y −3 dx = −2 · dz − 1 dx − cot x · z = cosec x 2 dz dx + 2 cot x · z = −2 cosec x Toc Back Solutions to exercises 31 cos x sin x dx Integrating factor: IF = e2 ∴ i.e. i.e. Use z = 1 y2 : ≡ e2 f (x) dx f (x) = e2 ln(sin x) = sin2 x. sin2 x · d dx dz dx + 2 sin x · cos x · z = −2 sin x sin2 x · z = −2 sin x z sin2 x = (−2) · (− cos x) + C sin2 x y2 = 2 cos x + C sin2 x 2 cos x+C i.e. y2 = . Return to Exercise 9 Toc Back