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Bernoulli Equations |
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1. Theory
A Bernoulli differential equation can be written in the following standard form:
| dydx + P(x) y = Q(x) yn |
- where n ≠ 1. The equation is thus non-linear.
To find the solution, change the dependent variable from y to z, where z = y1−n. This gives a differential equation in x and z that is linear, and can therefore be solved using the integrating factor method.
Dividing the above standard form by yn gives:
| 1yn dydx + P(x) y1−n | = | Q(x) | |
| i.e. | 1(1 − n) dzdx + P(x) y1−n | = | Q(x) |
- where we have used: dzdx = (1 − n) y−n dydx
2. Exercises
Click on questions to reveal their solutions
Exercise 1:
The general form of a Bernoulli equation is: dydx + 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 = y1−n reduces the equation to one that is linear in z (and hence solvable using the integrating factor method).
Solution:
| DIVIDE by yn: | 1yn dydx + P(x) y1−n = Q(x) | ||
| SET z = y1−n: | i.e. | dzdx = (1 − n) y1−n−1 dydx | |
| i.e. | 11 − n dzdx = 1yn dydx | ||
| SUBSTITUTE | 11 − n dzdx + P(x) z = Q(x) | ||
| i.e. | dz dx + P1(x) z = Q1(x) | linear in x | |
| where: | P1(x) = (1 − n)P(x) | ||
| Q1(x) = (1 − n)Q(x) |
Solve the following Bernoulli differential equations:
Exercise 2:
Solution:
The equation is of the form: dydx + P(x) y = Q(x) yn with P(x) = − 1x , Q(x) = x and n = 2
| DIVIDE by yn: | i.e. | 1y2 dydx − 1x y−1 = x |
| SET z = y1 −n = y−1: | i.e. | dzdx = −y−2 dydx = −1y2 dydx |
| ∴ | −dzdx − 1xz = x | |
| i.e. | dzdx + 1xz = −x | |
| INTEGRATING FACTOR | IF = | e∫1xdx = eln x = x |
| ∴ | xdzdx + z = −x2 | |
| i.e. | ddx [x⋅z] = −x2 | |
| i.e. | xz = −∫x2dx | |
| i.e. | xz = − x33 + C | |
| USE z = 1y | xy = − x33 + C | |
| i.e. | 1y = − x23 + Cx |
Exercise 3:
Solution:
The equation is of the form: dydx + P(x) y = Q(x) yn with P(x) = 1x, Q(x) = 1 and n = 2
| DIVIDE by yn: | i.e. | 1y2 dydx + 1x y−1 = 1 |
| SET z = y1 −n = y−1: | i.e. | dzdx = −1⋅y−2 dydx = −1y2 dydx |
| ∴ | −dzdx + 1xz = 1 | |
| i.e. | dzdx − 1xz = −1 | |
| INTEGRATING FACTOR | IF = | e−∫1xdx = e− ln x = 1x |
| ∴ | 1x dzdx −1x2z = −1x | |
| i.e. | ddx [1x⋅z] = −1x | |
| i.e. | 1x⋅z = −∫ dxx | |
| i.e. | zx = − ln x + C | |
| USE z = 1y | 1yx = C − ln x | |
| i.e. | 1y = − x(C − ln x) |
Exercise 4:
Solution:
The equation is of the form: dydx + P(x) y = Q(x) yn with P(x) = 13, Q(x) = ex and n = 4
| DIVIDE by yn: | i.e. | 1y4 dydx + 13 y−3 = ex |
| SET z = y1 −n = y−3: | i.e. | dzdx = −3y−4 dydx = −3y4 dydx |
| ∴ | −13 dzdx + 13z = ex | |
| i.e. | dzdx − z = −3 ex | |
| INTEGRATING FACTOR | IF = | e−∫dx = e−x |
| ∴ | e−xdzdx − e−xz = −3 e−x⋅ex | |
| i.e. | ddx [ e−x⋅z] = −3 | |
| i.e. | e−x⋅z = ∫ −3 dx | |
| i.e. | e−x⋅z = −3x + C | |
| USE z = 1y3 | e−x⋅1y3 = −3x + C | |
| i.e. | 1y3 = ex (C − 3x) |
Exercise 5:
Solution:
Bernoulli equation: dydx + yx = y3 with P(x) = 1x , Q(x) = 1 and n = 3
| DIVIDE by yn, i.e. y3: | 1y3 dydx + 1x y−2 = 1 | |
| SET z = y1 −n, i.e. y−2: | dzdx = −2y−3 dydx | |
| i.e. | −12 dzdx = 1y3 dydx | |
| ∴ | −12 dzdx + 1xz = 1 | |
| i.e. | dzdx − 2xz = −2 | |
| INTEGRATING FACTOR | IF = | e−2 ∫dxx = e−2 ln x = eln x−2 = 1x2 |
| ∴ | 1x2 dzdx − 2x3z = − 2x2 | |
| i.e. | ddx [ 1x2 ⋅z] = −2x2 | |
| i.e. | 1x2z = (−2)⋅(−1)1x + C | |
| i.e. | z = 2x + Cx2 | |
| USE z = 1y2: | y2 = 12x + Cx2 |
Exercise 6:
Solution:
The equation is of the form: dydx + P(x) y = Q(x) yn with P(x) = 2x, Q(x) = −x2 cos x and n = 2
| DIVIDE by yn: | i.e. | 1y2 dydx + 2x y−1 = −x2 cos x |
| SET z = y1 −n = y−1: | i.e. | dzdx = −1⋅y−2 dydx = −1y2 dydx |
| ∴ | −dzdx + 2xz = −x2 cos x | |
| i.e. | dzdx − 2xz = x2 cos x | |
| INTEGRATING FACTOR | IF = | e−∫2xdx = e−2 ∫dxx = e−2 ln x = eln x−2 = 1x2 |
| ∴ | 1x2 dzdx −2x3z = x2x2 cos x | |
| i.e. | ddx [1x2⋅z] = cos x | |
| i.e. | 1x2⋅z = ∫ cos xdx | |
| i.e. | 1x2 ⋅z = sin x + C | |
| USE z = 1y | 1x2y = sin x + C | |
| i.e. | 1y = x2(sin x + C) |
Exercise 7:
Solution:
Divide by 2 to get standard form:
This of the form: dydx + P(x) y = Q(x) yn with P(x) = 12 tan x, Q(x) = (4x + 5)22 cos x and n = 3
| DIVIDE by yn: | i.e. | 1y3 dydx + 12 tan x⋅y−2 = (4x + 5)22 cos x |
| SET z = y1 −n = y−2: | i.e. | dzdx = −2y−3 dydx = −2y3 dydx |
| ∴ | −12 dzdx + 12 tan x⋅z = (4x + 5)22 cos x | |
| i.e. | dzdx − tan x⋅z = (4x + 5)2 cos x | |
| INTEGRATING FACTOR | IF = | e∫ −tan x⋅dx = e∫ − sin xcos x dx [ ≡ e∫ − ƒ′(x)ƒ(x) dx] = eln cos x = cos x |
| ∴ | cos x dzdx − cos x tan x⋅z = cos x (4x + 5)2 cos x | |
| i.e. | cos xdzdx − sin x⋅z = (4x + 5)2 | |
| i.e. | ddx [ cos x⋅z ] = (4x + 5)2 | |
| i.e. | cos x⋅z = ∫ (4x + 5)2 dx | |
| i.e. | cos x⋅z = 14 ⋅ 13 (4x + 5)3 + C | |
| USE z = 1y2 | cos xy2 = 112(4x + 5)3 + C | |
| i.e. | 1y2 = 112 cos x (4x + 5)3 + Ccos x |
Exercise 8:
Solution:
Express in standard form: dydx + 1xy = (x ln x) y2 , so P(x) = 1x, Q(x) = x ln x and n = 2
| DIVIDE by y2: | i.e. | 1y2 dydx + 1x y−1 = x ln x |
| SET z = y−1: | i.e. | dzdx = −y−2 dydx = −1y2 dydx |
| ∴ | −dzdx + 1x z = x ln x | |
| i.e. | dzdx − 1x⋅z = −x ln x | |
| INTEGRATING FACTOR | IF = | e− ∫dxx = e− ln x = eln x−1 = 1x |
| ∴ | 1x dzdx − 1x2z = − ln x | |
| i.e. | ddx [ 1xz ] = − ln x | |
| i.e. | 1xz = ∫ ln x dx + C' | |
| Integrate by parts: | ∫ u dvdx = uv − ∫ v dudx, with u = ln x and dvdx = 1 | |
| i.e. | 1xz = − [x ln x − ∫ x 1x dx] + C | |
| USE z = 1y | 1xy = x(1 − ln x) + C |
Exercise 9:
Solution:
Express in standard form: dydx − cot x⋅y = cosec x⋅y
| DIVIDE by y3: | 1y3 dydx − cot x⋅y−2 = cosec x | |
| SET z = y−2: | i.e. | dzdx = − 2y−3 dydx = − 21y3 dydx |
| ∴ | − 12 dzdx − cot x⋅z = cosec x | |
| i.e. | dzdx + 2 cot x⋅z = cosec x | |
| INTEGRATING FACTOR | IF = | e2∫ cos xsin x dx ≡ e2∫ − ƒ′(x)ƒ(x) dx = e2 ln sin x = sin2 x |
| ∴ | sin2 x⋅dzdx + 2 sin x⋅cos x⋅z = −2 sin x | |
| i.e. | ddx [ sin2x⋅z ] = −2 sin x | |
| i.e. | z⋅sin2x = (−2)(−cos x) + C | |
| USE z = 1y2 | sin2xy2 = 2 cos x + C | |
| i.e. | y2 = sin2x2 cos x + C |
3. 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 dzdx 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:
| dzdx + P1(x) z = Q1(x) |
- where P1(x) and Q1(x) are functions of x, and in some cases may be constants.
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':
This factor is defined so that the equation becomes equivalent to:
Integrating both sides with respect to x gives:
Finally, division by the integrating factor (IF) gives z explicitly in terms of x, i.e. gives the solution to the equation.
4. Standard Integrals
| ƒ(x) | ∫ƒ(x) dx | ƒ(x) | ∫ƒ(x) dx | |
|---|---|---|---|---|
| xn | xn+1 n+1 (n ≠ −1) | [g(x)]n g'(x) | [g(x)]n+1 n+1 (n ≠ −1) | |
| 1x | ln x | g'(x)g(x) | ln g(x) | |
| ex | ex | ax | axln a (a > 0) | |
| sin x | −cos x | sinh x | cosh x | |
| cos x | sin x | cosh x | sinh x | |
| tan x | − ln cosx | tanh x | ln cosh x | |
| cosec x | ln tan x2 | cosech x | ln tanh x2 | |
| sec x | ln sec x + tan x | sec x | 2 tan−1ex | |
| sec2 x | tan x | sec2 x | tanh x | |
| cot x | ln sin x | cot x | ln sinh x | |
| sin2 x | x2 − sin 2x4 | sinh2 x | sinh 2x4 − x2 | |
| cos2 x | x2 + sin 2x4 | cosh2 x | sinh 2x4 + x2 | |
| 1 a2 + x2 | 1 a tan −1 x a (a > 0) | √a2 + x2 | a2 2 [ sinh−1( x a ) + x √a2−x2 a2 ] | |
| 1 a2 − x2 | 1 2a ln a + x a − x (0 < x < a) | √a2−x2 | a2 2 [ sin−1( x a ) + x √a2−x2 a2 ] | |
| 1 x2 − a2 | 1 2a ln x − a x + a (x > a > 0) | √x2 − a2 | a2 2 [−cosh−1( x a ) + x √x2−a2 a2 ] | |
| 1 √ a2 + x2 | ln x + √a2 + x2 a (a > 0) | |||
| 1 √ a2 − x2 | sin−1 x a (−a < x < a) | 1 √ x2 − a2 | ln x + √x2 − a2 a (x > a > 0) |