# 1. Theory

A Bernoulli differential equation can be written in the following standard form:

 dy/dx + P(x) y = Q(x) yn

- where n ≠ 1. The equation is thus non-linear.

Dividing the above standard form by yn gives:

 1/yn dy/dx + P(x) y1−n = Q(x) i.e. 1/(1 − n) dz/dx + P(x) y1−n = Q(x)

- where we have used: dz/dx = (1 − n) yndy/dx

# 2. Exercises

Click on questions to reveal their solutions

Exercise 1:

The general form of a Bernoulli equation is:  dy/dx + 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: 1/yn dy/dx + P(x) y1−n = Q(x) SET z = y1−n: i.e. dz/dx = (1 − n) y1−n−1 dy/dx i.e. 1/1 − n dz/dx = 1/yn dy/dx SUBSTITUTE 1/1 − n dz/dx + 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:

dy/dx1/xy = xy2

Solution:

The equation is of the form: dy/dx + P(x) y = Q(x) yn   with P(x) = − 1/x , Q(x) = x and n = 2

 DIVIDE by yn: i.e. 1/y2 dy/dx − 1/x  y−1 = x SET z = y1 −n = y−1: i.e. dz/dx = −y−2 dy/dx = −1/y2 dy/dx ∴ −dz/dx − 1/xz = x i.e. dz/dx + 1/xz = −x INTEGRATING FACTOR IF = e∫1/xdx = eln x = x ∴ xdz/dx + z = −x2 i.e. d/dx [x⋅z] = −x2 i.e. xz = −∫x2dx i.e. xz = − x3/3 + C USE z = 1/y x/y = − x3/3 + C i.e. 1/y = − x2/3 + C/x

Exercise 3:

dy/dx + y/x = y2

Solution:

The equation is of the form: dy/dx + P(x) y = Q(x) yn   with P(x) = 1/x, Q(x) = 1 and n = 2

 DIVIDE by yn: i.e. 1/y2 dy/dx + 1/x  y−1 = 1 SET z = y1 −n = y−1: i.e. dz/dx = −1⋅y−2 dy/dx = −1/y2 dy/dx ∴ −dz/dx + 1/xz = 1 i.e. dz/dx − 1/xz = −1 INTEGRATING FACTOR IF = e−∫1/xdx = e− ln x = 1/x ∴ 1/x dz/dx −1/x2z = −1/x i.e. d/dx [1/x⋅z] = −1/x i.e. 1/x⋅z = −∫ dx/x i.e. z/x = − ln x + C USE z = 1/y 1/yx = C − ln x i.e. 1/y = − x(C − ln x)

Exercise 4:

dy/dx + 1/3y = exy4

Solution:

The equation is of the form: dy/dx + P(x) y = Q(x) yn   with P(x) = 1/3, Q(x) = ex and n = 4

 DIVIDE by yn: i.e. 1/y4 dy/dx + 1/3  y−3 = ex SET z = y1 −n = y−3: i.e. dz/dx = −3y−4 dy/dx = −3/y4 dy/dx ∴ −1/3 dz/dx + 1/3z = ex i.e. dz/dx − z = −3 ex INTEGRATING FACTOR IF = e−∫dx = e−x ∴ e−xdz/dx − e−xz = −3 e−x⋅ex i.e. d/dx [ e−x⋅z] = −3 i.e. e−x⋅z = ∫ −3 dx i.e. e−x⋅z = −3x + C USE z = 1/y3 e−x⋅1/y3 = −3x + C i.e. 1/y3 = ex (C − 3x)

Exercise 5:

xdy/dx + y = xy3

Solution:

Bernoulli equation: dy/dx + y/x = y3   with P(x) = 1/x , Q(x) = 1 and n = 3

 DIVIDE by yn, i.e. y3: 1/y3 dy/dx + 1/x  y−2 = 1 SET z = y1 −n, i.e. y−2: dz/dx = −2y−3 dy/dx i.e. −1/2 dz/dx = 1/y3 dy/dx ∴ −1/2 dz/dx + 1/xz = 1 i.e. dz/dx − 2/xz = −2 INTEGRATING FACTOR IF = e−2 ∫dx/x = e−2 ln x = eln x−2 = 1/x2 ∴ 1/x2 dz/dx − 2/x3z = − 2/x2 i.e. d/dx [ 1/x2 ⋅z] = −2/x2 i.e. 1/x2z = (−2)⋅(−1)1/x + C i.e. z = 2x + Cx2 USE z = 1/y2: y2 = 1/2x + Cx2

Exercise 6:

dy/dx + 2/xy = −x2 cos xy2

Solution:

The equation is of the form: dy/dx + P(x) y = Q(x) yn   with P(x) = 2/x, Q(x) = −x2 cos x and n = 2

 DIVIDE by yn: i.e. 1/y2 dy/dx + 2/x  y−1 = −x2 cos x SET z = y1 −n = y−1: i.e. dz/dx = −1⋅y−2 dy/dx = −1/y2 dy/dx ∴ −dz/dx + 2/xz = −x2 cos x i.e. dz/dx − 2/xz = x2 cos x INTEGRATING FACTOR IF = e−∫2/xdx = e−2 ∫dx/x = e−2 ln x = eln x−2 = 1/x2 ∴ 1/x2 dz/dx −2/x3z = x2/x2 cos x i.e. d/dx [1/x2⋅z] = cos x i.e. 1/x2⋅z = ∫ cos xdx i.e. 1/x2 ⋅z = sin x + C USE z = 1/y 1/x2y = sin x + C i.e. 1/y = x2(sin x + C)

Exercise 7:

2dy/dx + tan xy = (4x + 5)2/cos x y3

Solution:

Divide by 2 to get standard form:

dy/dx + 1/2 tan xy = (4x + 5)2/2 cos x y3

This of the form: dy/dx + P(x) y = Q(x) yn   with P(x) = 1/2 tan x, Q(x) = (4x + 5)2/2 cos x and n = 3

 DIVIDE by yn: i.e. 1/y3 dy/dx + 1/2 tan x⋅y−2 = (4x + 5)2/2 cos x SET z = y1 −n = y−2: i.e. dz/dx = −2y−3 dy/dx = −2/y3 dy/dx ∴ −1/2 dz/dx + 1/2 tan x⋅z = (4x + 5)2/2 cos x i.e. dz/dx − tan x⋅z = (4x + 5)2/ cos x INTEGRATING FACTOR IF = e∫ −tan x⋅dx = e∫ − sin x/cos x dx [ ≡ e∫ − ƒ′(x)/ƒ(x) dx] = eln cos x = cos x ∴ cos x dz/dx − cos x tan x⋅z = cos x (4x + 5)2/ cos x i.e. cos xdz/dx − sin x⋅z = (4x + 5)2 i.e. d/dx [ cos x⋅z ] = (4x + 5)2 i.e. cos x⋅z = ∫ (4x + 5)2 dx i.e. cos x⋅z = 1/4 ⋅ 1/3 (4x + 5)3 + C USE z = 1/y2 cos x/y2 = 1/12(4x + 5)3 + C i.e. 1/y2 = 1/12 cos x (4x + 5)3 + C/cos x

Exercise 8:

xdy/ dx + y = y2x2 ln x

Solution:

Express in standard form:    dy/dx + 1/xy = (x ln x) y2 ,   so P(x) = 1/x, Q(x) = x ln x and n = 2

 DIVIDE by y2: i.e. 1/y2 dy/dx + 1/x y−1 = x ln x SET z = y−1: i.e. dz/dx = −y−2 dy/dx = −1/y2 dy/dx ∴ −dz/dx + 1/x z = x ln x i.e. dz/dx − 1/x⋅z = −x ln x INTEGRATING FACTOR IF = e− ∫dx/x = e− ln x = eln x−1 = 1/x ∴ 1/x dz/dx − 1/x2z = − ln x i.e. d/dx [ 1/xz ] = − ln x i.e. 1/xz = ∫ ln x dx + C' Integrate by parts: ∫ u dv/dx = uv − ∫ v du/dx,   with u = ln x and dv/dx = 1 i.e. 1/xz = − [x ln x − ∫ x 1/x dx] + C USE z = 1/y 1/xy = x(1 − ln x) + C

Exercise 9:

dy/dx = y cot x + y3 cosec x

Solution:

Express in standard form:    dy/dx − cot xy = cosec xy

 DIVIDE by y3: 1/y3 dy/dx − cot x⋅y−2 = cosec x SET z = y−2: i.e. dz/dx = − 2y−3 dy/dx = − 21/y3 dy/dx ∴ − 1/2 dz/dx − cot x⋅z = cosec x i.e. dz/dx + 2 cot x⋅z = cosec x INTEGRATING FACTOR IF = e2∫  cos x/sin x dx ≡ e2∫ − ƒ′(x)/ƒ(x) dx = e2 ln sin x = sin2 x ∴ sin2 x⋅dz/dx + 2 sin x⋅cos x⋅z = −2 sin x i.e. d/dx [ sin2x⋅z ] = −2 sin x i.e. z⋅sin2x = (−2)(−cos x) + C USE z = 1/y2 sin2x/y2 = 2 cos x + C i.e. y2 = sin2x/2 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 dz/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/dx + 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':

IF = eP1(x) dx

This factor is defined so that the equation becomes equivalent to:

d/dx (IF z) = IF Q1(x) dx

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.

# 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) 1/x ln x g'(x)/g(x) ln g(x) ex ex ax ax/ln 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  x/2 cosech x ln tanh  x/2 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 x/2 − sin 2x/4 sinh2 x sinh 2x/4 − x/2 cos2 x x/2 + sin 2x/4 cosh2 x sinh 2x/4 + x/2 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)