1. Theory

We consider here the following standard form of ordinary differential equation (o.d.e.):

P(x, y) dx + Q(x, y) dy = 0

If ∂P/∂y = ∂Q/∂x then the o.d.e is said to be exact.

This means that a function u(x, y) exists such that:

du = ∂u/∂xdx + ∂u/∂ydy
  = P dx + Q dy

One solves ∂u/∂x = P and ∂u/∂y = Q to find u(x, y).

Then du = 0 gives u(x, y) = C, where C is a constant.

This last equation gives the general solution of P dx + Q dy = 0.

2. Exercises

Show that each of the following differential equations is exact and use that property to find the general solution:

Click on questions to reveal their solutions

Exercise 1:

1/xdy − y/x2dx = 0

Solution:

Standard Form: P(x, y) dx + Q(x, y) dy = 0

i.e.  P(x, y) = −y/x2   and     Q(x, y) = −1/x

The equation is exact if ∂P/∂y = ∂Q/∂x

Check this:   ∂P/∂y = −1/x2 = ∂Q/∂x    ∴ o.d.e. is exact.

Since the equation is exact, u(x, y) exists such that:

du = ∂u/∂xdx + ∂u/∂ydy
  = P dx + Q dy

The equation has solution u = C   - where C is a constant.

∂u/∂x = P gives (i) ∂u/∂x = y/x2
∂u/∂y = Q gives (ii) ∂u/∂y = 1/x

Integrate (i) partially with respect to x: u = y/x + φ(y)    - where φ(y) is an arbitrary function of y

Differentiate with respect to y:

∂u/∂y = 1/x + ∂φ/∂y = 1/x + dφ/dy    - since φ = φ(y) only

Compare with equation (ii):

  1/x + dφ/dy = 1/x
i.e. dφ/dy = 0
so: φ = C'   - where C' is a constant
and u = y/x + C'

du = 0 implies u = C, where C is a constant.

∴ y/x = A    - where A = C − C' is another constant.

Exercise 2:

2xydy/dx + y2 − 2x = 0

Solution:

Standard Form: (y2 − 2x) dx + 2xy dy = 0

Exact if ∂P/∂y = ∂Q/∂x, where P(x, y) = y2 − 2x and Q(x, y) = 2xy

∂P/∂y = 2y = ∂Q/∂x   so the o.d.e. is exact.

∴ u(x, y) exists such that:

du = ∂u/∂xdx + ∂u/∂ydy
  = P dx + Q dy = 0

giving: (i) ∂u/∂x = y2 − 2x  and   (ii) ∂u/∂y = 2xy

Integrate (i): u = xy2 − x2 + φ(y)    - where φ(y) is an arbitrary function.

Differentiate and compare with (ii):

∂u/∂y = 2xy + dφ/dy = 2xy     ∴ dφ/dy = 0     ⇒ φ = C'
∴ u = xy2 − x2 + C'    and    du = 0 implies u = C
∴ xy2 − x2 = A, where A = C − C'

Exercise 3:

2(y + 1) ex dx + 2(ex − 2y) dy = 0

Solution:

P(x, y) dx + Q(x, y) dy = 0    - where P(x, y) = 2(y + 1) ex   and   Q(x, y) = 2(ex − 2y)

∂P/∂y = 2 ex = ∂Q/∂x    ∴ the o.d.e. is exact.

∴ u(x, y) exists such that:

du = ∂u/∂xdx + ∂u/∂ydy
  = P dx + Q dy = 0

giving: (i) ∂u/∂x = 2(y+ 1)  and   (ii) ∂u/∂y = 2(ex − 2y)

Integrate (i): u = 2(y + 1)ex + φ(y)

Differentiate and compare with (ii):

∂u/∂y = 2ex + dφ/dy = 2(ex − 2y)     ∴ dφ/dy = −4y     i.e.  ∫ φ = −4 ∫ y dy    ⇒   φ = −2y2 + C'
∴ u = 2(y + 1) ex + C'    and    du = 0 implies u = C
∴ (y + 1) ex − y2 = A = (C − C') ⁄ 2

Exercise 4:

(2xy + 6x) dx + (x2 + 4y3) dy = 0

Solution:

P(x, y) dx + Q(x, y) dy = 0    - where P(x, y) = 2xy + 6x   and    Q(x, y) = 2x2 + 4y3

∂P/∂y = 2x = ∂Q/∂x    ∴ the o.d.e. is exact.

∴ u(x, y) exists such that:

du = ∂u/∂xdx + ∂u/∂ydy
  = P dx + Q dy = 0

giving: (i) ∂u/∂x = 2xy + 6x  and   (ii) ∂u/∂y = x2 + 4y3

Integrate (i): u = x2y + 3x2 + φ(y)

Differentiate and compare with (ii):

∂u/∂y = x2 + dφ/dy = x2 + 4y3     ∴ dφ/dy = −4y3     i.e.  ∫ φ = 4 ∫ y3 dy    ⇒   φ = y4 + C'
∴ u = x2y + 3x3 + y4 + C'    and    du = 0 implies u = C
∴ x2y + 3x3 + y4 = A = C − C'

Exercise 5:

(8y − x2y) dy / dx + x − xy2

Solution:

(x − xy2) dx + (8y − x2y) dy = 0    ⇒   P(x, y) = x − xy2   and   Q(x, y) = 8y − x2y

∂P/∂y = −2xy = ∂Q/∂x    ∴ the o.d.e. is exact.

∴ u(x, y) exists such that:

du = ∂u/∂xdx + ∂u/∂ydy
  = P dx + Q dy = 0

Giving: (i) ∂u/∂x = x − xy2   and   (ii) ∂u/∂y = 8y − x2y

Integrate (i): u = ½x2(1 − y2 + φ(y)

Differentiate and compare with (ii):

∂u/∂y = −½x2⋅2y + dφ/dy = 8y − x2y     ∴ dφ/dy = 8y     i.e.  ∫ φ = 8 ∫ y dy    ⇒   φ = 4y2 + C'
∴ u = ½x2(1 − y2) + 4y2 + C'    and    du = 0 implies u = C
∴ ½x2(1 − y2) + 4y2 = A = C − C'

Exercise 6:

(e4x + 2xy2) dx + (cos y + 2x2y) dy = 0

Solution:

P(x, y) dx + Q(x, y) dy = 0    ⇒  P(x, y) = e4x + 2xy2   and  Q(x, y) = cos y + 2x2y

∂P/∂y = 4xy = ∂Q/∂x    ∴ the o.d.e. is exact.

∴ u(x, y) exists such that:

du = ∂u/∂xdx + ∂u/∂ydy
  = P dx + Q dy = 0

Giving: (i) ∂u/∂x = e4x + 2xy2   and   (ii) ∂u/∂y = cos y + 2x2y

Integrate (i): u = 1/4 e4x + x2y2 + φ(y)

Differentiate and compare with (ii):

∂u/∂y = 2x2y + dφ/dy = cos y + 2x2y     ∴ dφ/dy = cos y     i.e.  ∫ φ =  ∫ cos y dy    ⇒   φ = sin y + C'
∴ u = 1/4 e4x + x2y2 + sin y + C'    and    du = 0 implies u = C
∴ 1/4 e4x + x2y2 + sin y = A = C − C'

Exercise 7:

(3x2 + y cos x)dx + (sin x − 4y3) dy = 0

Solution:

P(x, y) dx + Q(x, y) dy = 0    ⇒   P(x, y) = 3x2 + y cos x   and   Q(x, y) = sin x − 4y3

∂P/∂y = cos x = ∂Q/∂x    ∴ the o.d.e. is exact.

∴ u(x, y) exists such that:

du = ∂u/∂xdx + ∂u/∂ydy = 0
  = P dx + Q dy = 0

Giving: (i) ∂u/∂x = 3x2 + y cos x   and   (ii) ∂u/∂y = sin x − 4y3

Integrate (i): u = x3 + y sin x + φ(y)

Differentiate and compare with (ii):

∂u/∂y = sin x + dφ/dy = sin x − 4y3     ∴ dφ/dy = −4y3    i.e.  ∫ φ = −4 ∫ y3 dy    ⇒   φ = −y4 + C'
∴ u = x3 + y sin x − y4 + C'    and    du = 0 implies u = C
∴ x3 + y sin x −y4 = A = C − C'

Exercise 8:

x tan−1 y⋅dx + x2 / 2(1 + y2)⋅dy = 0

Solution:

P(x, y) dx + Q(x, y) dy = 0    ⇒   P(x, y) = x tan−1 y   and   Q(x, y) = x2/2(1 + y2)

∂P/∂y = x/(1 + y2) = ∂Q/∂x    ∴ the o.d.e. is exact.

∴ u(x, y) exists such that:

du = ∂u/∂xdx + ∂u/∂ydy = 0
  = P dx + Q dy = 0

Giving: (i) ∂u/∂x = x tan−1 y   and   (ii) ∂u/∂y = x2/2(1 + y2)

Integrate (i): u = x2/2  tan−1 y + φ(y)

Differentiate and compare with (ii):

∂u/∂y = x2/2 1/(1 + y2) + dφ/dy = x2/(1 + y2)     ∴ dφ/dy = 0    i.e.  φ = C'
∴ u = x2/2  tan−1y + C'    and    du = 0 implies u = C
∴ x2/2  tan−1y = A = C − C'

Exercise 9:

(2x + x2y3) dx + (x3y2 + 4y3) dy = 0

Solution:

P(x, y) dx + Q(x, y) dy = 0    ⇒   P(x, y) = 2x + x2y3   and   Q(x, y) = x3y2 + 4y3

∂P/∂y = 3x2y2 = ∂Q/∂x    ∴ the o.d.e. is exact.

∴ u(x, y) exists such that:

du = ∂u/∂xdx + ∂u/∂ydy = 0
  = P dx + Q dy = 0

Giving: (i) ∂u/∂x = 2x + x2y3   and   (ii) ∂u/∂y = x3y2 + 4y3

Integrate (i): u = x2 + x3y3/3 + φ(y)

Differentiate and compare with (ii):

∂u/∂y = x3y2 + dφ/dy = x3y2 + 4y3     ∴ dφ/dy = 4y3    i.e.  ∫ φ = 4 ∫ y3 dy    ⇒   φ = y4 + C'
∴ u = x2/2  tan−1y + C'    and    du = 0 implies u = C
∴ x2 + x3y3/3 + y4 = A = C − C'

Exercise 10:

(2x3 − 3x2y + y3) dy/ dx = 2x3 − 6x2y + 3xy2

Solution:

(2x3 − 6x2y + 3xy2) dx + (−2x3 + 3x2y − y3) dy = 0
⇒   ∂P/∂y = −6x2 + 6xy = ∂Q/∂x    ∴ the o.d.e. is exact.

∴ u(x, y) exists such that:

du = ∂u/∂xdx + ∂u/∂ydy = 0
  = P dx + Q dy = 0

Giving: (i) ∂u/∂x = 2x3 − 6x2y + 3xy3   and   (ii) ∂u/∂y = −2x3 + 3x2y − y3

Integrate (i):  u = x4/4 − 2x3y + 3/2 x2y2 + φ(y) 

Differentiate and compare with (ii):

∂u/∂y = −2x3 + 3x2y + + dφ/dy = −2x3 3x2y − y3     ∴ dφ/dy = −y3    i.e.  ∫ φ = − ∫ y3 dy    ⇒   φ = − 1/4 y4 + C'
∴ u = x4/2 − 2x3y + 3/2 x2y2 − y4/4 + C'    and    du = 0 implies u = C
∴ x4/2 − 2x3y + 3/2 x2y2 − y4/4 = A = C − C'

Exercise 11:

(y2 cos x − sin x) dx + (2y sin x + 2) dy = 0

Solution:

P(x, y) dx + Q(x, y) dy = 0    ⇒   P(x, y) = y2 − sin x   and   Q(x, y) = 2y sin x + 2

∂P/∂y = 2y cos x = ∂Q/∂x    ∴ the o.d.e. is exact.

∴ u(x, y) exists such that:

du = ∂u/∂xdx + ∂u/∂ydy = 0
  = P dx + Q dy = 0

Giving: (i) ∂u/∂x = y2 cos x − sin x   and   (ii) ∂u/∂y = 2y sin x + 2

Integrate (i): u = y2 sin x + cos x + φ(y)

Differentiate and compare with (ii):

∂u/∂y = 2y sin x + dφ/dy = 2y sin x + 2     ∴ dφ/dy = 2    i.e.  ∫ φ = 2 ∫ dy    ⇒   φ = 2y + C'
∴ u = y2 sin x + cos x + 2y + C'    and    du = 0 implies u = C
∴ y2 sin x + cos x + 2y = A = C − C'

3. 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)

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