Exact Equations |
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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.
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:
Solution:
Standard Form: P(x, y) dx + Q(x, y) dy = 0
The equation is exact if ∂P∂y = ∂Q∂x
Check this: ∂P∂y = −1x2 = ∂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 = yx2 |
∂u∂y = Q | gives | (ii) | ∂u∂y = 1x |
Integrate (i) partially with respect to x: u = yx + φ(y) - where φ(y) is an arbitrary function of y
Differentiate with respect to y:
Compare with equation (ii):
1x + dφdy | = | 1x | |
i.e. | dφdy | = | 0 |
so: | φ | = | C' - where C' is a constant |
and | u | = | yx + C' |
du = 0 implies u = C, where C is a constant.
Exercise 2:
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
∴ 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):
Exercise 3:
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):
Exercise 4:
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):
Exercise 5:
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):
Exercise 6:
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 = 14 e4x + x2y2 + φ(y)
Differentiate and compare with (ii):
Exercise 7:
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):
Exercise 8:
Solution:
P(x, y) dx + Q(x, y) dy = 0 ⇒ P(x, y) = x tan−1 y and Q(x, y) = x22(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 = x22(1 + y2)
Integrate (i): u = x22 tan−1 y + φ(y)
Differentiate and compare with (ii):
Exercise 9:
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 + x3y33 + φ(y)
Differentiate and compare with (ii):
Exercise 10:
Solution:
∴ 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 = x44 − 2x3y + 32 x2y2 + φ(y)
Differentiate and compare with (ii):
Exercise 11:
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):
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 | |
1a2 + x2 | 1a tan −1 xa (a > 0) | √a2 + x2 | a22 [ sinh−1( xa ) + x √a2−x2 a2 ] | |
1a2 − x2 | 12a ln a + xa − x (0 < x < a) | √a2−x2 | a22 [ sin−1( xa ) + x √a2−x2 a2 ] | |
1x2 − a2 | 12a ln x − ax + a (x > a > 0) | √x2 − a2 | a22 [−cosh−1( xa ) + x √x2−a2 a2 ] | |
1 √ a2 + x2 | ln x + √a2 + x2 a (a > 0) | |||
1 √ a2 − x2 | sin−1 xa (−a < x < a) | 1 √ x2 − a2 | ln x + √x2 − a2 a (x > a > 0) |