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Homogeneous Functions |
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PPLATO / Tutorials / Differential Equations |
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1. Theory
M(x, y) = 3 × 2 + xy is a homogeneous function since the sum of the powers of x and y in each term is the same (i.e. x2 is x to power 2 and xy = x1y1 giving total power of 1 + 1 = 2).
The degree of this homogeneous function is 2.
Here, we consider differential equations with the following standard form:
| dydx = M(x, y)N(x, y) |
- where M(x, y) and N(x, y) are homogenous functions of the same degree.
To find the solution, change the dependent variable from y to v, where:
The LHS of this equation becomes:
- using the product rule for differentiation.
Solve the resulting equation by separating the variables v and x. Finally, re-express the solution in terms of x and y.
Note: This method also works for equations of the form:
2. Exercises
Click on questions to reveal their solutions
Exercise 1:
Find the general solution of dydx = xy + y2x2
Solution:
RHS = quotient of homogenous functions of same degree (=2)
| Set y = vx: | i.e. | ddx(vx) | = | xvx + v2x2 x2 |
| i.e. | xdvdx +v | = | v + v2 | |
| Rearrange to separate variables: | xdvdx | = | v2 (subtracted v from both sides) | |
| i.e. | dvv2 | = | dxx | |
| Integrate: | ∫ dvv2 | = | ∫ dxx | |
| i.e | −1v | = | ln x + C | |
| Re-express in terms of x, y: | − xy | = | ln x + C | |
| i.e. | y | = | −xln x + C |
Exercise 2:
Solve 2xydydx = x2 + y2, given that y = 0 at x = 1
Solution:
Express equation in standard form: dydx = x2 + y22xy, i.e. quotient of homogenous functions that have same degree.
| Set y = xv: | i.e. | ddx(xv) | = | x2 + x2v2 2x⋅xv |
| i.e. | xdvdx + dxdxv | = | x2(1 + v2) 2x2v | |
| i.e. | xdvdx + v | = | (1 + v2) 2v | |
| Rearrange to separate variables: | xdvdx | = | (1 + v2) 2v − v(2v) (2v) | |
| i.e. | xdvdx | = | 1 − v2 2v | |
| Integrate: | ∫ 2v(1 − v2) dv | = | ∫ dxx | |
| i.e. | −ln (1 − v2) | = | ln x + ln C | |
| i.e. | ln [(1 − v2)−1] | = | ln x + ln C | |
| i.e. | 11 − v2 | = | Cx | |
| Re-express in terms of x, y | 11 − y2x2 | = | Cx | |
| i.e. | x2 x2 − y2 | = | Cx | |
| i.e. | x C | = | x2 − y2 |
Particular solution: y = 0 when x = 1:
| gives: | 1C | = | 1 − 0 |
| ∴ | C | = | 1 |
| gives: | x2 − y2 | = | x |
Exercise 3:
Solve dydx = x + yx and find the particular solution when y(1) = 1.
Solution:
| Set y = xv: | xdvdx + v | = | x + xv x | |
| = | x x (1 + v) = (1 + v) | |||
| i.e. | xdvdx | = | 1 | |
| Separate variables and integrate: | ∫dv | = | ∫ dxx | |
| i.e. | v | = | ln x + ln k (ln k = constant) | |
| i.e. | v | = | ln (kx) | |
| Re-express in terms of x, y: | yx | = | ln (kx) | |
| i.e. | y | = | x ln (kx) |
Particular solution with y = 1 when x = 1:
| 1 | = | ln (k) | |
| i.e. | k | = | e1 = e |
| i.e. | y | = | x ln (ex) |
| = | x[ln e + ln x) | ||
| = | x[1 + ln x] | ||
| i.e. | y | = | x + x ln x |
Exercise 4:
Solve xdydx = x − y and find the particular solution when y(2) = 12
Solution:
Express equation in standard form: dydx = x − yx
| Set y = vx: | i.e. | xdvdx + v | = | x + xv x = 1 − v |
| i.e. | xdvdx | = | 1 − 2v | |
| Separate variables and integrate: | ∫ dv1 − 2v | = | ∫ dxx | |
| i.e. | − 12 ln (1 − 2v) | = | ln x + ln k | |
| i.e. | ln [(1 − 2v)−½] − ln x | = | ln (k) | |
| i.e. | ln [ 1(1 − 2v)−½x ] | = | ln (k) | |
| i.e. | 1 | = | kx(1 − 2v)½ | |
| Re-express in terms of x, y: | 1 | = | kx(1 − 2yx)½ | |
| i.e. | 1 | = | kx(x − 2y x)½ | |
| (square both sides) | 1 | = | K x2 (x − 2y x), where K = k2 | |
| i.e. | 1 | = | K x(x − 2y) |
Particular solution with y = 1 2 when x = 2:
| 1 | = | K⋅2 (2 − 2(1 2)) | |
| = | K⋅2⋅1 | ||
| i.e. | K | = | 12 |
| ∴ | 2 | = | x2 − 2xy |
Exercise 5:
Solve dydx = x − 2yx and find the particular solution when y(1) = −1
Solution:
| Set y = xv: | xdvdx + v | = | x − 2xv x | |
| = | 1 − 2v | |||
| i.e. | xdvdx | = | 1 − 3v | |
| Separate variables and integrate: | ∫ dv1 − 3v | = | ∫ dxx | |
| i.e. | 1(−3) ln (1 − 3v) | = | ln x + ln k (ln k = constant) | |
| i.e. | ln (1 − 3v) | = | −3 ln x − ln k | |
| i.e. | ln (1 − 3v) + ln x3 | = | −3 ln k | |
| i.e. | ln [x3(1 − 3v)] | = | −3 ln k | |
| i.e. | x3(1 − 3v) | = | K where K = constant | |
| Re-express in terms of x, y: | x3 (1 − 3yx) | = | K | |
| i.e. | x3 (x − 3yx) | = | K | |
| i.e. | x2 (x − 3y) | = | K |
Particular solution with y(1) = −1:
| 1(1 + 3) | = | K | |
| i.e. | K | = | 4 |
| ∴ | x2 (x − 3y) | = | 4 |
Exercise 6:
Given that dydx = x + yx − y, prove that tan−1 ( yx ) = 12 ln (x2 + y2) + A, where A is an arbitrary constant.
Solution:
The equation is already in standard form, with quotient of two first degree homogeneous functions.
| Set y = xv: | xdvdx + v | = | x + vx x − vx | |
| i.e. | xdvdx | = | x(1 + v) x(1 − v) − v | |
| = | (1 + v − v(1 − v) 1 − v | |||
| i.e. | xdvdx | = | 1 + v2 (1 − v) | |
| Separate variables and integrate: | ∫ 1 − v (1 + v2) dv | = | ∫ dxx | |
| i.e. | ∫ dv (1 + v2) − 12 ∫ 2v (1 + v2)dv | = | ∫ dxx | |
| i.e. | tan−1 v − 12 ln (1 + v2) | = | ln x + A | |
| Re-express in terms of x, y: | tan−1 (yx) − 12 ln (1 + y2x2) | = | ln (x) + A | |
| tan−1 (yx) | = | 12 ln (x2 + y2x2) + 12 ln (x2) + A | ||
| = | 12 ln [ (x2 + y2x2)⋅x2 ] + A | |||
| ∴ tan−1 (yx) | = | 12 ln (x2 + y2) + A |
Exercise 7:
Find the general solution of 2x2dydx = x2 + y2
Solution:
Express equation in standard form: dydx = x2 + y22x2
| Set y = xv: | i.e. | xdvdx + v | = | x2 + x2v2 2x2 |
| = | 1 + v22 | |||
| i.e. | xdvdx + v | = | 1 + v22 − 2v2 | |
| = | 1 + v2 − 2v2 | |||
| Separate variables and integrate: | ∫ dv 1 − 2v + v2 | = | 12 ∫ dxx | |
| i.e. | ∫ dv (1 − v)2 | = | 12 ∫ dxx | |
| [ Note: 1 − v is a linear function of v, therefore we use a Standard Integral and divide by the coefficient of v. In other words, let w = 1 − v, so: dwdv = −1. | ||||
| and | ∫ dv (1 − v)2 | = | 1 (−1) ∫ dw w2 ] | |
| i.e. | − ∫ dw w2 | = | 12 ∫ dxx | |
| i.e. | − ( − 1 w ) | = | 12 ln x + C | |
| i.e. | 1 1 − v | = | 12 ln x + C | |
| Re-express in terms of x, y: | 1 1 − y x | = | 12 ln x + C | |
| i.e. | x x − y | = | 12 ln x + C | |
| i.e. | 2x | = | (x − y)(ln x + C') (C' = 2C) | |
Exercise 8:
Find the general solution of (2x − y)dydx = 2y − x
Solution:
Express equation in standard form: dydx = 2y − x2x − y
| Set y = vx: | i.e. | xdvdx + v | = | 2v − 1 2 − v |
| ∴ | xdvdx | = | 2v − 1 − v(2 − v) 2 − v | |
| = | v2 − 1 2 − v | |||
| Separate variables and integrate: | ∫ 2 − vv2 − 1 dv | = | ∫ dxx | |
| Use Partial Fractions to help solve LHS: 2 − vv2 − 1 = Av − 1 + Bv + 1 = A(v + 1) + B(v − 1) v2 − 1
∴ A + B = −1, A − B = 2
⇒ A = 12 and B = −32
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| i.e. | 12 ∫ ( 1v − 1 − 3v + 1 ) dv | = | ∫ dxx | |
| i.e. | 12 ln (v − 1) − 32 ln (v + 1) | = | ln x + ln k | |
| i.e. | ln [ (v − 1)1⁄2 (v + 1)3⁄2; ] | = | ln (k) | |
| i.e. | v − 1 (v + 1 )3x2 | = | k2 | |
| Re-express in terms of x, y: | (yx − 1) (yx + 1)3x2 | = | k2 | |
| i.e. | ((y − x)x) (y + xx)3x2 | = | k2 | |
| i.e. | y − x | = | K(y + x)3 | |
Note: The key to solving the next three equations is to recognise that each equation can be written in the form:
Exercise 9:
Find the general solution of dydx = yx + tan ( yx )
Solution:
RHS is only a function of v = dydx, so substitute and separate variables.
| Set y = xv: | xdvdx + v | = | v + tan v | |
| i.e. | xdvdx | = | tan x | |
| Separate variables and integrate: | ∫ dvtan v | = | ∫ dxx | |
| { Note: | ∫ cos vsin v dv | ≡ | ∫ ƒ′(v)ƒ(v) dv = ln [ƒ(v)] + C } | |
| i.e. | ln [sin v] | = | ln x + ln k (ln k = constant) | |
| i.e. | ln [ sin vx ] | = | ln k | |
| i.e. | sin vx | = | k | |
| i.e. | sin v | = | kx | |
| Re-express in terms of x, y: | sin (yx) | = | kx |
Exercise 10:
Find the general solution of xdydx = y + x eyx
Solution:
Rewrite equation so that RHS is a function of yx only:
| dydx | = | (yx) + e yx | ||
| Set y = xv: | xdvdx + v | = | v + ev | |
| Separate variables and integrate: | ∫e−vdv | = | ∫ dxx | |
| i.e. | −ev | = | ln x + ln k | |
| = | ln (kx) | |||
| i.e. | e−v | = | − ln (kx) | |
| Re-express in terms of x, y: | e−yx | = | − ln (kx) | |
| i.e. | −yx | = | ln [− ln (kx)] | |
| i.e. | y | = | − x ln [−ln (kx)] |
Exercise 11:
Find the general solution of xdydx = y + √x2 + y2
Solution:
Rewrite equation so that RHS is a function of yx only:
| dydx | = | yx + 1x √x2 + y2 | ||
| = | yx + √1 + (yx)2 | |||
| Set y = xv: | xdvdx + v | = | v + √1 + v2 | |
| i.e. | xdvdx | = | √1 + v2 | |
| Separate variables and integrate: | ∫ dv √1 + v2 | = | ∫ dxx | |
| { Use Standard Integral: | ∫ dv √1 + v2 | = | sinh−1 (v) + C } | |
| i.e. | sinh−1 (v) | = | ln (x) + A | |
| Re-express in terms of x, y: | sinh−1 (yx) | = | ln (x) + A |
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) | |
| 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) |