# 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:

 dy/dx = 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:

y = vx

The LHS of this equation becomes:

dy/dx = xdv/dx + v

- 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:

dy/dx = ƒ ( y/x )

# 2. Exercises

Click on questions to reveal their solutions

Exercise 1:

Find the general solution of  dy/dx = xy + y2/x2

Solution:

RHS = quotient of homogenous functions of same degree (=2)

 Set y = vx: i.e. d/dx(vx) = xvx + v2x2 / x2 i.e. xdv/dx +v = v + v2 Rearrange to separate variables: xdv/dx = v2    (subtracted v from both sides) i.e. dv/v2 = dx/x Integrate: ∫ dv/v2 = ∫ dx/x i.e −1/v = ln x + C Re-express in terms of x, y: − x/y = ln x + C i.e. y = −x/ln x + C

Exercise 2:

Solve 2xydy/dx = x2 + y2, given that y = 0 at x = 1

Solution:

Express equation in standard form: dy/dx = x2 + y2/2xy, i.e. quotient of homogenous functions that have same degree.

 Set y = xv: i.e. d/dx(xv) = x2 + x2v2 / 2x⋅xv i.e. xdv/dx + dx/dxv = x2(1 + v2)/ 2x2v i.e. xdv/dx + v = (1 + v2)/ 2v Rearrange to separate variables: xdv/dx = (1 + v2)/ 2v − v(2v)/ (2v) i.e. xdv/dx = 1 − v2/ 2v Integrate: ∫ 2v/(1 − v2) dv = ∫ dx/x i.e. −ln (1 − v2) = ln x + ln C i.e. ln [(1 − v2)−1] = ln x + ln C i.e. 1/1 − v2 = Cx Re-express in terms of x, y 1/1 − y2/x2 = Cx i.e. x2/ x2 − y2 = Cx i.e. x/ C = x2 − y2

Particular solution: y = 0 when x = 1:

 gives: 1/C = 1 − 0 ∴ C = 1 gives: x2 − y2 = x

Exercise 3:

Solve dy/dx = x + y/x and find the particular solution when y(1) = 1.

Solution:

 Set y = xv: xdv/dx + v = x + xv/ x = x/ x (1 + v) = (1 + v) i.e. xdv/dx = 1 Separate variables and integrate: ∫dv = ∫ dx/x i.e. v = ln x + ln k (ln k = constant) i.e. v = ln (kx) Re-express in terms of x, y: y/x = 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 xdy/dx = xy and find the particular solution when y(2) = 1/2

Solution:

Express equation in standard form: dy/dx = xy/x

 Set y = vx: i.e. xdv/dx + v = x + xv/ x = 1 − v i.e. xdv/dx = 1 − 2v Separate variables and integrate: ∫ dv/1 − 2v = ∫ dx/x i.e. − 1/2 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 − 2y/x)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 = 1/2 ∴ 2 = x2 − 2xy

Exercise 5:

Solve dy/dx = x − 2y/x and find the particular solution when y(1) = −1

Solution:

 Set y = xv: xdv/dx + v = x − 2xv/ x = 1 − 2v i.e. xdv/dx = 1 − 3v Separate variables and integrate: ∫ dv/1 − 3v = ∫ dx/x 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 − 3y/x) = K i.e. x3 (x − 3y/x) = 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 dy/dx = x + y/xy, prove that tan−1 ( y/x ) = 1/2 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: xdv/dx + v = x + vx/ x − vx i.e. xdv/dx = x(1 + v)/ x(1 − v) − v = (1 + v − v(1 − v)/ 1 − v i.e. xdv/dx = 1 + v2/ (1 − v) Separate variables and integrate: ∫ 1 − v/ (1 + v2) dv = ∫ dx/x i.e. ∫ dv/ (1 + v2) − 1/2 ∫ 2v/ (1 + v2)dv = ∫ dx/x i.e. tan−1 v − 1/2 ln (1 + v2) = ln x + A Re-express in terms of x, y: tan−1 (y/x) − 1/2 ln (1 + y2/x2) = ln (x) + A tan−1 (y/x) = 1/2 ln (x2 + y2/x2) + 1/2 ln (x2) + A = 1/2 ln [ (x2 + y2/x2)⋅x2 ] + A ∴  tan−1 (y/x) = 1/2 ln (x2 + y2) + A

Exercise 7:

Find the general solution of 2x2dy/dx = x2 + y2

Solution:

Express equation in standard form: dy/dx = x2 + y2/2x2

 Set y = xv: i.e. xdv/dx + v = x2 + x2v2/ 2x2 = 1 + v2/2 i.e. xdv/dx + v = 1 + v2/2 − 2v/2 = 1 + v2 − 2v/2 Separate variables and integrate: ∫ dv/ 1 − 2v + v2 = 1/2 ∫ dx/x i.e. ∫ dv/ (1 − v)2 = 1/2 ∫ dx/x [ 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: dw/dv = −1. and ∫ dv/ (1 − v)2 = 1/ (−1) ∫ dw/ w2 ] i.e. − ∫ dw/ w2 = 1/2 ∫ dx/x i.e. − ( − 1/ w ) = 1/2 ln x + C i.e. 1/ 1 − v = 1/2 ln x + C Re-express in terms of x, y: 1/ 1 − y/ x = 1/2 ln x + C i.e. x/ x − y = 1/2 ln x + C i.e. 2x = (x − y)(ln x + C')     (C' = 2C)

Exercise 8:

Find the general solution of (2xy)dy/dx = 2yx

Solution:

Express equation in standard form: dy/dx = 2yx/2xy

 Set y = vx: i.e. xdv/dx + v = 2v − 1/ 2 − v ∴ xdv/dx = 2v − 1 − v(2 − v)/ 2 − v = v2 − 1/ 2 − v Separate variables and integrate: ∫ 2 − v/v2 − 1 dv = ∫ dx/x Use Partial Fractions to help solve LHS: 2 − v/v2 − 1 = A/v − 1 + B/v + 1 = A(v + 1) + B(v − 1)/ v2 − 1 ∴ A + B = −1, A − B = 2   ⇒  A = 1/2 and B = −3/2 i.e. 1/2 ∫ ( 1/v − 1 − 3/v + 1 ) dv = ∫ dx/x i.e. 1/2 ln (v − 1) − 3/2 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: (y/x − 1) / (y/x + 1)3x2 = k2 i.e. ((y − x)/x) / (y + x/x)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:

dy/dx = ƒ ( y/x )ƒ(v)

Exercise 9:

Find the general solution of dy/dx = y/x + tan  ( y/x )

Solution:

RHS is only a function of v = dy/dx, so substitute and separate variables.

 Set y = xv: xdv/dx + v = v + tan v i.e. xdv/dx = tan x Separate variables and integrate: ∫ dv/tan v = ∫ dx/x { Note: ∫ cos v/sin 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 v/x ] = ln k i.e. sin v/x = k i.e. sin v = kx Re-express in terms of x, y: sin (y/x) = kx

Exercise 10:

Find the general solution of xdy/dx = y + x ey/x

Solution:

Rewrite equation so that RHS is a function of y/x only:

 dy/dx = (y/x) + e y/x Set y = xv: xdv/dx + v = v + ev Separate variables and integrate: ∫e−vdv = ∫ dx/x i.e. −ev = ln x + ln k = ln (kx) i.e. e−v = − ln (kx) Re-express in terms of x, y: e−y/x = − ln (kx) i.e. −y/x = ln [− ln (kx)] i.e. y = − x ln [−ln (kx)]

Exercise 11:

Find the general solution of xdy/dx = y + x2 + y2

Solution:

Rewrite equation so that RHS is a function of y/x only:

 dy/dx = y/x + 1/x √x2 + y2 = y/x + √1 + (y/x)2 Set y = xv: xdv/dx + v = v + √1 + v2 i.e. xdv/dx = √1 + v2 Separate variables and integrate: ∫ dv/ √1 + v2 = ∫ dx/x { 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 (y/x) = 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) 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)