Homogeneous Differential Equations

Difficulty Level : Medium
Last Updated : 08 Jun, 2021

In mathematics, a differential equation is an equation that relates one or more functions and their derivatives. In this article, we are going to discuss homogeneous equations but before jumping to the topic let’s understand homogeneous function first.

Homogeneous Function

A function f(x, y) in x and y is said to be a homogeneous function of the degree of each term is p. For example: f(x, y) = (x² + y² – xy) is a homogeneous function of degree 2 where p = 2. Similarly, g(x, y) = (x³ – 3xy² + 3x²y + y³) is a homogeneous function of degree 3 where p = 3. In general, a homogeneous function ƒ(x, y) of degree n is expressible as:

ƒ(x, y) = xⁿƒ(y/x)

Homogeneous Differential Equation

An equation of the form dy/dx = f(x, y)/g(x, y), where both f(x, y) and g(x, y) are homogeneous functions of the degree n in simple word both functions are of the same degree, is called a homogeneous differential equation. For Example: dy/dx = (x² – y²)/xy is a homogeneous differential equation.

Solving a Homogeneous Differential Equation

Let dy/dx = f(x, y)/g(x, y) be a homogeneous differential equation. Now putting y = vx and dy/dx = (v + x dv/dx) in the given equation, we get

v + x dy/dx = F(v)

=> ∫dv/{F(v) – v} = ∫dx/x

=> ∫dv/{F(v) – v} = log|x| + C

Now, replace v by (y/x) to obtain the required solution. Lets look some examples.

Example 1: Solve dy/dx = y² – x²/2xy?

Solution:

Clearly, since each of the functions (y² – x²) and 2xy is a homogeneous function of degree 2, the given equation is homogeneous.
Putting y = vx and dy/dx = v + x dy/dx, the given equation becomes
v + x dv/dx = (v²x² – x²)/2vx²
=> v + x dv/dx = v² – 1/2v [after dividing (v²x²/2vx² – x²/2vx²)]
=> x dv/dx = ((v² – 1/2v) – v)
=> x dv/dx = -(1 + v²)/2v
=> 2v/(1 + v²)dv = -1/x dx
=> ∫2v/(1 + v²)dv = -∫1/x dx [Integrating both the sides]
=> log | 1 + v²| = -log | x | + log C
=> log | 1 + v²| + log | x | = log C
=> log | x(1 + v²) | = log C
=> x(1 + v²) = ±C
=> x(1 + v²) = C₁
=> x(1 + y²/x²) = C₁[Putting the original value of v = y/x]
=> (x² + y²) = xC₁, which is the required solution

Example 2: Solve (x√(x²+ y²) – y²)dx + xy dy = 0?

Solution:

The given equation may be written as
dy/dx = y²– x√(x²+ y²)/xy, which is clearly homogeneous
Putting y = vx and dy/dx = v + x dv/dx in it, we get
v + x dv/dx = {v²x² – x√(x²+ v²y²)}/vx²
=> x dv/dx = [{v² – √(1 + v²)}/v – v]
=> x dv/dx = -√(1 + v²)/v
=> ∫v/√(1 + v²)dv = -∫dx/xc [Integrating both the sides]
=> √(1 + v²) = -log | x | + C
=> √(x²+ y²) + x log | x | = Cx, which is the required solution after putting the value of v = y/x.

Example 3: Solve x dy/dx – y = √(x²+ y²)?

Solution:

The given equation may be written as dy/dx = {y + √(x²+ y²)}/x ,which is clearly homogeneous.
Putting y = vx and dy/dx = v + x dv/dx in it, we get
v + x dv/dx = {vx + √(x² + v²x²)}/x
=> v + x dv/dx = v + √(1+v²) [After dividing the {vx + √(x² + v²x²)}/x]
=> x dv/dx = √(1 + v²) [v on the both sides gets cancelled]
=> dv/√(1+v²) = 1/x dx [after rearranging]
=> ∫dv/√(1+v²) = ∫1/x dx [integrating both sides]
=> log | v | + √(1 + v²) | = log | x | + log C
=> log | {v + √(1 + v²)}/x | = log | C |
=> {v + √(1 + v²)}/x = ±C
=> v + √(1 + v²) = C₁x, where C₁ = ±C
=> y + √(x²+ y²) = C₁x², which is the required solution after putting the value of v = y/x

Example 4: Solve (x cos(y/x))(y dx + x dy) = y sin(y/x)(x dy – y dx)?

Solution:

The given equation may be written as
(x cos(y/x) + y sin(y/x))y – (y sin(y/x) – x cos (y/x)) x . dy/dx = 0
=> dy/dx = {x cos (y/x) + y sin(y/x)}y / {y sin(y/x) – x cos(y/x)}x
=> dy/dx = {cos (y/x) + (y/x)sin(y/x)}(y/x) / {(y/x)sin(y/x) – cos(y/x)} [Dividing numerator and denominator by x²], which is clearly homogeneous ,being a function of (y/x).
Putting y = vx and dy/dx = (v + x dv/dx) in it, we get
v + x dv/dx = v(cos v + sin v)/(v sin v – cos v)
=> x dv/dx = [v(cos v + sin v)/(v sin v – cos v) -v]
=> x dv/dx = 2vcos v/(v sin v – cos v)
=>∫{(v sin v – cos v)/2vcos v}dv = ∫x dx [Integrating both sides]
=> ∫tan v dv – ∫ dv/v = ∫ 2/x dx
=> -log | cos v | – log | v | + log C = 2 log | x |
=> log | cos v | + log | v | + 2log | x | = log | C |
=> log | x²v cos v | = log | C |
=> | x²v cos v | = C [After cancelling log on the both sides]
=> x²v cos v = ± C
=> x²v cos v = C₁ [here we taking ±C = C₁]
=> xy cos(y/x) = C_1,which is the required solution after putting the actual value of v = y/x