Examples on Homogeneous Differential Equations

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

Homogeneous Differential Equations are differential equations with homogenous functions. They are equations containing a differentiation operator, a function, and a set of variables. The general form of the homogeneous differential equation is f(x, y).dy + g(x, y).dx = 0, where f(x, y) and h(x, y) is a homogenous function. Homogenous functions are defined as functions in which the total power of all the terms of the function is constant. Before continuing with Homogeneous Differential Equations we should learn Homogeneous Functions first. In this article, we will learn about, Homongenous Functions, Homogeneous Differential Equations, their solutions, and others in detail.