Homogeneous Differential Equations: Definition, Formula & Examples   

1.0What is a Homogeneous Differential Equation?

Homogeneous Differential Equations Definition

A homogeneous differential equation is a first-order differential equation of the form:

$\frac{dy}{dx}=f(x,y)$

where the function f(x,y) is a homogeneous function of degree 0.

In other words, f(x, y) depends only on the ratio y/x (or x/y).

Mathematically, a function f(x, y) is homogeneous of degree n if:

$f(tx,ty)=t^{n}f(x,y),\forall t\ne 0$

So, if n = 0:

f(tx,ty) = f(x,y)

This ensures the equation can be solved by substitution.

General Form

The general form of a homogeneous differential equation is:

$\frac{dy}{dx}=\frac{F(x,y)}{G(x,y)}$

Or in differential form:

$M(x,y)dx+N(x,y)dy=0$

where both M(x,y) and N(x,y) are homogeneous functions of the same degree.

Condition for Homogeneity

For the equation:

$\frac{dy}{dx}=\frac{F(x,y)}{G(x,y)}$

to be homogeneous:

$F(tx,ty)=t^{n}F(x,y),G(tx,ty)=t^{n}G(x,y)$

This ensures both numerator and denominator have the same degree.

2.0Homogeneous Differential Equations Formula

The key formula to solve is:

$y=vx,\frac{dy}{dx}=v+x\frac{dv}{dx}$

$⟹\frac{dv}{dx}=\frac{f(v)-v}{x}$

Thus,

$\int \frac{dv}{f(v)-v}=\ln |x|+C$

3.0Steps to Solve Homogeneous Differential Equation

The standard substitution method makes solving these equations easier.

Substitution Method (y=vx)

We substitute:

$y=vx⟹v=\frac{y}{x}$

Differentiating:

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

This reduces the equation into a separable form.

Derivation of Formula

1. Start with: $\frac{dy}{dx}=f\left(\frac{y}{x}\right)$
2. Substituting y=vx: $\frac{dy}{dx}=v+x\frac{dv}{dx}$
3. Substitution yields: $v+x\frac{dv}{dx}=f(v)$
4. Rearranging: $x\frac{dv}{dx}=f(v)-v$
5. Separable form: $\frac{dv}{f(v)-v}=\frac{dx}{x}$
6. Integrate both sides: $\int \frac{dv}{f(v)-v}=\ln |x|+C$

General solution obtained.

4.0How to Solve a Homogeneous Differential Equation?

To solve a homogeneous differential equation, follow these systematic steps:

1. Check for Homogeneity:
   Ensure the equation can be written in the form $(\frac{dy}{dx}=F\left(\frac{y}{x}\right))$ or that both functions (M(x, y)) and (N(x, y)) are homogeneous of the same degree.
2. Use the Substitution (y = vx):
   Let (y = vx), where (v) is a function of (x). Then, $(\frac{dy}{dx}=v+x\frac{dv}{dx})$.
3. Transform the Equation:
   Substitute (y) and $(\frac{dy}{dx})$ in the original equation. This should result in an equation involving only (v) and (x).
4. Separate Variables:
   Rearrange the equation to group all (v) terms on one side and all (x) terms on the other, making it separable.
5. Integrate Both Sides:
   Integrate with respect to their respective variables. Use partial fractions or other integration techniques as needed.
6. Back-substitute:
   Replace (v) with (\frac{y}{x}) to return to the original variables.
7. General Solution:
   Simplify the result and include the constant of integration.

5.0Homogeneous Differential Equations Examples (Solved)

Example 1:

Solve

$\frac{dy}{dx}=\frac{x+y}{x-y}$

Solution:

$\frac{dy}{dx}=\frac{x+y}{x-y}$

Put y = vx, dy/dx = v + x dv/dx:

$v+x\frac{dv}{dx}=\frac{1+v}{1-v}$

$x\frac{dv}{dx}=\frac{1+v}{1-v}-v$

$x\frac{dv}{dx}=\frac{1+v-v(1-v)}{1-v}=\frac{1+v^2}{1-v}$

So,

$\frac{dv}{dx}=\frac{1+v^2}{x(1-v)}$

Rearranging:

$\frac{(1-v)}{1+v^2}dv=\frac{dx}{x}$

Integrating:

$\int \frac{1-v}{1+v^2}dv=\ln |x|+C$

Split:

$\int \frac{1}{1+v^2}dv-\int \frac{v}{1+v^2}dv=\ln |x|+C$

${\tan }^{-1}v-\frac{1}{2}\ln (1+v^2)=\ln |x|+C$

Back-substitute v=y/x:

${\tan }^{-1}\left(\frac{y}{x}\right)-\frac{1}{2}\ln (x^2+y^2)=\ln |x|+C$

Example 2:

Solve

$(x^2+y^2)dy=2xydx$

Solutions:

$\frac{dy}{dx}=\frac{2xy}{x^2+y^2}$

Put y=vx:

$v+x\frac{dv}{dx}=\frac{2v}{1+v^2}$

$x\frac{dv}{dx}=\frac{2v}{1+v^2}-v$

$x\frac{dv}{dx}=\frac{2v-v(1+v^2)}{1+v^2}=\frac{v-v^3}{1+v^2}$

$\frac{1+v^2}{v(1-v^2)}dv=\frac{dx}{x}$

6.0Practice Questions 

1. Solve: $\frac{dy}{dx}=\frac{x^2+y^2}{2xy}$
2. Solve: $(x^2-y^2)dy=2xydx$
3. Show and solve: $\frac{dy}{dx}=\frac{x+y}{x-y}$
4. Solve: $\frac{dy}{dx}=\frac{3x^2+y^2}{2xy}$
5. Evaluate the solution for: $\frac{dy}{dx}=\frac{x^2-xy}{y^2-xy}$