Homogeneous Differential Equations | Ch.9 Differential Equations

🔄 Homogeneous Functions

Definition — Homogeneous Function

A function $F(x, y)$ is homogeneous of degree $n$ if:

F(\lambda x,\ \lambda y) = \lambda^n\, F(x, y) \quad \text{for all } \lambda \neq 0

Examples

Function	Degree	Verification
$x^2 + y^2$	2	$(\lambda x)^2+(\lambda y)^2 = \lambda^2(x^2+y^2)$
$x^2 - xy + y^2$	2	Each term has degree 2
$\sin\dfrac{y}{x}$	0	$\sin\dfrac{\lambda y}{\lambda x} = \sin\dfrac{y}{x} = \lambda^0\cdot f$
$x + y + 1$	Not homogeneous	Constant term breaks homogeneity

📋 Homogeneous Differential Equation

Definition

A first-order DE $\dfrac{dy}{dx} = F(x, y)$ is homogeneous if $F(x,y)$ is a homogeneous function of degree zero, i.e., it can be expressed as a function of $\dfrac{y}{x}$ alone:

\frac{dy}{dx} = g\!\left(\frac{y}{x}\right)

Recognizing Homogeneous DEs

DE	Rewritten as $g(y/x)$?	Homogeneous?
$\frac{dy}{dx} = \frac{x+y}{x}$	$1 + \frac{y}{x}$	✅ Yes
$\frac{dy}{dx} = \frac{x^2+y^2}{xy}$	$\frac{x}{y}+\frac{y}{x}$	✅ Yes
$\frac{dy}{dx} = x + y$	Cannot write as $g(y/x)$	❌ No

🪜 Method of Solution — Substitution $y = vx$

Verify the DE is homogeneous.
Substitute $y = vx$, so $\dfrac{dy}{dx} = v + x\,\dfrac{dv}{dx}$.
Rewrite DE in terms of $v$ and $x$ only.
Separate variables: $v$ and $x$ go to different sides.
Integrate both sides.
Replace $v = \dfrac{y}{x}$ to get the solution in $x$ and $y$.

💡 Worked Examples

Example 1

Solve: $\dfrac{dy}{dx} = \dfrac{x + y}{x}$

Rewrite: $\dfrac{dy}{dx} = 1 + \dfrac{y}{x}$ — homogeneous ✓

Let $y = vx$: $v + x\dfrac{dv}{dx} = 1 + v$

$\Rightarrow x\dfrac{dv}{dx} = 1$

Separate: $dv = \dfrac{dx}{x}$

Integrate: $v = \ln|x| + C$

Back-substitute $v = \dfrac{y}{x}$:

\boxed{\frac{y}{x} = \ln|x| + C \implies y = x\ln|x| + Cx}

Example 2

Solve: $(x^2 + y^2)\,dx - 2xy\,dy = 0$

Rewrite: $\dfrac{dy}{dx} = \dfrac{x^2+y^2}{2xy}$

Divide by $x^2$: $\dfrac{dy}{dx} = \dfrac{1 + (y/x)^2}{2(y/x)}$ — homogeneous ✓

Let $y = vx$: $v + x\dfrac{dv}{dx} = \dfrac{1+v^2}{2v}$

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

Separate: $\dfrac{2v\,dv}{1-v^2} = \dfrac{dx}{x}$

Integrate: $-\ln|1-v^2| = \ln|x| + \ln|C|$

\boxed{x^2 - y^2 = Cx^3}

🔑 Key Takeaways

Homogeneous DE: $\frac{dy}{dx}$ can be expressed as $g\bigl(\frac{y}{x}\bigr)$.
Always verify homogeneity before applying the method.
Substitution: $y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}$.
After substituting, variables separate in $v$ and $x$.
Always back-substitute $v = \frac{y}{x}$ at the end.