Chapter 1 · First-Order ODEs

Homogeneous First-Order Equations

The v = y/x Substitution

Dr. Mohamed Mabrok · Qatar University

What is a Homogeneous Equation?

A homogeneous equation has a special property: the RHS depends only on the ratio of $y$ to $x$.

Definition (Form 1)

      A first-order ODE is homogeneous if it can be written as:
    
      $$\frac{dy}{dx} = F\left(\frac{y}{x}\right)$$

Equivalently (Form 2): $M(x,y)\,dx + N(x,y)\,dy = 0$ where $M,N$ are homogeneous functions of the same degree $n$.

Homogeneous Function

      A function $f(x,y)$ is homogeneous of degree $n$ if:
    
      $$f(tx, ty) = t^n \cdot f(x,y) \quad \text{for all } t > 0$$

Identifying Homogeneous Functions

Quick Test: Replace $x \to tx$ and $y \to ty$. If all $t$'s cancel, it's homogeneous.

Degree 2
$$x^2 + xy + y^2$$

        $f(tx,ty) = t^2(x^2 + xy + y^2)$ ✓
      

Degree 1
$$\sqrt{x^2 + y^2}$$

        $f(tx,ty) = t \sqrt{x^2 + y^2}$ ✓
      

Degree 0
$$\frac{y}{x}$$

        $f(tx,ty) = \frac{y}{x}$ ✓
      

      Key insight: If the RHS $F(y/x)$ is homogeneous of degree 0, the entire equation is homogeneous.
    

The Magic Substitution: $v = y/x$

The key is to use the substitution $v = y/x$ to convert this into a separable equation!

STEP 1 Substitute $v = y/x$, so $y = vx$

\frac{dy}{dx} = \frac{d}{dx}(vx) = v + x\frac{dv}{dx}

Product rule: $\frac{d(vx)}{dx} = v + x \frac{dv}{dx}$

STEP 2 Substitute into $\frac{dy}{dx} = F(v)$

v + x\frac{dv}{dx} = F(v)

STEP 3 Rearrange to separable form

\frac{dv}{F(v) - v} = \frac{dx}{x}

5-Step Solution Method for Homogeneous Equations

STEP 1
        Verify homogeneity

        Confirm RHS = $F(y/x)$ depends only on ratio
      

STEP 2
        Substitute $y = vx$

        Get $v + x(dv/dx) = F(v)$
      

STEP 3
        Separate and integrate

        $\int \frac{dv}{F(v)-v} = \int \frac{dx}{x} + C$
      

STEP 4
        Back-substitute $v = y/x$

        Express final solution in terms of $x,y$
      

STEP 5
        Check singular solutions

        Set $F(v) - v = 0$ — these give $y = v_0 x$
      

Worked Example 1

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

STEPS 1–2 Rewrite in form $F(v)$ and substitute

F\left(\frac{y}{x}\right) = \frac{1 + (y/x)^2}{y/x} = \frac{1 + v^2}{v}

STEP 2–3 Get separable equation

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

STEP 3–4 Separate, integrate, back-substitute

v\,dv = \frac{dx}{x} \quad \Rightarrow \quad \frac{v^2}{2} = \ln|x| + C \quad \Rightarrow \quad \boxed{y^2 = x^2(2\ln|x| + 2C)}

Worked Example 2

Solve: $xy' = y + x$, $y(1) = 1$

STEPS 1–2 Rewrite and identify

\frac{dy}{dx} = \frac{y}{x} + 1 = F(v), \quad \text{where } v = \frac{y}{x}

STEP 3 Substitute and separate

v + x\frac{dv}{dx} = v + 1 \quad \Longrightarrow \quad x\frac{dv}{dx} = 1 \quad \Longrightarrow \quad dv = \frac{dx}{x}

STEP 4–5 Integrate and back-substitute

v = \ln|x| + C \quad \Rightarrow \quad y = x(\ln x + C)

Apply $y(1) = 1$: $1 = 1 \cdot (0 + C)$, so $C = 1$

\boxed{y = x(\ln x + 1)}

Beautiful Geometry: Family of Circles

Example: $\dfrac{dy}{dx} = \dfrac{y^2 - x^2}{2xy}$

Solution Steps

      $F(v) = \dfrac{v^2-1}{2v}$, after substitution: $x\,\dfrac{dv}{dx} = -\dfrac{v^2+1}{2v}$
    
      $$\int \frac{2v\,dv}{v^2+1} = -\int \frac{dx}{x} \quad \Rightarrow \quad \ln(v^2+1) = -\ln|x| + C$$

Elegant Solution

      Substituting back and simplifying:
    
      $$x^2 + y^2 = Cx$$
    
      A family of circles passing through the origin!

      Complete the square: $(x - C/2)^2 + y^2 = (C/2)^2$

Common Pitfalls & Mistakes

Mistake 1

          Forgetting the product rule

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

Mistake 2

          Forgetting back-substitution

          Always replace $v = y/x$ at the end!

Mistake 3

          Ignoring singular solutions

          When $F(v) = v$, we get $x\,dv/dx = 0$

          This gives $v = $ constant, or $y = v_0 x$
        
          Always check these separately!

Mistake 4

          Forgetting to verify homogeneity first

          Test: Does RHS = $F(y/x)$? If yes, proceed!

Key Takeaways

🎯

        Test for homogeneity

        Check if RHS $= F(y/x)$ depends only on the ratio. If yes, it's homogeneous!

✂️

        $v = y/x$ always works

        The substitution converts any homogeneous equation into a separable one!

⚠️

        Check $F(v) = v$ solutions

        These singular solutions $y = v_0 x$ are easy to miss but geometrically important!

🌌

        Rotational symmetry

        Homogeneous equations have curves with rotational symmetry about the origin.

\text{Homogeneous} \;\Rightarrow\; v=y/x \;\Rightarrow\; \text{Separable} \;\Rightarrow\; \text{Integrate} \;\Rightarrow\; \text{Back-substitute}