Q.Which of the following is a homogeneous differential equation? (A) $(4x + 6y + 5)\, dy - (3y + 2x + 4)\, dx = 0$ (B) $(xy)\, dx - (x^3 + y^3)\, dy = 0$ (C) $(x^3 + 2y^2)\, dx + 2xy\, dy = 0$ (D) $y^2\, dx + (x^2 - xy - y^2)\, dy = 0$

Concept: Homogeneous Differential Equation — A first-order DE is homogeneous if it can be written as $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$ (or equivalently, if every term has the same total degree when expressed in $x$ and $y$).

Step 1: Check each option by rewriting in the form $M(x,y)\,dx + N(x,y)\,dy = 0$ and testing if $M$ and $N$ are homogeneous functions of the same degree.

• (A) $M = -(3y+2x+4)$, $N = 4x+6y+5$ — constant terms (+4, +5) break homogeneity. Not homogeneous.
• (B) $M = xy$ (degree 2), $N = -(x^3+y^3)$ (degree 3). Degrees differ. Not homogeneous.
• (C) $M = x^3+2y^2$ (mixed degrees 3 and 2), $N = 2xy$ (degree 2). Not homogeneous. …

The Core Idea: What Makes an Equation Homogeneous?

A first-order differential equation is called homogeneous if it can be written in the form

where the right-hand side depends only on the ratio $y/x$ (or equivalently, $x/y$). The classic test: replace $x$ with $tx$ and $y$ with $ty$ in the equation. If every term has the same total degree (the sum of the exponents of $x$ and $y$), then the equation is homogeneous.

Why does this matter? Because if an equation is homogeneous, the substitution $y = vx$ (or $x = vy$) turns it into a separable equation — a clean, solvable form. That’s the power of spotting homogeneity.

Let’s examine each option carefully.

1. Option (A): $(4x + 6y + 5)\, dy - (3y + 2x + 4)\, dx = 0$

Rewrite it as:

Now test homogeneity: replace $x$ with $tx$ and $y$ with $ty$:

The constants $4$ and $5$ do not have a factor of $t$. So the expression does not simplify to a function of $y/x$ alone. The presence of constant terms breaks homogeneity.

Conclusion: Not homogeneous.

2. Option (B): $(xy)\, dx - (x^3 + y^3)\, dy = 0$

Rewrite as:

Test homogeneity: replace $x$ with $tx$, $y$ with $ty$:

The factor $1/t$ remains — the expression is not a function of $y/x$ alone because it still depends on $t$. For homogeneity, the $t$ must cancel completely, leaving only the ratio.

Conclusion: Not homogeneous.

3. Option (C): $(x^3 + 2y^2)\, dx + 2xy\, dy = 0$

Rewrite as:

Degree check: numerator $x^3$ (degree 3) and $2y^2$ (degree 2) — they don’t even have the same degree within the numerator. Denominator $2xy$ has degree 2. So the expression cannot be a function of $y/x$ alone.

Test formally: replace $x$ with $tx$, $y$ with $ty$: …

Method: Testing whether a differential equation is homogeneous

To pick the homogeneous one from a list, apply the degree test to each candidate rather than solving anything.

Steps

Step 1: Write each equation as $M(x,y)\,dx+N(x,y)\,dy=0$.

Identify the coefficient of $dx$ and of $dy$.

Step 2: Find the total degree of every term.

The degree of a term is the sum of the exponents of $x$ and $y$ in it (a constant is degree $0$).

Step 3: Demand a single common degree. …

Common Mistakes

Mistake 1: Ignoring constant terms when judging homogeneity.

Why it's wrong: in option (A) the $+5$ and $+4$ are degree-$0$ terms sitting beside degree-$1$ terms, so the equation is not homogeneous even though $4x+6y$ "looks" uniform. Correct approach: every term, constants included, must share one total degree.

Mistake 2: Not checking that numerator and denominator have the same degree. …