Homogeneous equations

Some first-order differential equations look intimidating — a tangle of $x$ and $y$ on the right of $dy/dx$ — until you notice a hidden symmetry: the whole expression depends only on the ratio $y/x$ , never on $x$ and $y$ separately. That single observation is the entire trick. Divide top and bottom by the right power of $x$ , spot the ratio, and a substitution collapses the equation into something you already know how to solve: a variable separable equation. This article covers homogeneous differential equations as they appear in the ISC Class 12 Calculus unit — the second solving technique in the Differential Equations section, right after variable separable equations. You will learn how to test whether an equation is homogeneous, how the substitution $y = vx$ (or its mirror, $x = vy$ ) turns it separable, and how to carry a worked problem through to a particular solution using an initial condition. The method is short, mechanical, and — once the pattern is recognised — genuinely fast to execute under exam pressure.

The core insight

A first-order equation $dy/dx = F(y/x)$ is homogeneous, and the substitution $y = vx$ turns it into a variable separable equation in $v$ and $x$ .

The core idea

What "homogeneous" means for a function. A function $f(x,y)$ is called homogeneous of degree $n$ if scaling both variables by the same factor $\lambda$ scales the output by $\lambda^n$ :

f(\lambda x,\lambda y) = \lambda^n f(x,y)

where: $\lambda$ is any nonzero scaling factor, and $n$ is the degree — a fixed number for a given $f$ . For example, $f(x,y) = x^2+y^2$ satisfies $f(\lambda x,\lambda y)=\lambda^2(x^2+y^2)$ , so it is homogeneous of degree 2.

What "homogeneous" means for the differential equation. A first-order equation is homogeneous if it can be written so the right-hand side is a pure function of $y/x$ :

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

where: $F$ is any function of the single ratio $y/x$ — no bare $x$ or $y$ may survive on its own. In practice, if $\dfrac{dy}{dx} = \dfrac{g(x,y)}{h(x,y)}$ with $g$ and $h$ both homogeneous of the same degree, dividing numerator and denominator by that power of $x$ automatically produces this form.

Why the substitution works. Set $y = vx$ , treating $v$ as a new function of $x$ . By the product rule,

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

Substituting into $dy/dx = F(y/x) = F(v)$ gives $v + x\dfrac{dv}{dx} = F(v)$ , so

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

This is now variable separable in $v$ and $x$ — the $v$ 's go with $dv$ , the $x$ 's with $dx$ :

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

Integrating both sides and substituting $v = y/x$ back at the end returns the solution in the original variables. The figure below shows the whole pipeline as a chain of four transformations, which is worth memorising as a sequence rather than a set of isolated formulas.

Figure

A four stage flow diagram showing a homogeneous equation being tested then substituted then separated then integrated to reach the final solution

Choosing the mirror substitution. Either $y=vx$ or $x=vy$ works on any homogeneous equation, but one usually gives a cleaner separable integral than the other. If the equation reads more naturally as $dy/dx$ with $y$ dominating, try $y=vx$ first; if it reads more naturally as $dx/dy$ with $x$ dominating, $x=vy$ tends to be tidier. Both routes, carried through correctly, give equivalent solutions.

Derivation AReducing a homogeneous equation to a separable one

The goal of this derivation is to show, with nothing beyond the product rule and algebraic rearrangement, exactly why the substitution $y = vx$ always converts a first-order homogeneous equation $dy/dx = F(y/x)$ into a variable separable equation in $v$ and $x$ , so that the method is not memorised as a trick but understood as a forced consequence of the definitions.

Step 1. Start from the defining property. Because the equation is homogeneous, $dy/dx = F(y/x)$ for some function $F$ of the single ratio $y/x$ . This is the only fact we are given, and it is exactly what makes the next substitution productive rather than arbitrary.

Step 2. Introduce the new variable $v = y/x$ , so that $y = vx$ with $v$ itself a function of $x$ . Differentiate $y = vx$ with respect to $x$ using the product rule (treating $v(x)$ and $x$ as the two factors): $\dfrac{dy}{dx} = v\cdot 1 + x\cdot\dfrac{dv}{dx} = v + x\dfrac{dv}{dx}$ . This step is the one most often botched — the derivative of $vx$ is not simply $x\,dv/dx$ , because $v$ is not a constant.

Step 3. Substitute both pieces into the original equation. The left side $dy/dx$ becomes $v + x\,dv/dx$ , and the right side $F(y/x)$ becomes $F(v)$ , since $y/x = v$ by construction. The equation is now $v + x\dfrac{dv}{dx} = F(v)$ .

Step 4. Isolate the derivative term and separate. Subtracting $v$ from both sides gives $x\dfrac{dv}{dx} = F(v) - v$ . Provided $F(v) \neq v$ , divide both sides by $x\bigl(F(v)-v\bigr)$ to get $\dfrac{dv}{F(v)-v} = \dfrac{dx}{x}$ — every $v$ term is now on the left, every $x$ term on the right, which is precisely the variable separable form. Integrating both sides and finally replacing $v$ with $y/x$ delivers the general solution in the original variables. Verified: differentiating $y=vx$ gives $dy/dx = v+x\,dv/dx$ by the product rule; substituting into $dy/dx=F(v)$ and rearranging yields $dv/(F(v)-v)=dx/x$ , matching the standard separable form used in both worked examples below.

Worked example

Solve $\dfrac{dy}{dx} = \dfrac{x^2+y^2}{xy}$ , then find the particular solution given $y(1)=0$ is not imposed here (this is the general solution only); a full particular-solution version follows using the same equation family as Pass-1 Example 2.

Step 1. Check homogeneity. The numerator $x^2+y^2$ and denominator $xy$ are both homogeneous of degree 2 (scaling $x,y\to\lambda x,\lambda y$ scales each by $\lambda^2$ ), so the equation is homogeneous. Divide numerator and denominator by $x^2$ :

\frac{dy}{dx} = \frac{1+(y/x)^2}{y/x}

Step 2. Substitute $y=vx$ , so $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$ :

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

Step 3. Separate and integrate: $v\,dv = \dfrac{dx}{x}$ gives $\dfrac{v^2}{2} = \log|x|+C$ .

Step 4. Substitute back $v = y/x$ : $\dfrac{y^2}{2x^2} = \log|x|+C$ , so the general solution is $y^2 = 2x^2\left(\log|x|+C\right)$ .

Competitive stretch

Now solve the closely related equation $\left(x^2+y^2\right)dx - 2xy\,dy = 0$ with $y(1)=0$ . Rewriting gives $\dfrac{dy}{dx}=\dfrac{x^2+y^2}{2xy}$ , again homogeneous of degree 2. Substituting $y=vx$ : $v+x\dfrac{dv}{dx} = \dfrac{1+v^2}{2v}$ , so $x\dfrac{dv}{dx} = \dfrac{1-v^2}{2v}$ . Separating, $\dfrac{2v\,dv}{1-v^2}=\dfrac{dx}{x}$ , and integrating (recognising the left side as $-f'(v)/f(v)$ with $f(v)=1-v^2$ ) gives $-\log|1-v^2| = \log|x|+C_1$ , which rearranges to $x(1-v^2)=C$ . Substituting back $v=y/x$ : $x^2-y^2 = Cx$ . Applying $y(1)=0$ gives $1 = C$ , so the particular solution is $x^2-y^2=x$ .

Common traps

Misjudging homogeneity. Both the numerator and denominator (or the simplified right-hand side) must be homogeneous of the same degree; an equation like $\dfrac{dy}{dx}=\dfrac{x+y+1}{x-y}$ is not homogeneous because of the stray constant.
Dropping the product-rule term. $dy/dx$ becomes $v+x\,dv/dx$ under $y=vx$ , never just $x\,dv/dx$ — forgetting the lone $v$ is the single most common slip.
Not fully simplifying $F(v)-v$ before separating. Sign or algebra errors here (as in the competitive-stretch example) propagate through every later step.
Leaving the answer in $v$ . The final solution must be expressed back in $x$ and $y$ by substituting $v=y/x$ , not left half-finished.

Ask the tutor: Stuck on whether your equation is homogeneous, which substitution to pick, or the algebra after $y=vx$ — ask the AI tutor to walk through your specific equation step by step.

How do I quickly tell if an equation is homogeneous before starting the substitution?

Look at the right-hand side of $dy/dx = g(x,y)/h(x,y)$ and check whether replacing $x\to\lambda x$ , $y\to\lambda y$ everywhere scales $g$ and $h$ by the same power of $\lambda$ . In practice it is faster to just divide every term by the highest power of $x$ (or $y$ ) present and see whether the whole expression collapses into a function of $y/x$ alone, with no leftover bare $x$ or $y$ . If a constant term survives, like the $+1$ in $x+y+1$ , the equation is not homogeneous and this method does not apply.

Why does y = vx work, and could I use a different substitution instead?

It works because $v=y/x$ is exactly the quantity the equation's right-hand side depends on, so writing $y=vx$ turns every occurrence of that ratio into the single variable $v$ , collapsing two variables into one plus its derivative. The mirror substitution $x=vy$ works identically when the equation is more naturally read with $x$ as the dependent variable — same reasoning, roles swapped. Using an unrelated substitution generally will not separate the variables, since it is the specific structure $F(y/x)$ that makes $y=vx$ productive.

After I substitute back v = y/x, do I need to simplify the answer further?

Usually yes, at least enough to remove any obviously awkward fractions or logs of fractions, but ISC marking accepts any algebraically equivalent form as long as it is expressed purely in $x$ and $y$ (not $v$ ) and the arbitrary constant is clearly present. In the worked example, $\dfrac{y^2}{2x^2}=\log|x|+C$ and $y^2=2x^2(\log|x|+C)$ are the same solution written two ways; either is acceptable unless the question asks for a specific form.

Homogeneous equations

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads