Homogeneous differential equations

Not every first-order differential equation splits neatly into "a function of $x$ " times "a function of $y$ ." Homogeneous differential equations look stubborn at first glance — $\dfrac{dy}{dx}$ tangled up with both $x$ and $y$ in a way that resists direct separation — yet they hide a simple pattern: the right-hand side depends only on the ratio $y/x$ . That single observation is the key. The substitution $y = vx$ turns the ratio into a new variable $v$ , and the equation collapses into one that is separable. This is the second solving technique in the CBSE Class 12 Differential Equations chapter, building directly on variables-separable, and it is tested through a fixed rhythm of steps: check homogeneity, substitute, separate, integrate, substitute back.

The core insight

An equation $\dfrac{dy}{dx} = h\!\left(\dfrac{y}{x}\right)$ is homogeneous and becomes separable under $y = vx$ , since $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$ reduces it to $\dfrac{dv}{h(v)-v} = \dfrac{dx}{x}$ .

The core idea

A function $f(x,y)$ is homogeneous of degree $n$ if $f(\lambda x, \lambda y) = \lambda^n f(x,y)$ for every $\lambda \neq 0$ — scaling both inputs by $\lambda$ just scales the output by $\lambda^n$ . A first-order differential equation $\dfrac{dy}{dx} = F(x,y)$ is called homogeneous precisely when $F(x,y)$ can be rewritten as a function of $y/x$ alone:

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

where: $h$ is any function of the single ratio $t = y/x$ , with no leftover bare $x$ or $y$ . Equivalently, if $\dfrac{dy}{dx} = \dfrac{g(x,y)}{f(x,y)}$ , the equation is homogeneous exactly when $g$ and $f$ are homogeneous of the same degree $n$ — the $x^n$ factors cancel when you divide, leaving a pure function of $y/x$ . Two practical points carry most of the marks. First, recognising the form is a quick algebraic check: divide numerator and denominator by the highest power of $x$ and see whether only $y/x$ survives. Second, the substitution that unlocks the equation is always the same:

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

obtained by differentiating $y = vx$ with respect to $x$ via the product rule, treating $v$ as an unknown function of $x$ . This is the step most students fumble — dropping the $x\,dv/dx$ term and writing $dy/dx = v$ is the single most common error in the whole topic. The picture underneath is that a homogeneous right-hand side depends only on the slope of the ray from the origin through $(x,y)$ , not on how far along that ray you are. Move a point out along the same ray — scale both $x$ and $y$ by the same $\lambda$ — and the ratio $y/x$ , hence the slope $h(y/x)$ , does not change at all. So the differential equation "looks the same" along any ray $y = vx$ for fixed $v$ , which is exactly why the substitution that freezes $v$ turns the problem into one variable at a time: $v$ tags which ray you are on, and $x$ tags where along it you sit.

Figure

Rays from the origin at different slopes v illustrating that a homogeneous right hand side depends only on y over x

Solving it

Substituting $y = vx$ into $\dfrac{dy}{dx} = h(y/x) = h(v)$ gives

v + x\frac{dv}{dx} = h(v) \implies x\frac{dv}{dx} = h(v) - v,

which is separable in $v$ and $x$ . If the equation is more natural with $x$ as a function of $x/y$ , use $x = vy$ , $\dfrac{dx}{dy} = v + y\dfrac{dv}{dy}$ instead — the method is identical with the roles of $x$ and $y$ swapped.

Derivation AReducing a homogeneous equation to a separable one

The goal is to show, starting only from $\dfrac{dy}{dx} = h(y/x)$ and the product-rule fact $\dfrac{d}{dx}(vx) = v + x\dfrac{dv}{dx}$ , that the substitution $y = vx$ always converts a homogeneous first-order equation into a separable equation $\dfrac{dv}{h(v)-v} = \dfrac{dx}{x}$ whose two sides can be integrated independently, with no assumption on the specific form of $h$ .

Step 1. Write $y = vx$ , where $v$ is treated as an unknown function of $x$ (not a constant). Differentiate both sides with respect to $x$ using the product rule: $\dfrac{dy}{dx} = \dfrac{d}{dx}(v\cdot x) = \dfrac{dv}{dx}\cdot x + v\cdot 1 = v + x\dfrac{dv}{dx}$ .

Step 2. Substitute this into the given homogeneous equation $\dfrac{dy}{dx} = h\!\left(\dfrac{y}{x}\right)$ . Since $y/x = v$ by construction, the right side becomes simply $h(v)$ , so $v + x\dfrac{dv}{dx} = h(v)$ .

Step 3. Isolate the derivative term by subtracting $v$ from both sides: $x\dfrac{dv}{dx} = h(v) - v$ . Every $x$ that was tangled inside $h(y/x)$ has vanished — the right side is now a pure function of $v$ alone, and the left side is $x$ times $dv/dx$ .

Step 4. Divide both sides by $x\big(h(v)-v\big)$ , provided $h(v) \neq v$ , to separate the variables completely: $\dfrac{dv}{h(v)-v} = \dfrac{dx}{x}$ . The left side is entirely in $v$ , the right side entirely in $x$ , so each can now be integrated on its own, and back-substituting $v = y/x$ at the end returns the answer to the original variables.

Worked example

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

Step 1. Check homogeneity. Both $x^2+y^2$ and $xy$ are homogeneous of degree $2$ , so dividing top and bottom by $x^2$ gives $\dfrac{dy}{dx} = \dfrac{1+(y/x)^2}{y/x}$ — a function of $y/x$ alone. The equation is homogeneous.

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

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

Step 3. Isolate $x\,dv/dx$ :

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

Step 4. Separate and integrate:

v\,dv = \frac{dx}{x} \implies \int v\,dv = \int \frac{dx}{x} \implies \frac{v^2}{2} = \ln|x| + C_1.

Step 5. Substitute back $v = y/x$ and clear the fraction (multiply by $2x^2$ , write $C = 2C_1$ ):

\frac{y^2}{2x^2} = \ln|x| + C_1 \implies \boxed{y^2 = 2x^2\ln|x| + Cx^2.}

Competitive stretch

A close cousin, $(x^2+y^2)dx - 2xy\,dy = 0$ with $y(1)=1$ , reduces the same way to $\dfrac{2v}{1-v^2}dv = \dfrac{dx}{x}$ , integrating to $x = C(x^2-y^2)$ . Plugging in $x=1,y=1$ gives $1 = C\cdot 0$ , which is impossible — the point $(1,1)$ sits on the singular branch $x^2=y^2$ that got divided away during separation. Checking directly, $y=x$ satisfies the original equation and passes through $(1,1)$ , so that line — not the general family — is the particular solution. Losing a branch during separation and needing to recover it by direct substitution is a favourite twist in board and competitive papers alike.

Common traps

Dropping the product-rule term. $\dfrac{d}{dx}(vx) = v + x\dfrac{dv}{dx}$ , not just $v$ — omitting $x\,dv/dx$ is the most common single error here.
Misjudging homogeneity. Numerator and denominator (or all terms) must be homogeneous of the same degree; $\dfrac{dy}{dx} = \dfrac{x+y}{x^2+y}$ is not homogeneous because the parts have different degrees.
Forgetting to substitute back. After integrating in $v$ , the answer must be converted back to $x,y$ via $v = y/x$ ; leaving $v$ in the final answer loses marks.
Losing a solution branch. Dividing by an expression like $(1-v^2)$ during separation can silently discard a valid solution where that expression is zero — always check the answer against any given initial condition.
Confusing this method with variables-separable. A homogeneous equation is not directly separable in $x$ and $y$ ; the $y=vx$ substitution is what creates the separable equation in $v$ and $x$ — trying to separate $x$ and $y$ straight away, without substituting first, wastes time and rarely works.

Ask the tutor: if the algebra after substituting $y=vx$ gets messy, ask your AI tutor to slow down and show each simplification step for your specific equation.

How do I quickly tell if an equation is homogeneous without dividing everything out?

Check the degree of every term on both sides. In $\dfrac{dy}{dx} = \dfrac{g(x,y)}{f(x,y)}$ , if every term of $g$ has the same total power of $x$ and $y$ combined, and likewise for $f$ , and those two common degrees match, the equation is homogeneous. For example, in $x^2+y^2$ every term has degree $2$ , and in $xy$ the single term also has degree $2$ — matching degrees confirm homogeneity instantly, without doing the division.

Why does the substitution have to be y = vx and not something else?

Because $h(y/x)$ depends on the ratio $y/x$ , naming that ratio $v$ (so $y=vx$ ) turns the messy two-variable dependence into a one-variable function $h(v)$ . Any substitution that captures the ratio would work in principle, but $y=vx$ is the standard, cleanest choice because differentiating it by the product rule directly reproduces the $v + x\,dv/dx$ pattern that cancels the $v$ already sitting in $h(v)$ , leaving a clean separable equation.

What if I get stuck simplifying after substituting y = vx?

This is exactly where algebra errors creep in, usually from expanding $(vx)^2$ or forgetting a bracket. Redo the substitution one piece at a time: replace every $y$ with $vx$ first, simplify powers of $x$ so they cancel with the denominator, and only then subtract $v$ from both sides. If the algebra still looks tangled after that, ask your AI tutor to walk through each simplification step for your specific equation rather than guessing the next line.