What Are Homogeneous Differential Equations?
Homogeneous differential equations are first-order equations whose slope depends only on the ratio $y/x$. Concretely, they can be rearranged into the form
$$\frac{dy}{dx} = F\left(\frac{y}{x}\right),$$
where the whole right side collapses to a single function of $v = y/x$. An equivalent test uses the differential form $M(x,y),dx + N(x,y),dy = 0$: the equation is homogeneous when $M$ and $N$ are homogeneous functions of the same degree. A function $g$ is homogeneous of degree $n$ when scaling both inputs scales the output by $\lambda^n$:
$$g(\lambda x, \lambda y) = \lambda^n, g(x, y).$$
The two tests agree. If $M$ and $N$ share a degree, then $\frac{dy}{dx} = -\frac{M}{N}$ cancels down to a function of $y/x$ alone. This sits inside the wider family of ordinary differential equations, and its order and degree are both one.
A quick word on the name, because it trips up almost everyone: "homogeneous" is used for two different things in this subject, and this article means the ratio sense above, not the zero-right-hand-side sense. The section further down separates them honestly.
How Do You Solve Homogeneous Differential Equations?
Every homogeneous equation of the form $\frac{dy}{dx} = F(y/x)$ becomes solvable through one substitution. Set
$$v = \frac{y}{x} \quad\Longrightarrow\quad y = vx.$$
Because $v$ depends on $x$, differentiate $y = vx$ with the product rule:
$$\frac{dy}{dx} = v + x\frac{dv}{dx}.$$
Substitute both into $\frac{dy}{dx} = F(v)$:
$$v + x\frac{dv}{dx} = F(v) \quad\Longrightarrow\quad x\frac{dv}{dx} = F(v) - v.$$
That last equation is separable: every $v$ can go on one side and every $x$ on the other.
$$\frac{dv}{F(v) - v} = \frac{dx}{x}.$$
Integrate both sides, then replace $v$ with $y/x$ to get the answer in the original variables. The full method for the integration step lives in separable differential equations. The five steps, in order:
Check the equation is homogeneous (same-degree test, or the right side reduces to $F(y/x)$).
Substitute $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Simplify to $x\frac{dv}{dx} = F(v) - v$ and separate the variables.
Integrate both sides, keeping one constant $+C$.
Back-substitute $v = \frac{y}{x}$ to write the general solution in $x$ and $y$.
How Do You Tell If An Equation Is Homogeneous?
The fastest check is the scaling test. Replace $x$ with $\lambda x$ and $y$ with $\lambda y$ everywhere on the right side. If every $\lambda$ cancels and you are left with the original expression, the slope depends only on $y/x$, so the equation is homogeneous.
Table: Quick classification of common first-order equations.
Equation | Right side under $x \to \lambda x,\ y \to \lambda y$ | Homogeneous? |
|---|---|---|
$\frac{dy}{dx} = \frac{x + y}{x}$ | $\frac{\lambda x + \lambda y}{\lambda x} = \frac{x+y}{x}$ | Yes (degree 1 over degree 1) |
$\frac{dy}{dx} = \frac{x^2 + y^2}{xy}$ | $\frac{\lambda^2(x^2+y^2)}{\lambda^2 xy} = \frac{x^2+y^2}{xy}$ | Yes (degree 2 over degree 2) |
$\frac{dy}{dx} = \frac{x + y + 1}{x + y}$ | $\frac{\lambda x + \lambda y + 1}{\lambda x + \lambda y}$ | No (the $+1$ breaks the scaling) |
$\frac{dy}{dx} = x + y$ | $\lambda x + \lambda y = \lambda(x+y)$ | No (degree 1, not degree 0) |
The third and fourth rows are the honest warning: a stray constant or a right side that scales by $\lambda$ rather than staying fixed is not homogeneous in this sense, and the $v = y/x$ substitution will not separate cleanly. Those need other tools, such as the integrating factor method or a shift of variables.
What Are Some Worked Examples Of Homogeneous Differential Equations?
Each example is fully stepped, and every general solution is checked by substituting it back into the original equation.
Example 1: Solve $\frac{dy}{dx} = \frac{x + y}{x}$.
The right side is $1 + \frac{y}{x}$, a function of $y/x$, so the equation is homogeneous. Substitute $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$:
$$v + x\frac{dv}{dx} = 1 + v \quad\Longrightarrow\quad x\frac{dv}{dx} = 1.$$
Separate and integrate:
$$dv = \frac{dx}{x} \quad\Longrightarrow\quad v = \ln|x| + C.$$
Back-substitute $v = \frac{y}{x}$:
$$\frac{y}{x} = \ln|x| + C \quad\Longrightarrow\quad y = x\big(\ln|x| + C\big).$$
Check: differentiate $y = x\ln|x| + Cx$ to get $\frac{dy}{dx} = \ln|x| + 1 + C$. The original right side is $1 + \frac{y}{x} = 1 + \ln|x| + C$, which matches.
Final answer: $y = x\big(\ln|x| + C\big)$.
Example 2: Solve $\frac{dy}{dx} = \frac{x^2 + y^2}{xy}$.
Divide top and bottom by $x^2$ to expose the ratio: the right side becomes $\frac{1 + v^2}{v}$ with $v = y/x$. The equation is homogeneous (degree 2 over degree 2). Substitute:
$$v + x\frac{dv}{dx} = \frac{1 + v^2}{v} = \frac{1}{v} + v \quad\Longrightarrow\quad x\frac{dv}{dx} = \frac{1}{v}.$$
Separate and integrate:
$$v,dv = \frac{dx}{x} \quad\Longrightarrow\quad \frac{v^2}{2} = \ln|x| + C.$$
Back-substitute $v = \frac{y}{x}$ and clear the fraction:
$$\frac{y^2}{2x^2} = \ln|x| + C \quad\Longrightarrow\quad y^2 = x^2\big(2\ln|x| + C_1\big).$$
Check: differentiating $y^2 = x^2(2\ln|x| + C_1)$ implicitly gives $2y\frac{dy}{dx} = 2x(2\ln|x| + C_1) + 2x$, so $y\frac{dy}{dx} = x(2\ln|x| + C_1 + 1)$. Since $x^2 + y^2 = x^2(2\ln|x| + C_1 + 1)$, the right side $\frac{x^2+y^2}{xy}$ equals $\frac{x(2\ln|x|+C_1+1)}{y}$, which matches $\frac{dy}{dx}$.
Final answer: $y^2 = x^2\big(2\ln|x| + C_1\big)$.
Example 3: Solve $\frac{dy}{dx} = \frac{x - y}{x + y}$.
Dividing by $x$ gives $\frac{1 - v}{1 + v}$ with $v = y/x$, so the equation is homogeneous. Substitute:
$$v + x\frac{dv}{dx} = \frac{1 - v}{1 + v} \quad\Longrightarrow\quad x\frac{dv}{dx} = \frac{1 - v}{1 + v} - v = \frac{1 - 2v - v^2}{1 + v}.$$
Separate the variables:
$$\frac{(1 + v),dv}{1 - 2v - v^2} = \frac{dx}{x}.$$
The numerator $1 + v$ is $-\tfrac{1}{2}$ times the derivative of the denominator $1 - 2v - v^2$, so the left side integrates to $-\tfrac{1}{2}\ln|1 - 2v - v^2|$:
$$-\tfrac{1}{2}\ln\big|1 - 2v - v^2\big| = \ln|x| + C.$$
Multiply by $-2$, exponentiate, and write $1 - 2v - v^2 = \frac{C_2}{x^2}$. Back-substitute $v = \frac{y}{x}$ and multiply through by $x^2$:
$$x^2 - 2xy - y^2 = C.$$
Check: differentiating implicitly gives $2x - 2y - 2x\frac{dy}{dx} - 2y\frac{dy}{dx} = 0$, so $\frac{dy}{dx} = \frac{x - y}{x + y}$, the original equation.
Final answer: $x^2 - 2xy - y^2 = C$.
What Is The Difference Between Homogeneous And Homogeneous Linear Equations?
This is the single biggest source of confusion, so here it is plainly. The word "homogeneous" carries two unrelated meanings in differential equations.
The ratio sense (this article). A first-order equation is homogeneous when $\frac{dy}{dx} = F(y/x)$, that is, when its slope depends only on $y/x$. You solve it with the $v = y/x$ substitution. The right side is usually not zero.
The linear sense. A linear differential equation such as $y'' + p(x)y' + q(x)y = g(x)$ is called homogeneous when the forcing term $g(x)$ is zero, giving $y'' + p(x)y' + q(x)y = 0$. Here "homogeneous" means "no source term," and the solution method is completely different.
The two ideas share a name because both come from the notion of a homogeneous function, but they describe different structures. A first-order equation can be homogeneous in the ratio sense while its linear cousin is nonhomogeneous, and vice versa. When a textbook says "homogeneous," read the chapter title: a first-order substitution chapter means the ratio sense, while a linear or constant-coefficient chapter means the zero-right-hand-side sense.
Why Does The Substitution Work?
The substitution is not a trick pulled from nowhere. It works because of what "homogeneous" actually says about the geometry.
The slope is constant along each ray from the origin. If the slope depends only on $y/x$, then along any straight line through the origin (where $y/x$ is fixed) the direction field points the same way. The variable $v = y/x$ names that ray, so switching to $v$ trades two coupled variables for the one quantity the equation genuinely depends on.
A scale-free equation deserves a scale-free variable. Homogeneity means the equation looks identical after zooming in or out about the origin. The ratio $v = y/x$ is exactly the quantity that zooming leaves unchanged, so writing the equation in $v$ removes the redundancy and collapses it to a separable form.
Separation is what "one real variable" looks like. Once the equation truly involves only $v$ and $x$ in independent roles, the variables come apart, and integration finishes the job. The geometry (rays from the origin) and the algebra (a separable equation) are the same fact seen two ways.
That is why the method never fails for a genuinely homogeneous equation: the substitution is matched to the one symmetry the equation has.
Who Discovered Homogeneous Differential Equations?
The substitution that tames these equations is older than most of calculus notation, and it came from one of the two people who invented the subject.
Two more figures shaped the ideas behind the method:
Leonhard Euler (1707–1783, Switzerland) gave homogeneous functions their precise definition and proved the theorem on them ($x g_x + y g_y = n g$) that explains why the same-degree test works.
The Bernoulli family, Leibniz's collaborators in Basel, pushed these first-order substitution methods into the standard toolkit that every course still teaches.
Where Are Homogeneous Differential Equations Used In The Real World?
Because these equations model any process where only a ratio matters, they appear wherever a situation is scale-free or self-similar.
Geometry and orthogonal trajectories: finding a family of curves whose tangent direction depends only on the angle from the origin, such as curves cutting a pencil of lines at a fixed angle, produces a homogeneous equation.
Pursuit problems: a dog running always toward a moving target, or a guided craft steering by bearing alone, gives a slope set by the direction to the target, which is a ratio.
Economics: production functions with constant returns to scale are homogeneous of degree one, and models built on them lead to homogeneous rate equations where only capital-to-labour ratios drive the dynamics.
Fluid flow and heat: self-similar problems, where a profile keeps its shape as it spreads, are reduced to a single ratio variable in the same spirit as $v = y/x$, converting a partial differential equation into an ordinary one.
Population and mixing models: when a growth or concentration rate depends on the proportion between two quantities rather than their absolute sizes, the governing equation is homogeneous.
One structural idea, "only the ratio matters," links map-making geometry, chase problems, and economic growth, which is why the substitution keeps reappearing far from where it was first found.
What Are The Most Common Mistakes With Homogeneous Differential Equations?
These three errors account for most lost marks, and each matches a question real students ask on r/learnmath, r/calculus, and NCERT practice forums.
Confusing the ratio sense with the zero-right-hand-side sense.
Where it slips in:
A student sees "homogeneous," assumes it means the right side is zero, and tries to write down a characteristic equation for $\frac{dy}{dx} = \frac{x+y}{x}$.
Don't do this:
Do not import the linear "$g(x) = 0$" method into a first-order ratio equation. They are different meanings of the same word.
The correct way:
Confirm which sense applies. If the slope reduces to $F(y/x)$, use the $v = y/x$ substitution; the zero-right-hand-side method belongs to linear equations only.
Substituting without checking homogeneity first.
Where it slips in:
A student jumps straight to $y = vx$ on $\frac{dy}{dx} = \frac{x + y + 1}{x + y}$, then cannot separate the variables because of the leftover constant.
Don't do this:
Do not assume every messy first-order equation is homogeneous. The stray $+1$ makes the right side scale-dependent.
The correct way:
Run the scaling test $x \to \lambda x,\ y \to \lambda y$ first. If a $\lambda$ survives, the equation is not homogeneous and needs a shift of variables or an exact differential equations check instead.
Forgetting to back-substitute $v = y/x$.
Where it slips in:
A student integrates correctly to get $\frac{v^2}{2} = \ln|x| + C$ and stops, leaving the answer in $v$.
Don't do this:
Do not report a solution still written in the helper variable $v$. The original problem is in $x$ and $y$.
The correct way:
Replace $v$ with $\frac{y}{x}$ as the final step, then simplify, so $\frac{v^2}{2} = \ln|x| + C$ becomes $y^2 = x^2(2\ln|x| + C_1)$.
Practice Problems On Homogeneous Differential Equations
Work each one, then check against the verified answer.
Solve $\frac{dy}{dx} = \frac{x + 2y}{x}$.
(Answer: $y = C x^2 - x$. Check: $\frac{dy}{dx} = 2Cx - 1 = 1 + \frac{2y}{x}$.)Solve $\frac{dy}{dx} = \frac{y}{x}$.
(Answer: $y = Cx$, the straight lines through the origin.)Solve $\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$. (Answer: $x^2 - y^2 = Cx$.)
Which substitution reduces $\frac{dy}{dx} = F(y/x)$ to a separable equation?
(Answer: $y = vx$, giving $\frac{dy}{dx} = v + x\frac{dv}{dx}$.)Is $\frac{dy}{dx} = \frac{2x + 3y}{x - y}$ homogeneous?
(Answer: Yes, numerator and denominator are both degree 1, so it reduces to $\frac{2 + 3v}{1 - v}$.)Is $\frac{dy}{dx} = \frac{x + y + 1}{x + y}$ homogeneous in the ratio sense?
(Answer: No; the constant $1$ breaks the same-degree scaling, so use the shift $z = x + y$.)
Where Should You Go Next After Homogeneous Differential Equations?
Homogeneous equations are one branch of a larger family, and several natural doors open from here.
Separable differential equations. The engine the substitution feeds into, and the integration skills every step above relies on.
Linear differential equations. The other meaning of "homogeneous," plus the integrating-factor method for first-order linear equations.
Initial value problems. Pin down the constant $C$ using a given point, turning a general solution into a specific one.
If your child is meeting homogeneous differential equations for the first time, a live Bhanzu trainer teaches them from the ratio picture up, so the substitution feels inevitable rather than memorised, in the Bhanzu math program.
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