1.6 Homogeneous First-Order Equations
Homogeneous first-order differential equations form an elegant class of nonlinear ODEs that can be solved by a clever change of variables. They appear naturally whenever the rate of change depends only on the ratio \(y/x\) rather than on \(x\) and \(y\) separately — a situation common in geometry, fluid flow along radial lines, and physical problems with scale invariance.
Learning Objectives
Standard Form of a Homogeneous Equation
A first-order ODE is called homogeneous if it can be written in either of the equivalent forms:
— or, in differential form —
where \(M\) and \(N\) are homogeneous functions of the same degree \(n\); that is, \(M(tx,ty) = t^n M(x,y)\) and \(N(tx,ty) = t^n N(x,y)\) for every \(t\).
Why homogeneous equations are important:
- Geometric symmetry: The slope at any point depends only on direction from the origin, so solutions exhibit a scaling/dilation symmetry.
- Reducible to separable: A single substitution \(v = y/x\) always converts a homogeneous equation into a separable one — a guaranteed solution path.
- Bridge between methods: They sit between separable and exact equations and often appear in problems where neither of those methods applies directly.
Applications
Pursuit Curves: When one object pursues another with velocity always pointing toward the target, the resulting trajectory satisfies a homogeneous ODE. The substitution \(v = y/x\) reveals the pursuit path explicitly.
Radial Heat Flow: In a long cylindrical body with heat flowing outward, the temperature \(T(r)\) often satisfies an equation depending only on \(T/r\), making it homogeneous.
Mixing with Proportional Rates: When the rate of change of a substance is proportional to a ratio of two scaled quantities (e.g., concentration ratios), the resulting model is homogeneous.
Leonhard Euler systematically studied homogeneous functions in the 1750s, identifying the substitution \(y = vx\) as the key to reducing this class of equations. His insight — that scale-invariant problems should be analyzed in scale-invariant variables — remains a guiding principle in mathematical physics today.
Theory: Homogeneous Functions and the Substitution
Homogeneous Functions of Degree \(n\)
A function \(f(x,y)\) is called homogeneous of degree \(n\) if, for every real number \(t \neq 0\),
Examples:
- \(f(x,y) = x^2 + xy + y^2\) is homogeneous of degree \(2\).
- \(f(x,y) = \sqrt{x^2 + y^2}\) is homogeneous of degree \(1\).
- \(f(x,y) = \dfrac{y}{x}\) is homogeneous of degree \(0\).
- \(f(x,y) = x^2 + y\) is not homogeneous (the terms have different degrees).
Why the Substitution \(v = y/x\) Works
Starting from \(\displaystyle \frac{dy}{dx} = F\!\left(\frac{y}{x}\right)\), let
Differentiating both sides of \(y = vx\) with respect to \(x\) (using the product rule):
Substituting into the original equation:
Solving for \(\dfrac{dv}{dx}\) gives a separable equation in \(v\) and \(x\):
From Differential Form to Standard Form
If the equation is given as \(M(x,y)\,dx + N(x,y)\,dy = 0\), we can rewrite it as:
Because \(M\) and \(N\) are homogeneous of the same degree \(n\), the ratio \(M/N\) is homogeneous of degree \(0\) — meaning it depends only on \(y/x\). So we can write \(-M/N = F(y/x)\), and the standard substitution applies.
Two Equivalent Substitutions
Either substitution works for a homogeneous equation. Choose the one that makes the resulting integral simpler.
Standard: \(v = y/x\), so \(y = vx\) and \(\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}\).
Alternative: \(u = x/y\), so \(x = uy\) and \(\dfrac{dx}{dy} = u + y\dfrac{du}{dy}\). This is preferred when \(F(x/y)\) is simpler than \(F(y/x)\).
Quick Test for Homogeneity
To check whether \(\dfrac{dy}{dx} = G(x,y)\) is homogeneous, replace \(x \to tx\) and \(y \to ty\). If the \(t\)'s cancel completely (i.e., \(G(tx,ty) = G(x,y)\)), then \(G\) is homogeneous of degree \(0\) and the equation is homogeneous.
Step-by-Step Solution Method
Rewrite \(\dfrac{dy}{dx}\) so the right-hand side depends only on \(y/x\). If given in differential form \(M\,dx + N\,dy = 0\), confirm \(M\) and \(N\) are homogeneous of the same degree.
Compute \(\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}\) and substitute into the equation to obtain a relation between \(v\), \(x\), and \(\dfrac{dv}{dx}\).
Rearrange to \(\displaystyle \frac{dv}{F(v) - v} = \frac{dx}{x}\) and integrate both sides. Apply standard separable-equation techniques (partial fractions, substitution, etc.) as needed.
Replace \(v\) in your implicit or explicit solution by \(y/x\). Simplify, and apply any initial conditions to determine the constant of integration.
If you divided by \(F(v) - v\) during separation, the values of \(v\) where \(F(v) - v = 0\) correspond to linear solutions of the form \(y = v_0\,x\). These should be reported separately if they are not captured by the general solution.
Worked Examples
Given equation: \(\displaystyle \frac{dy}{dx} = \frac{x + y}{x} = 1 + \frac{y}{x}\)
Step 1: Verify homogeneity. The right-hand side depends only on \(y/x\), so the equation is homogeneous with \(F(v) = 1 + v\).
Step 2: Substitute \(y = vx\). Then \(\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}\), and the equation becomes: $$v + x\frac{dv}{dx} = 1 + v \quad\Longrightarrow\quad x\frac{dv}{dx} = 1.$$
Step 3: Separate and integrate. $$dv = \frac{dx}{x} \quad\Longrightarrow\quad v = \ln|x| + C.$$
Step 4: Back-substitute \(v = y/x\). $$\frac{y}{x} = \ln|x| + C \quad\Longrightarrow\quad \boxed{y = x\bigl(\ln|x| + C\bigr)}.$$
Verification: \(\dfrac{dy}{dx} = \ln|x| + C + x \cdot \dfrac{1}{x} = \ln|x| + C + 1\). Also \(\dfrac{y}{x} + 1 = \ln|x| + C + 1\). ✓
Given equation: \(\displaystyle \frac{dy}{dx} = \frac{x^2 + y^2}{xy}\)
Step 1: Verify homogeneity. Both numerator and denominator are homogeneous of degree \(2\). Dividing top and bottom by \(x^2\): $$\frac{dy}{dx} = \frac{1 + (y/x)^2}{y/x}.$$ So \(F(v) = \dfrac{1 + v^2}{v}\).
Step 2: Substitute \(y = vx\). $$v + x\frac{dv}{dx} = \frac{1 + v^2}{v} = \frac{1}{v} + v \quad\Longrightarrow\quad x\frac{dv}{dx} = \frac{1}{v}.$$
Step 3: Separate and integrate. $$v\,dv = \frac{dx}{x} \quad\Longrightarrow\quad \frac{v^2}{2} = \ln|x| + C_1 \quad\Longrightarrow\quad v^2 = 2\ln|x| + C.$$
Step 4: Back-substitute. $$\left(\frac{y}{x}\right)^2 = 2\ln|x| + C \quad\Longrightarrow\quad \boxed{y^2 = x^2\bigl(2\ln|x| + C\bigr)}.$$
Verification: Differentiating implicitly: \(2y\,y' = 4x\ln|x| + 2x + 2Cx\), so \(y' = \dfrac{2x\ln|x| + x + Cx}{y}\). Substituting \(y^2 = x^2(2\ln|x| + C)\) into the original right-hand side gives the same expression. ✓
Given equation: \(\displaystyle \frac{dy}{dx} = \frac{y^2 - x^2}{2xy}\)
Step 1: Verify homogeneity. Both \(y^2 - x^2\) and \(2xy\) are homogeneous of degree \(2\). Dividing by \(x^2\): \(F(v) = \dfrac{v^2 - 1}{2v}\).
Step 2: Substitute \(y = vx\). $$v + x\frac{dv}{dx} = \frac{v^2 - 1}{2v} \quad\Longrightarrow\quad x\frac{dv}{dx} = \frac{v^2 - 1}{2v} - v = \frac{v^2 - 1 - 2v^2}{2v} = -\frac{v^2 + 1}{2v}.$$
Step 3: Separate and integrate. $$\frac{2v}{v^2 + 1}\,dv = -\frac{dx}{x} \quad\Longrightarrow\quad \ln(v^2 + 1) = -\ln|x| + C_1.$$ Exponentiating: \(v^2 + 1 = \dfrac{C}{x}\), where \(C = e^{C_1}\) is a new constant.
Step 4: Back-substitute. $$\frac{y^2}{x^2} + 1 = \frac{C}{x} \quad\Longrightarrow\quad y^2 + x^2 = Cx.$$ Completing the square: \(\bigl(x - \tfrac{C}{2}\bigr)^2 + y^2 = \bigl(\tfrac{C}{2}\bigr)^2\).
Solution: \(\boxed{x^2 + y^2 = Cx}\) — a one-parameter family of circles passing through the origin with centers on the \(x\)-axis.
Given equation: \((x^2 + 3y^2)\,dx + 2xy\,dy = 0\)
Step 1: Verify homogeneity. \(M = x^2 + 3y^2\) and \(N = 2xy\) are both homogeneous of degree \(2\). Solving for \(\dfrac{dy}{dx}\): $$\frac{dy}{dx} = -\frac{x^2 + 3y^2}{2xy}.$$ Dividing by \(x^2\): \(F(v) = -\dfrac{1 + 3v^2}{2v}\).
Step 2: Substitute \(y = vx\). $$v + x\frac{dv}{dx} = -\frac{1 + 3v^2}{2v} \quad\Longrightarrow\quad x\frac{dv}{dx} = -\frac{1 + 3v^2}{2v} - v = -\frac{1 + 5v^2}{2v}.$$
Step 3: Separate and integrate. $$\frac{2v}{1 + 5v^2}\,dv = -\frac{dx}{x}.$$ Let \(u = 1 + 5v^2\), so \(du = 10v\,dv\) and \(2v\,dv = \tfrac{1}{5}du\): $$\frac{1}{5}\ln|1 + 5v^2| = -\ln|x| + C_1 \quad\Longrightarrow\quad \ln(1 + 5v^2) = -5\ln|x| + C_2.$$ Exponentiating: \(1 + 5v^2 = \dfrac{C}{x^5}\).
Step 4: Back-substitute. $$1 + \frac{5y^2}{x^2} = \frac{C}{x^5} \quad\Longrightarrow\quad x^2 + 5y^2 = \frac{C}{x^3} \quad\Longrightarrow\quad \boxed{x^3(x^2 + 5y^2) = C}.$$
Verification: Differentiate \(x^5 + 5x^3 y^2 = C\) implicitly: $$5x^4 + 15x^2 y^2 + 10x^3 y\,y' = 0 \quad\Longrightarrow\quad y' = -\frac{5x^4 + 15x^2 y^2}{10x^3 y} = -\frac{x^2 + 3y^2}{2xy}. \checkmark$$
Given equation: \(xy' = y + x\) with \(y(1) = 1\).
Step 1: Standard form. Divide by \(x\): $$\frac{dy}{dx} = \frac{y}{x} + 1 = F\!\left(\frac{y}{x}\right) \text{ with } F(v) = v + 1.$$
Step 2: Substitute \(y = vx\). $$v + x\frac{dv}{dx} = v + 1 \quad\Longrightarrow\quad x\frac{dv}{dx} = 1.$$
Step 3: Separate and integrate. $$dv = \frac{dx}{x} \quad\Longrightarrow\quad v = \ln|x| + C.$$
Step 4: Back-substitute and apply IC. $$\frac{y}{x} = \ln|x| + C \quad\Longrightarrow\quad y = x\bigl(\ln|x| + C\bigr).$$ At \(x = 1\): \(1 = 1 \cdot (0 + C) \Rightarrow C = 1\).
Solution: \(\boxed{y(x) = x\bigl(\ln x + 1\bigr)}\) for \(x > 0\).
Verification: \(y(1) = 1(0 + 1) = 1\). ✓ \(y' = \ln x + 1 + 1 = \ln x + 2\), and \(\dfrac{y}{x} + 1 = \ln x + 1 + 1 = \ln x + 2\). ✓
Given equation: \(\displaystyle \frac{dy}{dx} = \frac{x - y}{x + y}\)
Step 1: Verify homogeneity. Both numerator and denominator are homogeneous of degree \(1\). Dividing by \(x\): $$F(v) = \frac{1 - v}{1 + v}.$$
Step 2: Substitute \(y = vx\). $$v + x\frac{dv}{dx} = \frac{1 - v}{1 + v} \quad\Longrightarrow\quad x\frac{dv}{dx} = \frac{1 - v}{1 + v} - v = \frac{1 - v - v(1+v)}{1 + v} = \frac{1 - 2v - v^2}{1 + v}.$$
Step 3: Separate and integrate. $$\frac{1 + v}{1 - 2v - v^2}\,dv = \frac{dx}{x}.$$ Notice that \(\dfrac{d}{dv}(1 - 2v - v^2) = -2 - 2v = -2(1 + v)\). So the left side becomes: $$-\frac{1}{2}\cdot\frac{d(1 - 2v - v^2)}{1 - 2v - v^2} = \frac{dx}{x}.$$ Integrating: $$-\frac{1}{2}\ln|1 - 2v - v^2| = \ln|x| + C_1 \quad\Longrightarrow\quad \ln|1 - 2v - v^2| = -2\ln|x| + C_2.$$ Exponentiating: \(1 - 2v - v^2 = \dfrac{C}{x^2}\).
Step 4: Back-substitute \(v = y/x\) and multiply by \(x^2\). $$x^2 - 2xy - y^2 = C.$$
Solution: \(\boxed{x^2 - 2xy - y^2 = C}\) (implicit form).
Verification: Differentiating implicitly: \(2x - 2y - 2x\,y' - 2y\,y' = 0\), so \(y' = \dfrac{2x - 2y}{2x + 2y} = \dfrac{x - y}{x + y}\). ✓
Given equation: \((x^2 + y^2)\,dx - xy\,dy = 0\)
Step 1: Standard form. Solve for \(\dfrac{dy}{dx}\): $$\frac{dy}{dx} = \frac{x^2 + y^2}{xy}.$$ This is identical to Example 2! \(F(v) = \dfrac{1 + v^2}{v}\).
Step 2: Substitute \(y = vx\). $$v + x\frac{dv}{dx} = \frac{1 + v^2}{v} \quad\Longrightarrow\quad x\frac{dv}{dx} = \frac{1}{v}.$$
Step 3: Separate and integrate. $$v\,dv = \frac{dx}{x} \quad\Longrightarrow\quad \frac{v^2}{2} = \ln|x| + C_1.$$
Step 4: Back-substitute. $$\frac{y^2}{2x^2} = \ln|x| + C_1 \quad\Longrightarrow\quad \boxed{y^2 = 2x^2\ln|x| + Cx^2}.$$
Given equation: \(\displaystyle \frac{dy}{dx} = \frac{y}{x} + \tan\!\left(\frac{y}{x}\right)\)
Step 1: Verify homogeneity. The right-hand side depends only on \(y/x\), so \(F(v) = v + \tan v\).
Step 2: Substitute \(y = vx\). $$v + x\frac{dv}{dx} = v + \tan v \quad\Longrightarrow\quad x\frac{dv}{dx} = \tan v.$$
Step 3: Separate and integrate. $$\frac{dv}{\tan v} = \frac{dx}{x} \quad\Longrightarrow\quad \int \cot v\,dv = \int \frac{dx}{x}.$$ $$\ln|\sin v| = \ln|x| + C_1 \quad\Longrightarrow\quad \sin v = Cx.$$
Step 4: Back-substitute. $$\boxed{\sin\!\left(\frac{y}{x}\right) = Cx}.$$
Singular solutions: \(\tan v = 0\) gives \(v = k\pi\), i.e., \(y = k\pi x\) for integer \(k\). These straight-line solutions are not captured by the general formula and should be reported separately.
📝 Exam-Style Practice Problems
These problems are similar in style and difficulty to past exam questions. Click each problem to reveal the step-by-step solution.
Given equation: \(\displaystyle \frac{dy}{dx} = \frac{2y - x}{x} = \frac{2y}{x} - 1\)
Step 1: Identify homogeneity. \(F(v) = 2v - 1\).
Step 2: Substitute \(y = vx\). $$v + x\frac{dv}{dx} = 2v - 1 \quad\Longrightarrow\quad x\frac{dv}{dx} = v - 1.$$
Step 3: Separate and integrate. $$\frac{dv}{v - 1} = \frac{dx}{x} \quad\Longrightarrow\quad \ln|v - 1| = \ln|x| + C_1 \quad\Longrightarrow\quad v - 1 = Cx.$$
Step 4: Back-substitute. $$\frac{y}{x} - 1 = Cx \quad\Longrightarrow\quad \boxed{y = x + Cx^2}.$$
Verification: \(y' = 1 + 2Cx\). And \(\dfrac{2y - x}{x} = \dfrac{2x + 2Cx^2 - x}{x} = 1 + 2Cx\). ✓
Given equation: \((x + y)\,dx - x\,dy = 0\)
Step 1: Standard form. $$\frac{dy}{dx} = \frac{x + y}{x} = 1 + \frac{y}{x}.$$ Both \(M = x + y\) and \(N = -x\) are homogeneous of degree \(1\). \(F(v) = 1 + v\).
Step 2: Substitute \(y = vx\). $$v + x\frac{dv}{dx} = 1 + v \quad\Longrightarrow\quad x\frac{dv}{dx} = 1.$$
Step 3: Separate and integrate. $$dv = \frac{dx}{x} \quad\Longrightarrow\quad v = \ln|x| + C.$$
Step 4: Back-substitute. $$\frac{y}{x} = \ln|x| + C \quad\Longrightarrow\quad \boxed{y = x\bigl(\ln|x| + C\bigr)}.$$
Given equation: \(\displaystyle \frac{dy}{dx} = \frac{x + 2y}{2x + y}\) with \(y(1) = 0\).
Step 1: Identify homogeneity. Dividing by \(x\): \(F(v) = \dfrac{1 + 2v}{2 + v}\).
Step 2: Substitute \(y = vx\). $$v + x\frac{dv}{dx} = \frac{1 + 2v}{2 + v}.$$ Compute \(F(v) - v\): $$\frac{1 + 2v}{2 + v} - v = \frac{1 + 2v - v(2 + v)}{2 + v} = \frac{1 - v^2}{2 + v} = \frac{(1-v)(1+v)}{2+v}.$$ So \(\displaystyle x\frac{dv}{dx} = \frac{(1-v)(1+v)}{2+v}\).
Step 3: Separate and integrate. $$\frac{(2 + v)\,dv}{(1-v)(1+v)} = \frac{dx}{x}.$$ Partial fractions: \(\displaystyle \frac{2 + v}{(1-v)(1+v)} = \frac{A}{1-v} + \frac{B}{1+v}\). Multiplying out: \(2 + v = A(1+v) + B(1-v)\). Setting \(v = 1\): \(3 = 2A \Rightarrow A = \tfrac{3}{2}\). Setting \(v = -1\): \(1 = 2B \Rightarrow B = \tfrac{1}{2}\). So: $$\int\!\left[\frac{3/2}{1-v} + \frac{1/2}{1+v}\right]dv = \int \frac{dx}{x}.$$ $$-\frac{3}{2}\ln|1-v| + \frac{1}{2}\ln|1+v| = \ln|x| + C_1.$$ Multiplying by 2 and combining: \(\ln\!\left|\dfrac{1+v}{(1-v)^3}\right| = 2\ln|x| + C_2\), so $$\frac{1 + v}{(1 - v)^3} = K\,x^2.$$
Step 4: Back-substitute and apply IC. With \(v = y/x\): $$\frac{1 + y/x}{(1 - y/x)^3} = K\,x^2 \quad\Longrightarrow\quad \frac{(x + y)/x}{(x - y)^3/x^3} = K\,x^2 \quad\Longrightarrow\quad \frac{x^2(x + y)}{(x - y)^3} = K\,x^2.$$ Simplifying: \(\dfrac{x + y}{(x - y)^3} = K\). Apply \(y(1) = 0\): \(\dfrac{1 + 0}{(1 - 0)^3} = K = 1\).
Solution: \(\boxed{x + y = (x - y)^3}\) (implicit form).
Interactive Slope Field Visualizer
Explore the slope field for various homogeneous equations. Click any point on the plot to trace a solution curve passing through it.
Homogeneous equations teach a profound lesson: changing your perspective — moving from \((x, y)\) to the ratio \(y/x\) — can turn an intractable problem into a simple one. So too in life: a humble shift in viewpoint can illuminate what seemed impossibly complex.
Practice Problems
Self-Assessment Quiz
Question 1: Identify Homogeneous Equations
Which of the following is a first-order homogeneous equation?
Question 2: Standard Substitution
For a first-order homogeneous equation \(\dfrac{dy}{dx} = F(y/x)\), what substitution converts it to a separable equation?
Question 3: Derivative After Substitution
If \(y = vx\) where \(v = v(x)\), what is \(\dfrac{dy}{dx}\)?
Question 4: General Solution
Find the general solution of \(\displaystyle \frac{dy}{dx} = \frac{y}{x} + 1\).
Question 5: Initial Value Problem
Solve \(xy' = y + x\) with \(y(1) = 2\). What is \(y(e)\)?
Question 6: Implicit Solution
The general solution of \(\displaystyle \frac{dy}{dx} = \frac{x^2 + y^2}{xy}\) is:
Quick Reference Card
\(\displaystyle \frac{dy}{dx} = F\!\left(\frac{y}{x}\right)\)
\(M(x,y)\,dx + N(x,y)\,dy = 0\) with \(M, N\) homogeneous of same degree
\(f(tx, ty) = t^n\,f(x, y)\) for all \(t \neq 0\)
\(v = y/x \;\Rightarrow\; y = vx\)
\(\displaystyle \frac{dy}{dx} = v + x\,\frac{dv}{dx}\)
\(\displaystyle \frac{dv}{F(v) - v} = \frac{dx}{x}\)
\(u = x/y\) when \(F(x/y)\) is simpler
Replace \(x \to tx\), \(y \to ty\). If \(t\) cancels, equation is homogeneous.
Roots of \(F(v) - v = 0\) give linear solutions \(y = v_0 x\)
Always back-substitute \(v = y/x\) at the end