Homogeneous Differential Equations
Introduction
A Homogeneous Differential Equation is a special type of first-order DE that can be solved using a specific substitution technique. These DEs have a unique property involving the degrees of terms.
Homogeneous Function
Definition: A function $f(x, y)$ is called homogeneous of degree $n$ if:
$$\boxed{f(\lambda x, \lambda y) = \lambda^n f(x, y)}$$
for all $\lambda \neq 0$.
Examples of Homogeneous Functions
Example 1: $f(x, y) = x^2 + xy + y^2$
Check: $f(\lambda x, \lambda y) = (\lambda x)^2 + (\lambda x)(\lambda y) + (\lambda y)^2$
$= \lambda^2 x^2 + \lambda^2 xy + \lambda^2 y^2 = \lambda^2(x^2 + xy + y^2) = \lambda^2 f(x, y)$
Homogeneous of degree 2 ✓
Example 2: $f(x, y) = \frac{x^3 - y^3}{x^2 + y^2}$
Numerator: $(\lambda x)^3 - (\lambda y)^3 = \lambda^3(x^3 - y^3)$ (degree 3)
Denominator: $(\lambda x)^2 + (\lambda y)^2 = \lambda^2(x^2 + y^2)$ (degree 2)
$f(\lambda x, \lambda y) = \frac{\lambda^3(x^3 - y^3)}{\lambda^2(x^2 + y^2)} = \lambda \cdot \frac{x^3 - y^3}{x^2 + y^2} = \lambda f(x, y)$
Homogeneous of degree 1 ✓
Example 3: $f(x, y) = x + y + 1$ (NOT homogeneous)
$f(\lambda x, \lambda y) = \lambda x + \lambda y + 1 \neq \lambda^n(x + y + 1)$
Not homogeneous ✗ (constant term breaks homogeneity)
Interactive Demo: Visualize Homogeneous DE Solutions
Plot solution curves for homogeneous differential equations.
Homogeneous Differential Equation
Definition: A differential equation of the form:
$$\boxed{\frac{dy}{dx} = \frac{f(x, y)}{g(x, y)}}$$is called homogeneous if both $f(x, y)$ and $g(x, y)$ are homogeneous functions of the same degree.
Alternative Form
A homogeneous DE can always be written as:
$$\boxed{\frac{dy}{dx} = F\left(\frac{y}{x}\right)}$$
where $F$ is some function of the ratio $\frac{y}{x}$.
Standard Forms
Form 1: Explicit Ratio
$$\boxed{\frac{dy}{dx} = f\left(\frac{y}{x}\right)}$$Form 2: Fraction of Homogeneous Functions
$$\boxed{\frac{dy}{dx} = \frac{ax + by}{cx + dy}}$$(degree 1 in both numerator and denominator)
Form 3: General Homogeneous
$$\boxed{(ax + by)dx + (cx + dy)dy = 0}$$where degrees match
Solution Method
Substitution Technique
Key Substitution:
$$\boxed{y = vx \quad \text{or} \quad v = \frac{y}{x}}$$where $v$ is a new variable.
Step-by-Step Procedure
Step 1: Verify the DE is homogeneous
Step 2: Substitute $y = vx$
Step 3: Differentiate: $\frac{dy}{dx} = v + x\frac{dv}{dx}$
Step 4: Substitute in the original DE
Step 5: Simplify to get a variable separable DE in $v$ and $x$
Step 6: Solve for $v$ using variable separable method
Step 7: Replace $v = \frac{y}{x}$ to get solution in terms of $x$ and $y$
Memory Tricks
🎯 Homogeneity Check
“All terms same degree”:
- Count powers in each term
- All terms should have same total degree
- Or check: Can you write as function of $\frac{y}{x}$?
Quick Test: Replace $x$ with $\lambda x$ and $y$ with $\lambda y$
- If you can factor out $\lambda^n$, it’s homogeneous of degree $n$
🎯 Substitution Memory
“V for Variable, Y over X”:
- $v = \frac{y}{x}$ (y over x)
- So $y = vx$
- Derivative: $\frac{dy}{dx} = v + x\frac{dv}{dx}$ ("V plus X times dv/dx")
🎯 After Substitution
“Homogeneous → Separable”:
- Homogeneous DE + Substitution = Variable Separable DE
- Then use standard separation technique
Common Mistakes to Avoid
❌ Mistake 1: Wrong Homogeneity Check
Wrong: $\frac{dy}{dx} = \frac{x^2 + y}{x}$ is homogeneous ✗
Correct: Numerator has degree 2 and 1 (not same), denominator has degree 1. Not all terms have same degree → Not homogeneous ✓
❌ Mistake 2: Incorrect Substitution Derivative
Wrong: If $y = vx$, then $\frac{dy}{dx} = v$ ✗
Correct: $\frac{dy}{dx} = \frac{d}{dx}(vx) = v + x\frac{dv}{dx}$ (product rule!) ✓
❌ Mistake 3: Forgetting to Replace $v$
Wrong: Final answer in terms of $v$ and $x$ ✗
Correct: Replace $v = \frac{y}{x}$ to get answer in $x$ and $y$ ✓
❌ Mistake 4: Wrong Degree Count
Wrong: In $x^2y + xy^2$, degree = 2 + 2 = 4 ✗
Correct: In $x^2y$, degree = 2 + 1 = 3; in $xy^2$, degree = 1 + 2 = 3 ✓ (Add powers within each term, don’t add across terms)
Solved Examples
Example 1: Basic Homogeneous (JEE Main)
Solve: $\frac{dy}{dx} = \frac{y}{x}$
Solution:
Check homogeneity: $\frac{dy}{dx} = \frac{y}{x}$ (degree 1 / degree 1) ✓ Homogeneous
Substitute $y = vx$:
$$\frac{dy}{dx} = v + x\frac{dv}{dx}$$Original DE becomes:
$$v + x\frac{dv}{dx} = \frac{vx}{x} = v$$ $$x\frac{dv}{dx} = 0$$ $$\frac{dv}{dx} = 0$$ $$v = C$$Replace $v = \frac{y}{x}$:
$$\boxed{y = Cx}$$Example 2: Standard Form (JEE Main)
Solve: $\frac{dy}{dx} = \frac{x + y}{x}$
Solution:
Rewrite: $\frac{dy}{dx} = 1 + \frac{y}{x}$
This is of the form $F(y/x)$ → Homogeneous ✓
Substitute $y = vx$:
$$v + x\frac{dv}{dx} = 1 + v$$ $$x\frac{dv}{dx} = 1$$ $$dv = \frac{dx}{x}$$Integrate:
$$v = \ln|x| + C$$Replace $v = \frac{y}{x}$:
$$\frac{y}{x} = \ln|x| + C$$ $$\boxed{y = x\ln|x| + Cx}$$Example 3: With Initial Condition (JEE Main)
Solve: $(x^2 + y^2)dx = 2xy dy$ with $y(1) = 0$
Solution:
Rewrite:
$$\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$$Check: Numerator degree = 2, denominator degree = 2 → Homogeneous ✓
Substitute $y = vx$:
$$v + x\frac{dv}{dx} = \frac{x^2 + v^2x^2}{2x \cdot vx} = \frac{x^2(1 + v^2)}{2vx^2} = \frac{1 + v^2}{2v}$$ $$x\frac{dv}{dx} = \frac{1 + v^2}{2v} - v = \frac{1 + v^2 - 2v^2}{2v} = \frac{1 - v^2}{2v}$$Separate variables:
$$\frac{2v dv}{1 - v^2} = \frac{dx}{x}$$Integrate:
$$-\ln|1 - v^2| = \ln|x| + C_1$$ $$\ln|1 - v^2| = -\ln|x| + C$$ $$\ln|1 - v^2| + \ln|x| = C$$ $$\ln|x(1 - v^2)| = C$$ $$x(1 - v^2) = A$$where $A = \pm e^C$
Replace $v = \frac{y}{x}$:
$$x\left(1 - \frac{y^2}{x^2}\right) = A$$ $$x - \frac{y^2}{x} = A$$ $$x^2 - y^2 = Ax$$Using $y(1) = 0$:
$$1 - 0 = A(1)$$, so $A = 1$
Solution: $\boxed{x^2 - y^2 = x}$
Example 4: Trigonometric Form (JEE Advanced)
Solve: $x\frac{dy}{dx} = y + x\sin\left(\frac{y}{x}\right)$
Solution:
Rewrite:
$$\frac{dy}{dx} = \frac{y}{x} + \sin\left(\frac{y}{x}\right) = F\left(\frac{y}{x}\right)$$Homogeneous ✓
Substitute $y = vx$:
$$v + x\frac{dv}{dx} = v + \sin v$$ $$x\frac{dv}{dx} = \sin v$$Separate:
$$\frac{dv}{\sin v} = \frac{dx}{x}$$ $$\csc v \, dv = \frac{dx}{x}$$Integrate:
$$\int \csc v \, dv = \int \frac{dx}{x}$$ $$\ln|\csc v - \cot v| = \ln|x| + C_1$$ $$\ln|\csc v - \cot v| - \ln|x| = C$$ $$\ln\left|\frac{\csc v - \cot v}{x}\right| = C$$Replace $v = \frac{y}{x}$:
$$\boxed{\ln\left|\frac{\csc(y/x) - \cot(y/x)}{x}\right| = C}$$or in simplified form:
$$\boxed{\csc\left(\frac{y}{x}\right) - \cot\left(\frac{y}{x}\right) = Cx}$$Example 5: Disguised Homogeneous (JEE Advanced)
Solve: $(x - y)dy = (x + y)dx$
Solution:
Rewrite:
$$\frac{dy}{dx} = \frac{x + y}{x - y}$$Check: Numerator degree = 1, denominator degree = 1 → Homogeneous ✓
Substitute $y = vx$:
$$v + x\frac{dv}{dx} = \frac{x + vx}{x - vx} = \frac{x(1 + v)}{x(1 - v)} = \frac{1 + v}{1 - v}$$ $$x\frac{dv}{dx} = \frac{1 + v}{1 - v} - v = \frac{1 + v - v(1 - v)}{1 - v} = \frac{1 + v - v + v^2}{1 - v} = \frac{1 + v^2}{1 - v}$$Separate:
$$\frac{(1 - v)dv}{1 + v^2} = \frac{dx}{x}$$ $$\frac{dv}{1 + v^2} - \frac{v \, dv}{1 + v^2} = \frac{dx}{x}$$Integrate:
$$\tan^{-1}v - \frac{1}{2}\ln(1 + v^2) = \ln|x| + C$$Replace $v = \frac{y}{x}$:
$$\boxed{\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2}\ln\left(1 + \frac{y^2}{x^2}\right) = \ln|x| + C}$$Simplify:
$$\boxed{\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2}\ln(x^2 + y^2) + \ln|x| = C}$$Practice Problems
Level 1: JEE Main Basics
Problem 1.1: Solve $x\frac{dy}{dx} = y$.
Solution
$\frac{dy}{dx} = \frac{y}{x}$ (Homogeneous)
$y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}$
$v + x\frac{dv}{dx} = v$
$x\frac{dv}{dx} = 0 \Rightarrow v = C$
$\frac{y}{x} = C$
Answer: $y = Cx$
Problem 1.2: Solve $\frac{dy}{dx} = \frac{x - y}{x + y}$.
Solution
Homogeneous (both degree 1)
$y = vx$: $v + x\frac{dv}{dx} = \frac{x - vx}{x + vx} = \frac{1 - v}{1 + v}$
$x\frac{dv}{dx} = \frac{1 - v}{1 + v} - v = \frac{1 - v - v(1 + v)}{1 + v} = \frac{1 - v - v - v^2}{1 + v} = \frac{1 - 2v - v^2}{1 + v}$
$\frac{(1 + v)dv}{1 - 2v - v^2} = \frac{dx}{x}$
Wait, let me recalculate: $\frac{1 - v}{1 + v} - v = \frac{1 - v - v - v^2}{1 + v} = \frac{1 - 2v - v^2}{1 + v}$
Note: $1 - 2v - v^2 = -(v^2 + 2v - 1)$
$\frac{(1 + v)dv}{-(v^2 + 2v - 1)} = \frac{dx}{x}$
This requires partial fractions or completing square. For JEE Main, numerical answer expected.
Problem 1.3: Check if $\frac{dy}{dx} = \frac{x^2 + y^2}{xy}$ is homogeneous.
Solution
Numerator: $x^2 + y^2$ has degree 2
Denominator: $xy$ has degree 2
Both same degree → Homogeneous ✓
Level 2: JEE Main Advanced
Problem 2.1: Solve $x dy - y dx = \sqrt{x^2 + y^2} dx$.
Solution
$x dy = (y + \sqrt{x^2 + y^2})dx$
$\frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x}$
Divide by $x$: $\frac{dy}{dx} = \frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}}$
Let $y = vx$: $v + x\frac{dv}{dx} = v + \sqrt{1 + v^2}$
$x\frac{dv}{dx} = \sqrt{1 + v^2}$
$\frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}$
$\sinh^{-1}(v) = \ln|x| + C$ or $\ln(v + \sqrt{1 + v^2}) = \ln|x| + C$
$v + \sqrt{1 + v^2} = Ax$
$\frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} = Ax$
$y + \sqrt{x^2 + y^2} = Ax^2$
Problem 2.2: Solve $(x^2 - y^2)dx + 2xy dy = 0$ with $y(1) = 1$.
Solution
$\frac{dy}{dx} = -\frac{x^2 - y^2}{2xy}$
Homogeneous (degree 2/2)
$y = vx$: $v + x\frac{dv}{dx} = -\frac{x^2 - v^2x^2}{2x \cdot vx} = -\frac{1 - v^2}{2v}$
$x\frac{dv}{dx} = -\frac{1 - v^2}{2v} - v = -\frac{1 - v^2 + 2v^2}{2v} = -\frac{1 + v^2}{2v}$
$\frac{2v \, dv}{1 + v^2} = -\frac{dx}{x}$
$\ln(1 + v^2) = -\ln|x| + C$
$\ln(1 + v^2) + \ln|x| = C$
$(1 + v^2)x = A$
$x + xv^2 = A$
$x + \frac{y^2}{x} = A$
$x^2 + y^2 = Ax$
Using $y(1) = 1$: $1 + 1 = A \Rightarrow A = 2$
Answer: $x^2 + y^2 = 2x$
Problem 2.3: Solve $\frac{dy}{dx} = \frac{y}{x} + \tan\left(\frac{y}{x}\right)$.
Solution
Homogeneous (function of $y/x$)
$y = vx$: $v + x\frac{dv}{dx} = v + \tan v$
$x\frac{dv}{dx} = \tan v$
$\frac{dv}{\tan v} = \frac{dx}{x}$
$\cot v \, dv = \frac{dx}{x}$
$\ln|\sin v| = \ln|x| + C$
$\sin v = Ax$
$\sin\left(\frac{y}{x}\right) = Ax$
Level 3: JEE Advanced
Problem 3.1: Solve $x\frac{dy}{dx} = y - x\cos^2\left(\frac{y}{x}\right)$.
Solution
$\frac{dy}{dx} = \frac{y}{x} - \cos^2\left(\frac{y}{x}\right)$
Homogeneous
$y = vx$: $v + x\frac{dv}{dx} = v - \cos^2 v$
$x\frac{dv}{dx} = -\cos^2 v$
$\frac{dv}{\cos^2 v} = -\frac{dx}{x}$
$\sec^2 v \, dv = -\frac{dx}{x}$
$\tan v = -\ln|x| + C$
$\tan\left(\frac{y}{x}\right) = -\ln|x| + C$
Problem 3.2: Solve $2xy dy = (x^2 + 3y^2)dx$ and find particular solution through $(1, 1)$.
Solution
$\frac{dy}{dx} = \frac{x^2 + 3y^2}{2xy}$
Homogeneous
$y = vx$: $v + x\frac{dv}{dx} = \frac{x^2 + 3v^2x^2}{2x \cdot vx} = \frac{1 + 3v^2}{2v}$
$x\frac{dv}{dx} = \frac{1 + 3v^2}{2v} - v = \frac{1 + 3v^2 - 2v^2}{2v} = \frac{1 + v^2}{2v}$
$\frac{2v \, dv}{1 + v^2} = \frac{dx}{x}$
$\ln(1 + v^2) = \ln|x| + C$
$(1 + v^2) = Ax$
$1 + \frac{y^2}{x^2} = Ax$
$x^2 + y^2 = Ax^3$
Using $(1, 1)$: $1 + 1 = A \Rightarrow A = 2$
Answer: $x^2 + y^2 = 2x^3$
Problem 3.3: Find the orthogonal trajectories of the family of curves $xy = c$.
Solution
Given family: $xy = c$
Differentiate: $y + x\frac{dy}{dx} = 0$
So: $\frac{dy}{dx} = -\frac{y}{x}$
For orthogonal trajectories, slope is negative reciprocal: $\frac{dy}{dx} = \frac{x}{y}$
This is separable: $y \, dy = x \, dx$
$\frac{y^2}{2} = \frac{x^2}{2} + C$
$y^2 - x^2 = K$
Orthogonal trajectories: Family of hyperbolas $y^2 - x^2 = K$
Recognition Checklist
A DE is homogeneous if:
- Can be written as $\frac{dy}{dx} = F(y/x)$
- Both numerator and denominator have same degree
- Replacing $x$ with $\lambda x$ and $y$ with $\lambda y$ gives $\lambda^n$ factor
- All terms have same total degree in $x$ and $y$
Solution Strategy Flowchart
Is DE homogeneous?
↓
Yes → Substitute y = vx
↓
Differentiate: dy/dx = v + x(dv/dx)
↓
Substitute in original DE
↓
Simplify → Get variable separable in v and x
↓
Solve using separation method
↓
Replace v = y/x
↓
Final solution
Cross-References
- Variable Separable: Core technique used after substitution → Variable Separable
- Linear DE: Alternative method for some DEs → Linear DE
- Integration: Required for solving → Integration
Quick Revision Points
- Homogeneous if same degree in numerator and denominator
- Or if can write as $F(y/x)$
- Substitute $y = vx$ always
- Remember: $\frac{dy}{dx} = v + x\frac{dv}{dx}$ (product rule!)
- After substitution → Variable separable
- Always replace $v = y/x$ at the end
Last updated: November 15, 2025