MathsCBSE Class 12 · Differential Equations

Homogeneous Differential Equations and the y = vx Substitution (Class 12)

Updated 15 October 2026 · Fact-checked

A first-order equation dy/dx = F(x, y) is homogeneous if F(λx, λy) = F(x, y), so F depends only on y/x. Put y = vx; then dy/dx = v + x dv/dx, the equation becomes separable in v and x, and after integrating you replace v by y/x.

Follow the arrows: put y = vx, separate v and x, integrate, then replace v by y/x; a constant term spoils homogeneity.

Illustration drawn for QuestionBuddy.

Understand Homogeneous Differential Equations

Homogeneous functions. A function $f (x, y)$ is homogeneous of degree $n$ if $f (λ x, λ y) = λ^{n} f (x, y)$ for every non-zero $λ$ . For example $x^{2} + x y$ has degree 2, and $\frac{x ^{2} + y ^{2}}{x y}$ has degree 0. A degree-0 function depends only on the ratio $\frac{y}{x}$ : $\frac{x ^{2} + y ^{2}}{x y} = \frac{1 + ( y / x ) ^{2}}{y / x}$ .

Homogeneous differential equations. The equation $\frac{d y}{d x} = F (x, y)$ is homogeneous when $F$ is homogeneous of degree 0, i.e. $F (x, y) = g (\frac{y}{x})$ . A quick test: if $\frac{d y}{d x} = \frac{P ( x , y )}{Q ( x , y )}$ and $P$ and $Q$ are homogeneous of the same degree, the equation is homogeneous. A constant term (such as $x + y + 1$ ) breaks homogeneity.

**The substitution $y = v x$ .** Since $F$ depends only on $v = \frac{y}{x}$ :

1. Put $y = v x$ , so $\frac{d y}{d x} = v + x \frac{d v}{d x}$ (product rule).
2. The equation becomes $v + x \frac{d v}{d x} = g (v)$ , i.e. $x \frac{d v}{d x} = g (v) - v$ .
3. Separate: $\frac{d v}{g ( v ) - v} = \frac{d x}{x}$ , and integrate both sides.
4. Replace $v$ by $\frac{y}{x}$ and simplify; use any condition for the particular solution.

The integral in step 3 may need partial fractions, a log, or an inverse trigonometric form, so revise those methods. Equations with trigonometric or exponential functions of $\frac{y}{x}$ , such as $\frac{d y}{d x} = \frac{y}{x} + sec \frac{y}{x}$ , simplify especially nicely because the $v$ terms cancel.

The other form. If the equation is more naturally written as $\frac{d x}{d y} = G (\frac{x}{y})$ , put $x = v y$ and $\frac{d x}{d y} = v + y \frac{d v}{d y}$ instead.

A geometric example. The equation $\frac{d y}{d x} = \frac{y ^{2} - x ^{2}}{2 x y}$ has solutions $x^{2} + y^{2} = C x$ : circles through the origin with centres on the x-axis. Through $(1, 1)$ the solution is $x^{2} + y^{2} = 2 x$ .

The syllabus restricts this to homogeneous equations of first order and first degree (NCERT Section 9.4.2). Homogeneous equations have been the most frequently asked differential-equation type in recent board papers, usually for 3 to 5 marks.

Homogeneous Differential Equations: formulas to remember

Homogeneous function of degree n: $f (λ x, λ y) = λ^{n} f (x, y)$; Test every term's total degree.
Homogeneous equation: $\frac{d y}{d x} = F (x, y), F (λ x, λ y) = F (x, y) ⟺ F = g (\frac{y}{x})$; F is homogeneous of degree 0.
Substitution: $y = v x, \frac{d y}{d x} = v + x \frac{d v}{d x}$; Do not forget the $x \frac{d v}{d x}$ term.
Separated form: $\int \frac{d v}{g ( v ) - v} = \int \frac{d x}{x} = lo g ∣ x ∣ + C$; Then replace v by $\frac{y}{x}$ .
Alternative form: $\frac{d x}{d y} = G (\frac{x}{y}) : x = v y, \frac{d x}{d y} = v + y \frac{d v}{d y}$; Use when x/y appears naturally.
Quick test for a ratio: $\frac{d y}{d x} = \frac{P ( x , y )}{Q ( x , y )}, de g P = de g Q (both homogeneous)$; Then the equation is homogeneous.

How to solve Homogeneous Differential Equations questions

The standard 3- to 5-mark layout for a homogeneous differential equation.

1Write the equation as $\frac{d y}{d x} = F (x, y)$ and show $F (λ x, λ y) = F (x, y)$ (or that P and Q have the same degree).
2Put $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$ ; simplify $F$ to a function of $v$ .
3Rearrange to $x \frac{d v}{d x} = g (v) - v$ and separate the variables.
4Integrate both sides (use partial fractions or standard integrals if needed); add one constant.
5Replace $v$ by $\frac{y}{x}$ , simplify to a neat form, and apply the given condition for a particular solution.

Homogeneous Differential Equations in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Solving a homogeneous equation for a design curve

Suppose a lamp designer wants a curve for a shade whose slope at every point $(x, y)$ is $\frac{x + y}{x}$ , and the curve must pass through $(1, 1)$ (in decimetres).

Find the equation of the curve and its height at $x = 2$ .

Try it first, then show the solutionSolution

$\frac{d y}{d x} = 1 + \frac{y}{x}$ is homogeneous. Put $y = v x$ : $v + x \frac{d v}{d x} = 1 + v$ .
$x \frac{d v}{d x} = 1 \Rightarrow d v = \frac{d x}{x} \Rightarrow v = lo g x + C$ , so $y = x lo g x + C x$ .
Through $(1, 1)$ : $1 = 0 + C$ , so $y = x lo g x + x$ .
At $x = 2$ : $y = 2 lo g 2 + 2 \approx 3.39$ dm.

Answer: $y = x lo g x + x$ ; height about 3.39 dm at $x = 2$ .

Memory hook: Slope depending only on y/x: put y = vx.

A particular solution through a point

Imagine a stream line in a model of fluid flow whose slope at $(x, y)$ is $\frac{y}{x} + \frac{x}{y}$ , and which passes through $(1, 1)$ .

Find the equation of the stream line.

Try it first, then show the solutionSolution

Put $y = v x$ : $v + x \frac{d v}{d x} = v + \frac{1}{v}$ , so $v d v = \frac{d x}{x}$ .
Integrate: $\frac{v ^{2}}{2} = lo g ∣ x ∣ + C$ , i.e. $y^{2} = 2 x^{2} lo g ∣ x ∣ + 2 C x^{2}$ .
Through $(1, 1)$ : $1 = 2 C$ , so $y^{2} = 2 x^{2} lo g x + x^{2}$ .

Answer: $y^{2} = 2 x^{2} lo g x + x^{2}$

Memory hook: When the v's cancel, separation is immediate.

Tricks to remember Homogeneous Differential Equations

Short memory aids. Say each one aloud once, then use it on a question the same day.

Same degree top and bottom
If every term on top and every term at the bottom has the same total degree, the equation is homogeneous.
Use it when: Quick test in MCQs.
v plus x dv by dx
Chant it: 'y equals vx, dy by dx equals v plus x dv by dx'.
Use it when: Never dropping the extra term.
Subtract v, then separate
From $v + x v^{'} = g (v)$ , move $v$ across: $x v^{'} = g (v) - v$ , then separate.
Use it when: Every homogeneous question.
Back to y/x at the end
The answer must be in $x$ and $y$ ; replace every $v$ by $\frac{y}{x}$ .
Use it when: Final mark.

Homogeneous Differential Equations: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: A first example · 3 marks · Short answerBoard level

Solve $\frac{d y}{d x} = \frac{x + y}{x}$ .

Try it first, then show the solutionSolution

Homogeneous: numerator and denominator have degree 1. Put $y = v x$ : $v + x \frac{d v}{d x} = 1 + v$ . (1 mark)
$x \frac{d v}{d x} = 1 \Rightarrow d v = \frac{d x}{x} \Rightarrow v = lo g ∣ x ∣ + C$ . (1 mark)
$\frac{y}{x} = lo g ∣ x ∣ + C$ , i.e. $y = x lo g ∣ x ∣ + C x$ . (1 mark)

Answer: $y = x lo g ∣ x ∣ + C x$

Example 2: Partial fractions in v · 3 marks · Short answerBoard level

Solve $x^{2} \frac{d y}{d x} = y^{2} + 2 x y$ .

Try it first, then show the solutionSolution

$\frac{d y}{d x} = \frac{y ^{2} + 2 x y}{x ^{2}}$ is homogeneous. With $y = v x$ : $v + x v^{'} = v^{2} + 2 v$ , so $x v^{'} = v^{2} + v$ . (1 mark)
$\frac{d v}{v ( v + 1 )} = \frac{d x}{x}$ , and $\frac{1}{v ( v + 1 )} = \frac{1}{v} - \frac{1}{v + 1}$ : $lo g \frac{v}{v + 1} = lo g ∣ x ∣ + lo g ∣ C ∣$ . (1 mark)
$\frac{v}{v + 1} = C x \Rightarrow \frac{y}{y + x} = C x$ . (1 mark)

Answer: $\frac{y}{x + y} = C x$

Example 3: A trigonometric function of y/x · 5 marks · Long answerHarder board level

Find the particular solution of $\frac{d y}{d x} = \frac{y}{x} + sec \frac{y}{x}$ , given $y = \frac{π}{6}$ when $x = 1$ .

Try it first, then show the solutionSolution

The right side is a function of $\frac{y}{x}$ , so the equation is homogeneous. Put $y = v x$ . (1 mark)
$v + x \frac{d v}{d x} = v + sec v \Rightarrow x \frac{d v}{d x} = sec v$ . (1 mark)
Separate: $cos v d v = \frac{d x}{x}$ ; integrate: $sin v = lo g ∣ x ∣ + C$ . (1 mark)
So $sin \frac{y}{x} = lo g ∣ x ∣ + C$ . At $(1, \frac{π}{6})$ : $\frac{1}{2} = 0 + C$ . (1 mark)
Particular solution: $sin \frac{y}{x} = lo g ∣ x ∣ + \frac{1}{2}$ . (1 mark)

Answer: $sin \frac{y}{x} = lo g ∣ x ∣ + \frac{1}{2}$

Example 4: A log of a quadratic in v · 3 marks · Short answerHarder board level

Solve $\frac{d y}{d x} = \frac{x - y}{x + y}$ .

Try it first, then show the solutionSolution

With $y = v x$ : $v + x v^{'} = \frac{1 - v}{1 + v}$ , so $x v^{'} = \frac{1 - 2 v - v ^{2}}{1 + v}$ . (1 mark)
$\frac{( 1 + v ) d v}{1 - 2 v - v ^{2}} = \frac{d x}{x}$ . Since $\frac{d}{d v} (1 - 2 v - v^{2}) = - 2 (1 + v)$ : $- \frac{1}{2} lo g ∣1 - 2 v - v^{2} ∣ = lo g ∣ x ∣ + C$ . (1 mark)
So $x^{2} (1 - 2 v - v^{2}) = A$ , i.e. $x^{2} - 2 x y - y^{2} = A$ . (1 mark)

Answer: $x^{2} - 2 x y - y^{2} = C$

Example 5: Solution curves that are circles · 5 marks · Long answerHOTS / competency

Find the curve through $(1, 1)$ satisfying $\frac{d y}{d x} = \frac{y ^{2} - x ^{2}}{2 x y}$ , and describe it.

Try it first, then show the solutionSolution

Homogeneous (degree 2 over degree 2). With $y = v x$ : $v + x v^{'} = \frac{v ^{2} - 1}{2 v}$ . (1 mark)
$x v^{'} = \frac{v ^{2} - 1 - 2 v ^{2}}{2 v} = - \frac{1 + v ^{2}}{2 v}$ , so $\frac{2 v d v}{1 + v ^{2}} = - \frac{d x}{x}$ . (1 mark)
$lo g (1 + v^{2}) = - lo g ∣ x ∣ + lo g C \Rightarrow x (1 + v^{2}) = C \Rightarrow x^{2} + y^{2} = C x$ . (1 mark)
Through $(1, 1)$ : $2 = C$ , so $x^{2} + y^{2} = 2 x$ . (1 mark)
$(x - 1)^{2} + y^{2} = 1$ : the circle with centre $(1, 0)$ and radius 1. (1 mark)

Answer: $x^{2} + y^{2} = 2 x$ , the circle with centre $(1, 0)$ and radius 1.

Example 6: Case study: a curve for a cycle track · 4 marks · Case-basedHOTS / competency

Suppose an engineer models a track so that its slope at any point $(x, y)$ with $x, y > 0$ is $\frac{d y}{d x} = \frac{y ^{3} + x ^{3}}{x y ^{2}}$ , and the track passes through $(1, 2)$ .
(i) Show that the equation is homogeneous. (1 mark)
(ii) Reduce it to variables separable form using $y = v x$ . (1 mark)
(iii) Find the general solution. (1 mark)
(iv) Find the equation of the track. (1 mark)

Try it first, then show the solutionSolution

(i) $F (λ x, λ y) = \frac{λ ^{3} ( y ^{3} + x ^{3} )}{λ ^{3} x y ^{2}} = F (x, y)$ : degree 0, so homogeneous. (1 mark)
(ii) $v + x v^{'} = \frac{v ^{3} + 1}{v ^{2}} = v + \frac{1}{v ^{2}}$ , so $v^{2} d v = \frac{d x}{x}$ . (1 mark)
(iii) $\frac{v ^{3}}{3} = lo g x + C$ , i.e. $y^{3} = 3 x^{3} lo g x + 3 C x^{3}$ . (1 mark)
(iv) Through $(1, 2)$ : $8 = 3 C$ , so $y^{3} = 3 x^{3} lo g x + 8 x^{3}$ . (1 mark)

Answer: (iii) $y^{3} = 3 x^{3} lo g x + K x^{3}$ (iv) $y^{3} = 3 x^{3} lo g x + 8 x^{3}$

Where students lose marks in Homogeneous Differential Equations

The slips examiners see most often, and what to write instead.

Mistake: Writing dy/dx = v after putting y = vx
v depends on x, so the product rule gives an extra term.
Wrong working
$y = v x \Rightarrow \frac{d y}{d x} = v$ .
Right working
$\frac{d y}{d x} = v + x \frac{d v}{d x}$ .
Do this: Differentiate $v x$ as a product every time.
Usually costs: The whole question.
Mistake: Not showing homogeneity
The first mark is usually for checking that the equation is homogeneous.
Wrong working
Directly put $y = v x$ .
Right working
First write $F (λ x, λ y) = F (x, y)$ , so the equation is homogeneous.
Do this: Add one line showing the degree-0 property.
Usually costs: 1 mark.
Mistake: Leaving the answer in v
The question is about x and y.
Wrong working
$sin v = lo g ∣ x ∣ + \frac{1}{2}$ .
Right working
$sin \frac{y}{x} = lo g ∣ x ∣ + \frac{1}{2}$ .
Do this: Replace v by $\frac{y}{x}$ before applying the condition.
Usually costs: Half to one mark.
Mistake: Applying the condition to v without converting
The condition is given for x and y.
Wrong working
$y (1) = \frac{π}{6}$ , so put $v = 1$ .
Right working
At $x = 1$ , $y = \frac{π}{6}$ , so $v = \frac{y}{x} = \frac{π}{6}$ .
Do this: Compute $v = \frac{y}{x}$ from the given point.
Usually costs: The particular-solution mark.

Exam tips

Homogeneous equations were asked in the 2023, 2024 and 2026 board papers and in the 2026 and 2027 sample papers, for 3 to 5 marks; follow the five-step layout.
Write ' $y = v x$ , $\frac{d y}{d x} = v + x \frac{d v}{d x}$ ' as a separate line.
Revise partial fractions and $\int \frac{f ^{'} ( v )}{f ( v )} d v$ ; most integrals in $v$ use them.
If $\frac{x}{y}$ appears naturally, use $x = v y$ and $\frac{d x}{d y}$ .
Leave the general solution in the simplest implicit form; solving for y is not required unless asked.

Practice questions on Homogeneous Differential Equations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which of these differential equations is homogeneous?
(a) $\frac{d y}{d x} = \frac{x ^{2} + y}{x}$
(b) $\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{x y}$
(c) $\frac{d y}{d x} = x + y + 1$
(d) $\frac{d y}{d x} = \frac{y + 1}{x}$
Hint
Check whether every term has the same total degree.
Answer
(b). Numerator and denominator are both homogeneous of degree 2.
Q2 · 2 marks · Very short answerBoard level
Show that $f (x, y) = \frac{x ^{3} + y ^{3}}{x + y}$ is a homogeneous function and find its degree.
Hint
Replace x by $λ x$ and y by $λ y$ .
Answer
$f (λ x, λ y) = \frac{λ ^{3} ( x ^{3} + y ^{3} )}{λ ( x + y )} = λ^{2} f (x, y)$ : homogeneous of degree 2.
Q3 · 3 marks · Short answerBoard level
Solve $x \frac{d y}{d x} = x + 2 y$ .
Hint
With $y = v x$ : $x v^{'} = 1 + v$ .
Answer
$\frac{d v}{1 + v} = \frac{d x}{x} \Rightarrow lo g ∣1 + v ∣ = lo g ∣ x ∣ + lo g ∣ C ∣ \Rightarrow 1 + \frac{y}{x} = C x \Rightarrow x + y = C x^{2}$ .
Q4 · 3 marks · Short answerHarder board level
Solve $x \frac{d y}{d x} = y + x e^{y / x}$ , given $y = 0$ when $x = 1$ .
Hint
After $y = v x$ , you get $e^{- v} d v = \frac{d x}{x}$ .
Answer
$v + x v^{'} = v + e^{v} \Rightarrow e^{- v} d v = \frac{d x}{x} \Rightarrow - e^{- v} = lo g ∣ x ∣ + C$ . At $(1, 0)$ : $C = - 1$ . So $e^{- y / x} = 1 - lo g x$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $\frac{d y}{d x} = \frac{x + y + 1}{x - y}$ is a homogeneous differential equation.
Reason (R): The numerator and denominator are polynomials of degree 1.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Is $x + y + 1$ homogeneous?
Answer
(d). The constant 1 has degree 0, so the numerator is not homogeneous and A is false; R is true.
Q6 · 5 marks · Long answerHarder board level
Find the particular solution of $2 x y \frac{d y}{d x} = x^{2} + 3 y^{2}$ , given $y (1) = 1$ .
Hint
With $y = v x$ : $x v^{'} = \frac{1 + v ^{2}}{2 v}$ .
Answer
$v + x v^{'} = \frac{1 + 3 v ^{2}}{2 v} \Rightarrow x v^{'} = \frac{1 + v ^{2}}{2 v} \Rightarrow \frac{2 v d v}{1 + v ^{2}} = \frac{d x}{x} \Rightarrow lo g (1 + v^{2}) = lo g ∣ x ∣ + lo g C \Rightarrow 1 + \frac{y ^{2}}{x ^{2}} = C x$ , i.e. $x^{2} + y^{2} = C x^{3}$ . With $y (1) = 1$ : $C = 2$ , so $x^{2} + y^{2} = 2 x^{3}$ .
Q7 · 3 marks · Short answerHOTS / competency
Solve $\frac{d y}{d x} = \frac{y}{x} + cos^{2} \frac{y}{x}$ , given $y = \frac{π}{4}$ when $x = 1$ .
Hint
$sec^{2} v d v = \frac{d x}{x}$ .
Answer
$x v^{'} = cos^{2} v \Rightarrow sec^{2} v d v = \frac{d x}{x} \Rightarrow tan v = lo g ∣ x ∣ + C$ . At $(1, \frac{π}{4})$ : $C = 1$ . So $tan \frac{y}{x} = lo g ∣ x ∣ + 1$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.When is $\frac{d y}{d x} = F (x, y)$ homogeneous?Show answer
When $F (λ x, λ y) = F (x, y)$ , i.e. F is a function of $\frac{y}{x}$ .
2.What substitution solves it?Show answer
$y = v x$ , $\frac{d y}{d x} = v + x \frac{d v}{d x}$ .
3.What is the degree of $x^{2} y + y^{3}$ ?Show answer
3.
4.Is $\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{x + y}$ homogeneous?Show answer
No: degree 2 over degree 1.
5.Solve $\frac{d y}{d x} = \frac{y}{x}$ .Show answer
$y = C x$ .
6.What substitution is used for $\frac{d x}{d y} = G (\frac{x}{y})$ ?Show answer
$x = v y$ .

Homogeneous Differential Equations: frequently asked questions

How do I check whether a differential equation is homogeneous?

Write it as dy/dx = F(x, y) and replace x by λx and y by λy. If F does not change, the equation is homogeneous. For a fraction, it is enough that the numerator and denominator are homogeneous of the same degree.

Why does y = vx work?

Because the right side depends only on y/x, which is v. The substitution turns the equation into one in v and x that can be separated.

Are non-homogeneous equations like dy/dx = (x + y + 1)/(x − y) in the syllabus?

No. The syllabus limits this topic to homogeneous equations of first order and first degree.

Which NCERT section covers homogeneous equations?

Section 9.4.2 (Homogeneous Differential Equations) of Chapter 9 in NCERT Mathematics Part II for Class 12.

Must I write y explicitly in the answer?

No. A neat implicit relation between x and y, such as x² − 2xy − y² = C, is a complete general solution.