MathsISC Class 12 · Differential Equations

Homogeneous Differential Equations: Identify, Substitute y = vx, Solve

Updated 15 October 2026 · Fact-checked

A first-order equation dy/dx = f(x, y) is homogeneous when f can be written as a function of y/x alone, that is f(λx, λy) = f(x, y). Substitute y = vx, so dy/dx = v + x dv/dx; the variables then separate. Integrate, replace v by y/x, and use any given condition for the particular solution.

Always test that λ cancels first, then put y = vx so the variables separate; use x = vy when terms like e^(x/y) appear.

Illustration drawn for QuestionBuddy.

Understand Homogeneous Differential Equations

What makes an equation homogeneous. A function $F (x, y)$ is homogeneous of degree $n$ if $F (λ x, λ y) = λ^{n} F (x, y)$ . For example $x^{2} + y^{2}$ , $x y$ and $x^{2} - 2 y^{2}$ are all homogeneous of degree 2, while $x^{2} + y$ is not homogeneous. A first-order, first-degree differential equation
$\frac{d y}{d x} = \frac{P ( x , y )}{Q ( x , y )}$
is homogeneous when $P$ and $Q$ are homogeneous of the same degree. Then the right side is unchanged when $x$ and $y$ are scaled together, so it is really a function of $\frac{y}{x}$ alone: $\frac{d y}{d x} = g (\frac{y}{x})$ . A quick test: replace $x$ by $λ x$ and $y$ by $λ y$ ; if every $λ$ cancels, the equation is homogeneous. The ISC 2024 and 2025 examiners both reported candidates who could not identify a homogeneous equation, so make this check your first line.

The substitution. Put $y = v x$ , where $v$ is a new function of $x$ . Then
$\frac{d y}{d x} = v + x \frac{d v}{d x} .$
The equation becomes $v + x \frac{d v}{d x} = g (v)$ , so $x \frac{d v}{d x} = g (v) - v$ , which separates:
$\int \frac{d v}{g ( v ) - v} = \int \frac{d x}{x} .$
After integrating, replace $v$ by $\frac{y}{x}$ and simplify.

**When $x = v y$ is better.** If the equation is naturally in the form $\frac{d x}{d y} = h (\frac{x}{y})$ , for instance when terms like $e^{x / y}$ appear, put $x = v y$ and $\frac{d x}{d y} = v + y \frac{d v}{d y}$ . ISC 2024 set exactly such an equation, $2 y e^{x / y} d x + (y - 2 x e^{x / y}) d y = 0$ ; the 2025 examiners advised practising both forms.

Equations in differential form. If the question gives $M d x + N d y = 0$ , rewrite it as $\frac{d y}{d x} = - \frac{M}{N}$ (or $\frac{d x}{d y} = - \frac{N}{M}$ ) before testing.

Integration after separating. The $v$ -integral usually needs a standard technique: a substitution ( $\int \frac{2 v}{1 - v ^{2}} d v = - lo g ∣1 - v^{2} ∣$ ), partial fractions, or a standard form such as $\int \frac{d v}{1 + v ^{2}} = lo g ∣ v + 1 + v^{2} ∣$ . The constant is usually best written as $lo g C$ , so that the logarithms combine neatly.

General and particular solutions. The answer with an arbitrary constant is the general solution, a family of curves. A condition such as " $y = 1$ when $x = 1$ " picks out one member, the particular solution. Both 2024 and 2025 examiners noted candidates who stopped before finding it. Substitute the condition, find $C$ , and write the final equation.

Geometric meaning. Homogeneous equations describe curves whose slope at a point depends only on the direction of that point from the origin, not on its distance. That is why so many curve-design and tangent-property questions lead to them.

Homogeneous Differential Equations: formulas to remember

Homogeneous function: $F (λ x, λ y) = λ^{n} F (x, y)$; Degree n.
Homogeneous equation: $\frac{d y}{d x} = g (\frac{y}{x})$; Numerator and denominator homogeneous of the same degree.
Substitution y = vx: $\frac{d y}{d x} = v + x \frac{d v}{d x}$; Then x dv/dx = g(v) − v.
Separated form: $\int \frac{d v}{g ( v ) - v} = \int \frac{d x}{x}$; Variables are now separated; integrate both sides.
Substitution x = vy: $\frac{d x}{d y} = v + y \frac{d v}{d y}$; For dx/dy = h(x/y), such as terms in e^{x/y}.
Useful integral: $\int \frac{d v}{1 + v ^{2}} = lo g v + 1 + v^{2} + C$; Appears when √(x² + y²) is present.

How to solve Homogeneous Differential Equations questions

Use this sequence for every homogeneous differential equation question.

1Rewrite the equation as $\frac{d y}{d x} = \frac{P ( x , y )}{Q ( x , y )}$ (or $\frac{d x}{d y}$ if that is more natural).
2Test: replace $x, y$ by $λ x, λ y$ . If $λ$ cancels completely, write "the equation is homogeneous".
3Put $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$ (or $x = v y$ and $\frac{d x}{d y} = v + y \frac{d v}{d y}$ ).
4Simplify to $x \frac{d v}{d x} = g (v) - v$ and separate the variables.
5Integrate both sides, using substitution or partial fractions on the $v$ -side; write the constant as $lo g C$ if logarithms appear.
6Replace $v$ by $\frac{y}{x}$ and simplify to get the general solution.
7If a condition is given, substitute it, find $C$ , and state the 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.

Homogeneous equation for a curve design

Suppose a garden designer wants a curved path, starting at the point $(1, 1)$ on a plan (in metres), whose slope at every point $(x, y)$ is $\frac{x + y}{x}$ .

Find the equation of the path, and the value of $y$ where $x = 3$ .

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 $y = x lo g ∣ x ∣ + C x$ . At $(1, 1)$ : $1 = 0 + C$ , so $C = 1$ .
Path: $y = x lo g x + x$ (for $x > 0$ ). At $x = 3$ : $y = 3 lo g 3 + 3 \approx 6.30$ .

Answer: $y = x lo g x + x$ ; at $x = 3$ , $y \approx 6.30$ m.

Memory hook: A slope that depends only on y/x is a homogeneous equation: substitute y = vx.

Particular solution of a homogeneous equation

Imagine a robot that is programmed to steer so that at every point $(x, y)$ of its path the slope is $\frac{y}{x} + sin \frac{y}{x}$ . It starts at $(1, \frac{π}{2})$ .

Find its path.

Try it first, then show the solutionSolution

Put $y = v x$ : $v + x \frac{d v}{d x} = v + sin v$ , so $cosec v d v = \frac{d x}{x}$ .
Integrate: $lo g tan \frac{v}{2} = lo g ∣ x ∣ + lo g C$ , so $tan \frac{y}{2 x} = C x$ .
At $(1, \frac{π}{2})$ : $tan \frac{π}{4} = C$ , so $C = 1$ .

Answer: $tan \frac{y}{2 x} = x$ , that is $y = 2 x tan^{- 1} x$ .

Memory hook: Steering by angle from the origin gives a homogeneous path equation.

Tricks to remember Homogeneous Differential Equations

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

Scale test
Multiply x and y by λ. If every λ cancels, it is homogeneous.
Use it when: Identifying the type before choosing a method.
v plus x dv
y = vx gives dy/dx = v + x dv/dx: "v plus x dv by dx".
Use it when: Writing the substitution correctly.
e^{x/y}? Use x = vy
If x/y appears, especially in an exponent, put x = vy instead.
Use it when: Questions like ISC 2024 Q13(i).
Finish with C
A condition given? Then a particular solution is wanted: find C.
Use it when: Avoiding the commonest lost mark.

Homogeneous Differential Equations: solved examples

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

Example 1: Identifying a homogeneous equation · 1 mark · MCQBoard level

Which of these differential equations is homogeneous?
(a) $\frac{d y}{d x} = \frac{x + y ^{2}}{x}$
(b) $\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{x y}$
(c) $\frac{d y}{d x} = x^{2} + y$
(d) $\frac{d y}{d x} = \frac{y}{x + 1}$

Try it first, then show the solutionSolution

In (b), numerator and denominator are both of degree 2, so $λ$ cancels: $\frac{λ ^{2} ( x ^{2} + y ^{2} )}{λ ^{2} x y}$ .

Answer: (b).

Example 2: A first homogeneous equation · 3 marks · Short answerBoard level

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

Try it first, then show the solutionSolution

Homogeneous; put $y = v x$ : $v + x \frac{d v}{d x} = v + v^{2}$ . (1 mark)
$x \frac{d v}{d x} = v^{2} \Rightarrow \int \frac{d v}{v ^{2}} = \int \frac{d x}{x} \Rightarrow - \frac{1}{v} = lo g ∣ x ∣ + C$ . (1 mark)
$- \frac{x}{y} = lo g ∣ x ∣ + C$ , so $y = - \frac{x}{l o g ∣ x ∣ + C}$ . (1 mark)

Answer: $y = - \frac{x}{l o g ∣ x ∣ + C}$ (and also $y = 0$ , which was lost when dividing by $v^{2}$ ).

Example 3: Differential form with x² + y² · 5 marks · Long answerHarder board level

Solve $(x^{2} - y^{2}) d x + 2 x y d y = 0$ .

Try it first, then show the solutionSolution

$\frac{d y}{d x} = \frac{y ^{2} - x ^{2}}{2 x y}$ , homogeneous of degree 0. Put $y = v x$ : $v + x \frac{d v}{d x} = \frac{v ^{2} - 1}{2 v}$ . (1 mark)
$x \frac{d v}{d x} = \frac{v ^{2} - 1 - 2 v ^{2}}{2 v} = - \frac{1 + v ^{2}}{2 v}$ . (1 mark)
Separate: $\int \frac{2 v}{1 + v ^{2}} d v = - \int \frac{d x}{x}$ . (1 mark)
$lo g (1 + v^{2}) = - lo g ∣ x ∣ + lo g C \Rightarrow x (1 + v^{2}) = C$ . (1 mark)
$x (1 + \frac{y ^{2}}{x ^{2}}) = C \Rightarrow x^{2} + y^{2} = C x$ . (1 mark)

Answer: $x^{2} + y^{2} = C x$ : a family of circles through the origin with centres on the x-axis.

Example 4: A particular solution with tan(y/x) · 3 marks · Short answerHarder board level

Solve $x \frac{d y}{d x} = y + x tan \frac{y}{x}$ , given $y = \frac{π}{6}$ when $x = 1$ .

Try it first, then show the solutionSolution

$\frac{d y}{d x} = \frac{y}{x} + tan \frac{y}{x}$ . Put $y = v x$ : $x \frac{d v}{d x} = tan v$ , so $cot v d v = \frac{d x}{x}$ . (1 mark)
$lo g ∣ sin v ∣ = lo g ∣ x ∣ + lo g C \Rightarrow sin \frac{y}{x} = C x$ . (1 mark)
At $x = 1$ , $y = \frac{π}{6}$ : $sin \frac{π}{6} = C$ , so $C = \frac{1}{2}$ . Particular solution: $sin \frac{y}{x} = \frac{x}{2}$ . (1 mark)

Answer: $sin \frac{y}{x} = \frac{x}{2}$ , that is $y = x sin^{- 1} \frac{x}{2}$ .

Example 5: An equation with √(x² + y²) · 3 marks · Short answerHarder board level

Solve $x d y - y d x = x^{2} + y^{2} d x$ for $x > 0$ , and find the solution through $(1, 0)$ .

Try it first, then show the solutionSolution

$\frac{d y}{d x} = \frac{y + x ^{2} + y ^{2}}{x}$ . With $y = v x$ : $x \frac{d v}{d x} = 1 + v^{2}$ . (1 mark)
$\int \frac{d v}{1 + v ^{2}} = \int \frac{d x}{x} \Rightarrow lo g v + 1 + v^{2} = lo g x + lo g C$ , so $y + x^{2} + y^{2} = C x^{2}$ . (1 mark)
Through $(1, 0)$ : $0 + 1 = C$ , so $y + x^{2} + y^{2} = x^{2}$ . Solving for $y$ : $y = \frac{x ^{2} - 1}{2}$ . (1 mark)

Answer: General: $y + x^{2} + y^{2} = C x^{2}$ ; through $(1, 0)$ : $y = \frac{x ^{2} - 1}{2}$ .

Example 6: A logarithmic homogeneous equation · 5 marks · Long answerHOTS / competency

Solve $x \frac{d y}{d x} = y (lo g y - lo g x + 1)$ , given $y = e$ when $x = 1$ .

Try it first, then show the solutionSolution

$\frac{d y}{d x} = \frac{y}{x} (lo g \frac{y}{x} + 1)$ , a function of $\frac{y}{x}$ . Put $y = v x$ . (1 mark)
$v + x \frac{d v}{d x} = v (lo g v + 1) \Rightarrow x \frac{d v}{d x} = v lo g v$ . (1 mark)
$\int \frac{d v}{v l o g v} = \int \frac{d x}{x}$ . With $t = lo g v$ : $lo g ∣ lo g v ∣ = lo g ∣ x ∣ + lo g C$ , so $lo g v = C x$ . (1 mark)
General solution: $lo g \frac{y}{x} = C x$ . At $(1, e)$ : $lo g e = C$ , so $C = 1$ . (1 mark)
Particular solution: $lo g \frac{y}{x} = x$ , that is $y = x e^{x}$ . Check: $x y^{'} = x (e^{x} + x e^{x})$ and $y (lo g \frac{y}{x} + 1) = x e^{x} (x + 1)$ . (1 mark)

Answer: $y = x e^{x}$ .

Where students lose marks in Homogeneous Differential Equations

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

Mistake: Not identifying the equation as homogeneous
Both the 2024 and 2025 examiners reported this; candidates then tried a method that does not work.
Do this: Write the λ-test as your first line, and name the type.
Usually costs: The whole 5- or 6-mark question.
Mistake: Forgetting the v in dy/dx = v + x dv/dx
Writing dy/dx = x dv/dx loses a term and the equation never separates.
Wrong working
$y = v x \Rightarrow \frac{d y}{d x} = x \frac{d v}{d x}$ .
Right working
$\frac{d y}{d x} = v + x \frac{d v}{d x}$ (product rule).
Do this: Differentiate y = vx with the product rule every time.
Usually costs: Everything after that line.
Mistake: Stopping at the general solution
The examiners in both years noted candidates who did not obtain the particular solution when a condition was given.
Do this: Underline the condition in the question and use it at the end to find C.
Usually costs: 1 to 2 marks.
Mistake: Errors in the v-integral
The 2025 examiners noted errors in basic integration after the substitution.
Do this: Identify the integral type (substitution, partial fractions, standard form) before integrating, and check by differentiating.
Usually costs: 2 marks.

Exam tips

Write "the equation is homogeneous since $f (λ x, λ y) = f (x, y)$ " before substituting; it earns a mark.
Keep the constant as $lo g C$ when logarithms appear on both sides; combining them is then one step.
Replace $v$ by $\frac{y}{x}$ before applying the initial condition.
If the equation has $\frac{x}{y}$ in it, consider $x = v y$ with $\frac{d x}{d y}$ .
Present the final particular solution clearly, solved for $y$ if it is easy to do so.

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
To solve a homogeneous equation $\frac{d y}{d x} = g (\frac{y}{x})$ , the standard substitution is:
(a) $y = v + x$
(b) $y = v x$
(c) $y = \frac{v}{x}$
(d) $x = v + y$
Hint
It makes $\frac{y}{x}$ a single variable.
Answer
(b) $y = v x$ .
Q2 · 3 marks · Short answerBoard level
Solve $\frac{d y}{d x} = \frac{y}{x} + \frac{x}{y}$ , given $y = 1$ when $x = 1$ .
Hint
After $y = v x$ : $v d v = \frac{d x}{x}$ .
Answer
$\frac{v ^{2}}{2} = lo g ∣ x ∣ + C$ , so $y^{2} = 2 x^{2} lo g ∣ x ∣ + C x^{2}$ . At $(1, 1)$ : $C = 1$ . Particular solution $y^{2} = 2 x^{2} lo g x + x^{2}$ .
Q3 · 3 marks · Short answerHarder board level
Solve $\frac{d y}{d x} = \frac{x + y}{x - y}$ .
Hint
Separate into $\frac{1 - v}{1 + v ^{2}} d v$ .
Answer
$x \frac{d v}{d x} = \frac{1 + v ^{2}}{1 - v}$ , so $\int (\frac{1}{1 + v ^{2}} - \frac{v}{1 + v ^{2}}) d v = \int \frac{d x}{x}$ . $tan^{- 1} v - \frac{1}{2} lo g (1 + v^{2}) = lo g ∣ x ∣ + C$ , which simplifies to $tan^{- 1} \frac{y}{x} = \frac{1}{2} lo g (x^{2} + y^{2}) + C$ .
Q4 · 3 marks · Short answerHarder board level
Find the particular solution of $x^{2} d y + (x y + y^{2}) d x = 0$ , given $y = 1$ when $x = 1$ .
Hint
$x \frac{d v}{d x} = - v (v + 2)$ ; use partial fractions.
Answer
$\int \frac{d v}{v ( v + 2 )} = - \int \frac{d x}{x}$ gives $\frac{1}{2} lo g \frac{v}{v + 2} = - lo g ∣ x ∣ + c$ , so $\frac{y}{y + 2 x} = \frac{C}{x ^{2}}$ . At $(1, 1)$ : $C = \frac{1}{3}$ , so $3 x^{2} y = y + 2 x$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $\frac{d y}{d x} = \frac{x ^{3} + y ^{3}}{x y ^{2}}$ is a homogeneous differential equation.
Reason (R): A differential equation $\frac{d y}{d x} = f (x, y)$ is homogeneous if $f (λ x, λ y) = f (x, y)$ for all $λ \neq = 0$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Numerator and denominator are both of degree 3.
Answer
(a).
Q6 · 3 marks · Short answerHOTS / competency
Solve $y e^{x / y} d x = (x e^{x / y} + y) d y$ , given $x = 0$ when $y = 1$ .
Hint
Write $\frac{d x}{d y} = \frac{x}{y} + e^{- x / y}$ and put $x = v y$ .
Answer
$v + y \frac{d v}{d y} = v + e^{- v}$ , so $e^{v} d v = \frac{d y}{y}$ and $e^{v} = lo g ∣ y ∣ + C$ . That is $e^{x / y} = lo g ∣ y ∣ + C$ . At $x = 0$ , $y = 1$ : $C = 1$ . Particular solution $e^{x / y} = lo g y + 1$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose an engineer wants a reflector curve with this property: the tangent at any point $P (x, y)$ (with $x > 0$ ) cuts the y-axis at a height equal to the distance $O P$ of $P$ from the origin.
(i) Show that the curve satisfies $y - x \frac{d y}{d x} = x^{2} + y^{2}$ . (1 mark)
(ii) Show that this equation is homogeneous. (1 mark)
(iii) Solve it. (1 mark)
(iv) Find the curve through $(3, 4)$ and identify it. (1 mark)
Hint
The tangent at $P$ meets the y-axis at $y - x \frac{d y}{d x}$ .
Answer
(i) Setting $X = 0$ in $Y - y = y^{'} (X - x)$ gives the intercept $y - x y^{'}$ , which equals $x^{2} + y^{2}$ . (ii) $\frac{d y}{d x} = \frac{y - x ^{2} + y ^{2}}{x}$ ; scaling by $λ$ leaves it unchanged. (iii) $y = v x$ gives $\frac{d v}{1 + v ^{2}} = - \frac{d x}{x}$ , so $y + x^{2} + y^{2} = C$ . (iv) $4 + 5 = 9 = C$ ; squaring $x^{2} + y^{2} = 9 - y$ gives $x^{2} = 81 - 18 y$ , a parabola.

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 $\frac{d y}{d x} = \frac{P ( x , y )}{Q ( x , y )}$ homogeneous?Show answer
When $P$ and $Q$ are homogeneous functions of the same degree.
2.What is $\frac{d y}{d x}$ after the substitution $y = v x$ ?Show answer
$v + x \frac{d v}{d x}$ .
3.Is $x^{2} + x y$ homogeneous? Of what degree?Show answer
Yes, of degree 2.
4.Which substitution suits $\frac{d x}{d y} = h (\frac{x}{y})$ ?Show answer
$x = v y$ .
5.Solve $\frac{d y}{d x} = \frac{y}{x}$ .Show answer
$y = C x$ .
6.What turns a general solution into a particular one?Show answer
A given condition, used to find the constant $C$ .

Homogeneous Differential Equations: frequently asked questions

How do I know if a differential equation is homogeneous?

Write it as dy/dx = f(x, y). Replace x by λx and y by λy. If λ cancels completely, so that f(λx, λy) = f(x, y), the equation is homogeneous and can be solved with y = vx.

Why does the substitution y = vx work?

A homogeneous equation depends only on y/x. Setting v = y/x turns it into an equation in v and x in which the variables separate, so it can be integrated directly.

When should I use x = vy instead?

Use x = vy when the equation is more naturally written as dx/dy = h(x/y), for example when x/y appears in an exponent like e^{x/y}. The working mirrors y = vx with the roles of x and y swapped.

Do I need to give the general solution or the particular solution?

If the question gives a condition, such as y = 1 when x = 1, it wants the particular solution: find C and write the final equation. Otherwise, the general solution with C is the answer.

Are homogeneous differential equations in the ISC 2027 syllabus?

Yes. The CISCE 2027 syllabus includes homogeneous differential equations of first order and first degree, and they carried long questions in both the 2024 and 2025 ISC papers.