How do I know if a DE is homogeneous?

Chapter 12 of 18

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CBSE Class 12 Mathematics★ 6-8 marks · CBSE Class 12 Mathematics Chapter 9

Differential Equations

An equation involving DERIVATIVES is a differential equation — and it describes how things CHANGE. From population growth to radioactive decay, differential equations model the world. CBSE Class 12 Mathematics Chapter 9.

⏱ 18 min readHard difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Determine the order (highest derivative) and degree (power of highest derivative, when polynomial) of a differential equation
2Verify that a given function is a solution of a DE by substituting and checking
3Solve variable separable DEs: separate variables, integrate both sides
4Solve homogeneous DEs using the substitution y = vx
5Solve first-order linear DEs of the form dy/dx + P(x)y = Q(x) using integrating factor μ = e^∫P dx

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Why this chapter matters

Differential Equations is the chapter where integration meets word problems. Variable separable, homogeneous, and linear DEs with integrating factors are all tested. The board exam reliably includes one 5-6 mark DE problem. The most common source of errors is not applying the integrating factor correctly in linear DEs.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Differential Equations

"A differential equation is the mathematical form of a scientific law."

1. Chapter Overview

A DIFFERENTIAL EQUATION (DE) is an equation involving an unknown function and its derivatives. This chapter covers: order and degree of a DE, forming a DE (eliminate arbitrary constants), and solving first-order, first-degree DEs — variable separable, homogeneous, and linear differential equations.

2. Basic Concepts

Order: The HIGHEST derivative appearing in the DE. y'' + y = 0 → Order = 2.
Degree: The POWER of the highest derivative (after clearing radicals/fractions). Must be a positive integer.

3. Forming a Differential Equation

Given a family of curves with arbitrary constants. DIFFERENTIATE as many times as there are constants. Eliminate the constants. 'The number of arbitrary constants = the ORDER of the DE.'

4. Solving First-Order, First-Degree DEs

Variable Separable

dy/dx = f(x) × g(y) → Separate: (1/g(y)) dy = f(x) dx. Integrate BOTH sides.

Homogeneous DEs

dy/dx = F(y/x). Let y = vx → dy/dx = v + x(dv/dx). Substitute. Separate variables. Solve. Back-substitute v = y/x.

Linear DE

dy/dx + P(x)y = Q(x). Solution: y × IF = ∫ (Q × IF) dx + C, where IF (Integrating Factor) = e^∫P dx.

5. Applications

Population growth: dP/dt = kP → P = P₀eᵏᵗ (exponential growth)
Radioactive decay: dN/dt = -λN → N = N₀e⁻λᵗ
Newton's law of cooling: dT/dt = -k(T — Tₐ)

6. Exam Focus

Order and degree — definitions. Forming a DE from a family of curves.
Variable separable method.
Homogeneous DE — substitution y = vx.
Linear DE — IF = e^∫Pdx. Solution formula.

7. Conclusion

Differential equations are where calculus meets the REAL WORLD:

ORDER: How complex is the relationship?
SOLVING: Variable separable. Homogeneous. Linear. 'Each type has its method. Recognise the type → apply the method.'
APPLICATIONS: Population. Decay. Cooling. 'The universe speaks in differential equations. Calculus translates.'

'If you want to understand the universe, learn to solve differential equations. Everything that changes does so according to one.'

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Order, Degree, and General/Particular Solutions

ORDER: the order of the highest derivative in the DE. DEGREE: the power (exponent) of the highest-order derivative, AFTER the DE is made polynomial in all derivatives. Degree is defined only when the DE is a polynomial in derivatives. GENERAL SOLUTION: contains as many arbitrary constants as the order of the DE. PARTICULAR SOLUTION: obtained by applying initial conditions to determine the arbitrary constants. FORMATION: to form DE from a family of curves with n arbitrary constants → differentiate n times and eliminate the n constants.

DEGREE TRAP: The DE d²y/dx² + sin(dy/dx) = 0 has NO defined degree (because sin of a derivative is not a polynomial in the derivative). Degree is undefined when the DE contains transcendental functions of derivatives. Order is always defined. For d²y/dx² + x(dy/dx)³ = 0: order=2, degree=1 (the highest order derivative d²y/dx² appears to power 1).

Variable Separable Method

A DE is VARIABLE SEPARABLE if it can be written as f(y) dy = g(x) dx. PROCEDURE: (1) Rearrange to get all y-terms with dy on one side and all x-terms with dx on the other. (2) Integrate both sides: ∫f(y) dy = ∫g(x) dx + C. (3) Write the general solution. (4) If initial condition given, substitute to find C (particular solution). STANDARD FORM: dy/dx = f(x)·g(y) → separate as dy/g(y) = f(x) dx → integrate.

The constant of integration C appears on ONE side only (conventionally on the right with x). Do NOT write C on both sides. When ln appears from integration, absorb the constant cleanly: ∫1/y dy = ln|y| + C₁ and ∫1/x dx = ln|x| + C₂ → ln|y| = ln|x| + C (where C = C₂ − C₁). Write the answer in the form y = f(x) with explicit C if possible.

Homogeneous Differential Equations

A DE dy/dx = F(x, y) is HOMOGENEOUS if F(x, y) = F(tx, ty) for all t (i.e., F is a homogeneous function of degree 0). Equivalently: F(x, y) can be written as g(y/x). SUBSTITUTION: Let y = vx, so dy/dx = v + x·dv/dx. Substitute: v + x·dv/dx = g(v). Separate variables: x·dv/dx = g(v) − v → dv/[g(v)−v] = dx/x. Integrate both sides → general solution in v. Back-substitute v = y/x.

Identify homogeneous DE: every term in numerator and denominator has the SAME total degree. Example: dy/dx = (x²+y²)/(xy) — numerator degree 2, denominator degree 2 → homogeneous. KEY STEP: After substitution y = vx, the DE always becomes variable separable. Never skip the back-substitution v → y/x at the end.

Linear Differential Equations and Integrating Factor

Standard LINEAR DE: dy/dx + P(x)·y = Q(x) (P and Q are functions of x only). INTEGRATING FACTOR (IF): μ = e^∫P(x)dx. SOLUTION: Multiply both sides by μ: d/dx[μ·y] = μ·Q(x). Integrate: μ·y = ∫μ·Q(x) dx + C. General solution: y = (1/μ)·[∫μ·Q(x) dx + C]. FOR dx/dy FORM: dx/dy + P(y)·x = Q(y) → IF = e^∫P(y)dy → solution: x = (1/IF)[∫IF·Q(y) dy + C].

CRITICAL CHECK before applying the linear formula: the coefficient of dy/dx must be 1. If the DE is given as a·dy/dx + by = Q, divide EVERY term by a first. Example: x·dy/dx + 2y = x² → divide by x → dy/dx + (2/x)y = x. Now P = 2/x, Q = x. IF = e^∫(2/x)dx = e^(2ln|x|) = x². The IF must be applied to BOTH sides.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Not dividing by the coefficient of dy/dx before computing the integrating factor

✓ The linear DE formula requires the form dy/dx + Py = Q with coefficient of dy/dx equal to 1. If you have x·dy/dx + 2y = sin x, you MUST divide by x first: dy/dx + (2/x)y = sin x/x. Then P = 2/x, Q = sin x/x. Computing IF from the original (undivided) equation gives a wrong IF and wrong solution.

WATCH OUT

✗ Forgetting to back-substitute v = y/x at the end of a homogeneous DE

✓ In homogeneous DE, after solving for v in terms of x, you MUST replace v with y/x to express the solution in terms of the original variables x and y. The general solution must be in x and y (or y = f(x)), not in v. Write the step 'Substituting v = y/x:' explicitly.

WATCH OUT

✗ Confusing order and degree when derivatives appear inside trig/log functions

✓ When a DE contains sin(d²y/dx²) or ln(dy/dx), the degree is UNDEFINED (the DE is not polynomial in its derivatives). The order is still defined (= order of the highest derivative appearing). Never assign a degree to such DEs — state 'Degree: not defined.'

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex s 9.1–9.6

Exercises 9.1–9.6

Order/degree identification; verifying solutions; variable separable; homogeneous (with y=vx); linear DEs with integrating factor; formation of DEs; word problems (growth/decay, mixture)

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· variable-separable

Solve the differential equation: dy/dx = (1 + y²) / (1 + x²), given y = 0 when x = 0.

▶ Show solution

Separate variables: dy/(1+y²) = dx/(1+x²). Integrate both sides: tan⁻¹(y) = tan⁻¹(x) + C. Apply initial condition y = 0, x = 0: tan⁻¹(0) = tan⁻¹(0) + C → 0 = 0 + C → C = 0. PARTICULAR SOLUTION: tan⁻¹(y) = tan⁻¹(x) → y = x.

Q2MEDIUM· homogeneous-DE

Solve: x·dy/dx − y = √(x² + y²).

▶ Show solution

Rewrite: dy/dx = [y + √(x²+y²)] / x. Check homogeneity: f(tx,ty) = [ty + √(t²x²+t²y²)]/(tx) = t[y + √(x²+y²)]/(tx) = [y + √(x²+y²)]/x. Homogeneous of degree 0. ✓ SUBSTITUTION: Let y = vx. Then dy/dx = v + x(dv/dx). Substituting: v + x(dv/dx) = [vx + √(x²+v²x²)]/x = v + √(1+v²). x(dv/dx) = √(1+v²). Separate: dv/√(1+v²) = dx/x. Integrate: ln|v + √(1+v²)| = ln|x| + ln|C| = ln|Cx|. v + √(1+v²) = Cx. Back-substitute v = y/x: y/x + √(1+y²/x²) = Cx → y/x + √(x²+y²)/x = Cx → y + √(x²+y²) = Cx². GENERAL SOLUTION: y + √(x²+y²) = Cx².

Q3HARD· linear-DE

Solve: (1 + x²) dy/dx + 2xy = cos x. Also find the particular solution when y = 0 at x = 0.

▶ Show solution

Divide by (1+x²): dy/dx + [2x/(1+x²)]·y = cos x/(1+x²). This is linear: P = 2x/(1+x²), Q = cos x/(1+x²). INTEGRATING FACTOR: IF = e^∫P dx = e^∫[2x/(1+x²)] dx = e^(ln(1+x²)) = (1+x²). Multiply both sides by IF: d/dx[(1+x²)y] = (1+x²)·cos x/(1+x²) = cos x. Integrate: (1+x²)y = ∫cos x dx = sin x + C. GENERAL SOLUTION: y = (sin x + C)/(1+x²). PARTICULAR SOLUTION: y=0 when x=0: 0 = (sin 0 + C)/(1+0) = C. So C = 0. PARTICULAR SOLUTION: y = sin x/(1+x²).

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Order = highest derivative order. Degree = power of highest derivative (only if polynomial in derivatives; else undefined).
•General solution has n arbitrary constants for order n DE.
•Variable separable: rearrange as f(y)dy = g(x)dx, then integrate both sides.
•Homogeneous: dy/dx = g(y/x). Substitution y = vx → x(dv/dx) = g(v) − v → separable.
•Linear DE: dy/dx + Py = Q. IF = e^∫Pdx. Solution: y·IF = ∫Q·IF dx + C.
•After solving homogeneous: back-substitute v = y/x — never leave in v.
•Before linear IF: ensure coefficient of dy/dx is 1 — divide if needed.
•dy/dt = ky → y = Ce^(kt). Growth: k > 0. Decay: k < 0.
•Particular solution: apply given initial conditions (x₀, y₀) to find C.

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 6-8 marks

Question type	Marks each	Typical count	What it tests
Short — Order/Degree or Variable Separable	2-3	1	Identify order and degree; solve simple separable DE; find particular solution given IC
Long — Homogeneous or Linear DE	5-6	1	Solve homogeneous DE using y=vx; solve linear DE using integrating factor; word problem (exponential growth/decay, mixture)

Prep strategy

Identify DE type FIRST: (1) Can I separate f(y)dy = g(x)dx? → Variable separable. (2) Can I write dy/dx = g(y/x)? → Homogeneous (use y=vx). (3) Is it dy/dx + Py = Q form? → Linear (use IF = e^∫Pdx). Choosing the wrong method and proceeding wastes all working marks.
For linear DEs: show these four steps explicitly: (1) Rewrite in standard form dy/dx + Py = Q. (2) Write IF = e^∫Pdx and evaluate. (3) Write d/dx[IF·y] = IF·Q. (4) Integrate and write general solution. Each step is a mark.
For word problems involving growth/decay: the DE is typically dy/dt = ky. Solution: y = y₀·eᵏᵗ. For population problems, k > 0 (growth); for radioactive decay, k < 0 (decay). Always state the model DE before solving.

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Epidemiological Modelling (SIR Model)

The spread of infectious diseases is modelled by a system of differential equations: dS/dt = −βSI (susceptible), dI/dt = βSI − γI (infected), dR/dt = γI (recovered). This SIR model — the mathematical backbone of COVID-19 projections — is a system of first-order DEs. The exponential growth phase (when I is small) follows the separable DE dI/dt ≈ (β−γ)I, giving exponential growth I = I₀e^((β−γ)t). Every epidemic projection by WHO or government health departments uses DE-based compartmental models derived directly from this chapter.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

State the type of DE explicitly before solving: 'This DE is linear / homogeneous / variable separable.' Then apply the standard procedure. If you name the wrong type, you lose all method marks. Correct identification earns marks even if subsequent arithmetic is wrong.
For linear DEs: simplify IF = e^∫Pdx carefully — common values are e^(ln f(x)) = f(x) and e^(2ln x) = x². Write each simplification step explicitly. Examiners look for d/dx[IF·y] = IF·Q as the key line.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Study the LAPLACE TRANSFORM: it converts a linear DE into an algebraic equation (in s-domain), solves it, then inverts back. For example, d²y/dt² + y = 0 becomes (s²+1)Y(s) = ... → Y(s) = 1/(s²+1) → y = sin t. This transforms calculus problems into algebra — the engine behind control systems, signal processing, and circuit analysis
Explore the LOGISTIC EQUATION dy/dt = ry(1−y/K): a more realistic population model than dy/dt = ky, incorporating a carrying capacity K. The solution y = K/(1 + Ae^(−rt)) produces the characteristic S-shaped (sigmoid) curve seen in population biology, market adoption curves, and the activation function in neural networks

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 12 Board (Mathematics)High

JEE Main (Differential Equations)High

CUET (Mathematics)Medium

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Questions students ask

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A DE dy/dx = F(x,y) is homogeneous if F(tx, ty) = F(x, y) for all t > 0 (i.e., F has degree 0). The quick test: write F(x,y) as a fraction of polynomials in x and y. If every term in the numerator and every term in the denominator has the SAME total degree, it's homogeneous. Example: (x²+y²)/(xy) — numerator degree 2, denominator degree 2 → homogeneous. Shortcut: if you can write dy/dx as a function of (y/x) alone (by dividing numerator and denominator by xⁿ), it's homogeneous.

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