Looking for the Differential Equations Class 12 chapter? This page covers the Differential Equations chapter (Chapter 3) of the NCERT Class 12 Mathematics Part II textbook — a 38-page chapter that moves from recognising a differential equation to classifying it and solving it three ways.
The official PDF for the current rationalised edition is one link below, and the sections after it explain what the chapter teaches, using the book’s own section numbers.
Download the Differential Equations Class 12 NCERT chapter PDF
The link below opens the NCERT textbook portal at ncert.nic.in, which is where the official chapter file is published — it is the same PDF schools work from in the current rationalised edition, so nothing here is an older or abridged version. Open it once and keep it beside you; everything below follows the book’s own section numbers.
Open the Differential Equations Class 12 NCERT chapter PDF from the official NCERT portal, then use this page as a guide to the chapter’s sections and exercises.
| What the chapter holds | Count | Where it is used |
|---|---|---|
| Printed pages | 38 | |
| Sections in the chapter | 6 | |
| Worked examples | 1 | solved step by step in our NCERT Solutions |
| Exercise questions | 78 | answered in our NCERT Solutions |
| Official NCERT PDF | Download the chapter PDF | the chapter exactly as NCERT publishes it |
Chapter at a glance
The table below shows what the chapter file holds — its sections, worked examples and exercise questions — so you can see the size of the job before you start.
The chapter is organised in a straight line, and each part has a single job:
- 9.1 Introduction — frames the chapter as the reverse of integration and lists where differential equations appear in science and economics.
- 9.2 Basic Concepts — the definition, ordinary versus partial, the notation, then order (9.2.1) and degree (9.2.2).
- 9.3 General and Particular Solutions — what solving means here: a function, not a number.
- 9.4 Methods of Solving First Order, First Degree Differential Equations — the three methods: variables separable (9.4.1), homogeneous (9.4.2), linear (9.4.3).
- Miscellaneous Examples and Miscellaneous Exercise — mixed problems that combine everything.
- Summary and Historical Note — the chapter’s results in compact form, then a short history.
Each solving method has its own exercise: Exercise 9.3 for separable, Exercise 9.4 for homogeneous, Exercise 9.5 for linear. They sit after Exercise 9.1 (order and degree) and Exercise 9.2 (verifying solutions), so the chapter is really a chain — classify, verify, solve, apply.
What this chapter covers
This section is a map: it shows where each idea fits, so the definitions and formulas later have a home. The chapter moves in five steps — recognise a differential equation, classify it, decide what a solution is, solve it, apply it.
Step 1 — 9.1 Introduction. The chapter opens where Integral Calculus stopped. There you found a function whose derivative is given; here that task is written as \( \frac{dy}{dx} = g(x) \) and given its proper name.
The book points out that such equations come from physics, chemistry, biology, anthropology, geology and economics — which is why the chapter treats solving them as a serious skill.
Step 2 — 9.2 Basic Concepts. Section 9.2 asks a classification question: which equations count, and how do we talk about them? It contrasts ordinary algebraic equations such as \( x^2 – 3x + 3 = 0 \) with \( x\frac{dy}{dx} + y = 0 \), defines an ordinary differential equation, introduces the shorthand \( y’, y”, y”’ \) and \( y_n \) for derivatives, and then defines order (9.2.1) and degree (9.2.2).
Step 3 — 9.3 General and Particular Solutions. Section 9.3 changes the meaning of solve. For \( x^2 + 1 = 0 \) the answer is a number; for a differential equation the answer is a function \( \phi \). A solution that carries arbitrary constants is the general solution; fixing the constants gives a particular solution.
Step 4 — 9.4 the three solving methods. Section 9.4 deals only with first order, first degree equations and teaches three recognisable shapes: variables separable (9.4.1), homogeneous (9.4.2) and linear (9.4.3). Each shape gets its own exercise, so you can practise one method at a time.
Step 5 — mixed problems and closing. The Miscellaneous Examples (numbered 19 to 22) and the Miscellaneous Exercise mix every method, and the chapter closes with a Summary and a Historical Note. Applications are woven through the exercises: curves found from a given tangent slope, bank interest, bacteria growth and a balloon being inflated.
Key concepts: what each idea means
This section explains what a differential equation is and why the chapter classifies equations before solving them. Once the definitions make sense, the formulas stop looking arbitrary.
A differential equation versus an algebraic equation. The test is whether a derivative of the dependent variable appears. The equations \( x^2 – 3x + 3 = 0 \), \( \sin x + \cos x = 0 \) and \( x + y = 7 \) in §9.2 contain only variables. Equation (4) of the same section, \( x\frac{dy}{dx} + y = 0 \), contains \( \frac{dy}{dx} \), so it is a differential equation (NCERT, §9.2).
Why ordinary? When derivatives involve only one independent variable, the equation is an ordinary differential equation. Equations with derivatives with respect to more than one independent variable are partial differential equations, and this chapter studies only the ordinary kind. From §9.2 onward, the book uses the single word differential equation for them.
Order. The order is the order of the highest order derivative present: \( \frac{dy}{dx} \) gives order 1, \( \frac{d^2y}{dx^2} \) gives order 2, \( \frac{d^3y}{dx^3} \) gives order 3 (NCERT, §9.2.1). Find the tallest derivative first; its order is the order of the equation.
Degree. The degree is the highest power of the highest order derivative — but only if the equation is a polynomial in derivatives. That condition is the whole point of §9.2.2: an equation such as \( \frac{dy}{dx} + \sin\left(\frac{dy}{dx}\right) = 0 \) has no degree at all. Order and degree, when the degree exists, are always positive integers (NCERT, §9.2.2, note).
A solution is a function. A solution of a differential equation is a function \( \phi \) that satisfies it: substitute \( \phi \) for \( y \), and the left side becomes the right side. The graph \( y = \phi(x) \) is the solution curve, also called the integral curve (NCERT, §9.3).
Because the general solution carries arbitrary constants, it represents a whole family of curves; a particular solution picks one member of that family.
The three methods as shapes. All three methods in §9.4 answer the same question — how to solve \( \frac{dy}{dx} = F(x, y) \) — and each is keyed to the shape of \( F \).
It is separable when \( F \) splits into a product \( g(x) \cdot h(y) \); homogeneous when \( F \) depends only on the ratio \( y/x \); linear when \( y \) and its derivative appear only to the first power, arrangeable as \( \frac{dy}{dx} + Py = Q \). Recognising the shape before solving is the skill the exercises drill.
Figures in the chapter (and what to draw yourself)
This section covers what the chapter’s printed images do and do not show, and the two diagrams worth sketching by hand. Students often search this chapter for a diagram to memorise — there is not one.
The chapter is text-based: it has no concept diagrams carrying information you must reproduce. The only images are decorative — D. Hilbert’s quotation at the chapter’s opening, and the portrait of Henri Poincaré beside section 9.2 Basic Concepts. Neither needs to be learned.
Two visual ideas do matter, and the book expects you to hold them in your head:
The solution curve makes a family. A general solution like \( y = a\sin(x + b) \) in §9.3 is not one curve but a family — one curve for each choice of the constants \( a \) and \( b \). Sketch several members of the family filling the plane; a condition such as \( y = 2 \) when \( x = 0 \) selects the single member you want.
That picture explains why the particular solution is a curve, not a point.
The homogeneous substitution chain. For a homogeneous equation, the substitution \( y = vx \), together with \( \frac{dy}{dx} = v + x\frac{dv}{dx} \), turns \( \frac{dy}{dx} = g\left(\frac{y}{x}\right) \) into a separable equation in \( v \) and \( x \).
After integrating you must walk the chain backwards, replacing \( v \) by \( y/x \) to state the answer in \( x \) and \( y \) (NCERT, §9.4.2). Draw the chain as: original equation → substitute \( y = vx \) → separate and integrate → replace \( v \) by \( y/x \) → final answer.
The (Why?) in Example 10. Midway through Example 10 the book moves \( -\log|x| + C_1 \) across the equation and it reappears as \( +\frac{1}{2}\log x^2 \), with a (Why?) asking you to justify it. The reason: \( \log|x| = \frac{1}{2}\log x^2 \), because \( x^2 = |x|^2 \) gives \( \log x^2 = 2\log|x| \).
Writing the term as \( \frac{1}{2}\log x^2 \) puts both logarithms under the same coefficient, so they merge into one logarithm — \( \frac{1}{2}\log|v^2+v+1| + \frac{1}{2}\log x^2 = \frac{1}{2}\log|(v^2+v+1)x^2| \), which becomes \( \log|x^2 + xy + y^2| \) after \( v = y/x \).
Definitions every student must write correctly
Exam answers are judged on the precise meaning of these terms, so each one is given below in plain words. The wording is yours to use; the meaning matches the book (NCERT, §9.2–§9.4.3).
| Term | Plain definition | Where it appears |
|---|---|---|
| Differential equation | An equation that involves the derivative (or derivatives) of the dependent variable with respect to the independent variable. | §9.2 |
| Ordinary differential equation | A differential equation whose derivatives involve only one independent variable. | §9.2 |
| Order of a differential equation | The order of the highest order derivative present in the equation. | §9.2.1 |
| Degree of a differential equation | The highest power (a positive integer) of the highest order derivative — defined only when the equation is a polynomial in its derivatives. | §9.2.2 |
| Solution of a differential equation | A function \( \phi \) that satisfies the equation when substituted for \( y \): the left side becomes the right side. | §9.3 |
| General solution (primitive) | The solution containing as many arbitrary constants as the order of the equation. | §9.3 |
| Particular solution | The solution free of arbitrary constants, obtained by fixing the constants of the general solution. | §9.3 |
| Solution curve (integral curve) | The graph \( y = \phi(x) \) of a solution function. | §9.3 |
| Variable separable equation | An equation of the form \( \frac{dy}{dx} = h(y) \cdot g(x) \), where terms in \( y \) can be kept with \( dy \) and terms in \( x \) with \( dx \). | §9.4.1 |
| Homogeneous function of degree n | A function \( F(x, y) \) for which \( F(\lambda x, \lambda y) = \lambda^n F(x, y) \) for any nonzero constant \( \lambda \). | §9.4.2 |
| Homogeneous differential equation | An equation \( \frac{dy}{dx} = F(x, y) \) in which \( F \) is a homogeneous function of degree zero — so it depends only on \( y/x \). | §9.4.2 |
| Linear differential equation | An equation \( \frac{dy}{dx} + Py = Q \) with \( P, Q \) constants or functions of \( x \) only, or \( \frac{dx}{dy} + P_1x = Q_1 \) with \( P_1, Q_1 \) functions of \( y \) only. | §9.4.3 |
| Integrating factor | The multiplier \( e^{\int P\,dx} \) that turns the left side of a linear equation into the derivative of \( y \cdot (\text{the factor}) \). | §9.4.3 |
The formula sheet: order, degree and the three methods
Below is the chapter’s formula list — every solving formula it uses, with the meaning of each symbol, so a formula is never a string of letters you cannot read back.
| Formula | What each symbol means | Used in |
|---|---|---|
| \( \frac{dy}{dx} = g(x) \) | The opening problem of the chapter: find \( y \) from its derivative. \( g(x) \) is a known function of \( x \). | §9.1 |
| \( \frac{dy}{dx} = h(y) \cdot g(x) \) | Variable separable form: the right side is a product of a function of \( y \) and a function of \( x \). | §9.4.1 |
| \( \frac{dy}{h(y)} = g(x)\,dx \), then \( H(y) = G(x) + C \) | Separation step: \( H \) is an antiderivative of \( 1/h(y) \), \( G \) of \( g(x) \), and \( C \) is the arbitrary constant. | §9.4.1, Exercise 9.3 |
| \( y = vx \), \( \frac{dy}{dx} = v + x\frac{dv}{dx} \) | Homogeneous substitution: turns \( g(y/x) \) into \( g(v) \), making the equation separable in \( v \) and \( x \). | §9.4.2, Exercise 9.4 |
| \( \int \frac{dv}{g(v) – v} = \int \frac{dx}{x} \), then replace \( v \) by \( y/x \) | The homogeneous solution template; converting back to \( x, y \) is compulsory. If the equation is \( \frac{dx}{dy} = h\left(\frac{x}{y}\right) \), substitute \( x = vy \) instead. | §9.4.2 |
| \( \frac{dy}{dx} + Py = Q \), I.F. \( = e^{\int P\,dx} \) | Linear form with \( y \) as the dependent variable: \( P, Q \) are constants or functions of \( x \) only. | §9.4.3, Exercise 9.5 |
| \( y \cdot \text{I.F.} = \int (Q \times \text{I.F.})\,dx + C \) | The solution formula for the linear equation, written in one line. | §9.4.3 |
| \( \frac{dx}{dy} + P_1x = Q_1 \), I.F. \( = e^{\int P_1\,dy} \) | The matching linear form with \( x \) as the dependent variable; the factor is integrated with respect to \( y \). | §9.4.3 |
Why the integrating factor works. The chapter derives it in §9.4.3, and the reasoning is one line. We want the left side of \( \frac{dy}{dx} + Py = Q \) to become a single derivative. Multiply by an unknown function \( g(x) \); the product rule on \( y \cdot g(x) \) gives \( g\frac{dy}{dx} + g’y \).
Comparing with the multiplied equation \( g\frac{dy}{dx} + Pgy = Qg \) forces \( g’ = Pg \), that is, \( \frac{g’}{g} = P \). Integrating both sides gives \( \log g = \int P\,dx \), so \( g = e^{\int P\,dx} \). With that choice the left side collapses into \( \frac{d}{dx}(y \cdot g) \), and one integration finishes the problem.
How to choose the right method: a decision guide
The method is decided by the shape of \( \frac{dy}{dx} \)** before any solving starts. Run these tests in order — the first one that fits decides the method.
| Inspect the right side of \( \frac{dy}{dx} = F(x, y) \) | Type | What to do |
|---|---|---|
| \( F \) factors as \( g(x) \cdot h(y) \) — a product of something in \( x \) and something in \( y \) | Variable separable (§9.4.1) | Write \( \frac{dy}{h(y)} = g(x)\,dx \), integrate, add \( C \). |
| \( F \) depends only on the ratio \( y/x \) — replacing \( x, y \) by \( \lambda x, \lambda y \) leaves it unchanged | Homogeneous (§9.4.2) | Substitute \( y = vx \), separate in \( v \) and \( x \), then replace \( v \) by \( y/x \). |
| Can be rearranged to \( \frac{dy}{dx} + Py = Q \) with \( P, Q \) only in \( x \) | Linear, \( y \)-form (§9.4.3) | Find I.F. \( = e^{\int P\,dx} \), then \( y \cdot \text{I.F.} = \int (Q \times \text{I.F.})\,dx + C \). |
| Arranged as \( \frac{dx}{dy} + P_1x = Q_1 \) with \( P_1, Q_1 \) only in \( y \) | Linear, \( x \)-form (§9.4.3) | Use I.F. \( = e^{\int P_1\,dy} \), then \( x \cdot \text{I.F.} = \int (Q_1 \times \text{I.F.})\,dy + C \). |
Worked example 1 — variable separable. Solve \( \frac{dy}{dx} = \frac{2x + 3}{y + 1} \), given \( y(0) = 0 \).
Step 1 — recognise the shape.
The right side is \( (2x + 3) \cdot \frac{1}{y+1} \), a product of a function of \( x \) and a function of \( y \).
That is the variable separable form (NCERT, §9.4.1).
Step 2 — separate the variables.
Multiply both sides by \( y + 1 \) and by \( dx \):
\[ (y + 1)\,dy = (2x + 3)\,dx \]
Step 3 — integrate both sides.
\[ \int (y+1)\,dy = \int (2x+3)\,dx \;\Rightarrow\; \frac{y^2}{2} + y = x^2 + 3x + C \]
Step 4 — fix the constant.
Use \( y(0) = 0 \): both sides are \( 0 \) at \( x = 0 \), so \( C = 0 \).
Step 5 — solve for \( y \).
Multiply by 2 and complete the square:
\[ y^2 + 2y = 2x^2 + 6x \;\Rightarrow\; (y+1)^2 = 2x^2 + 6x + 1 \]
\[ y = -1 + \sqrt{2x^2 + 6x + 1} \]
Check the branch.
The squared form allows \( y = -1 \pm \sqrt{2x^2 + 6x + 1} \), but only the plus branch gives \( y(0) = -1 + 1 = 0 \), so the minus branch is rejected.
Final answer: \( y = -1 + \sqrt{2x^2 + 6x + 1} \).
Worked example 2 — homogeneous. Solve \( \frac{dy}{dx} = \frac{y}{x} + \left(\frac{y}{x}\right)^2 \), given \( y(1) = 1 \).
Step 1 — test homogeneity.
Replace \( x, y \) by \( \lambda x, \lambda y \): \( \frac{\lambda y}{\lambda x} + \left(\frac{\lambda y}{\lambda x}\right)^2 = \frac{y}{x} + \left(\frac{y}{x}\right)^2 \), unchanged.
The right side depends only on \( y/x \), so the equation is homogeneous of degree zero (NCERT, §9.4.2).
Step 2 — substitute \( y = vx \).
Then \( \frac{dy}{dx} = v + x\frac{dv}{dx} \), and the equation becomes:
\[ v + x\frac{dv}{dx} = v + v^2 \;\Rightarrow\; x\frac{dv}{dx} = v^2 \]
Step 3 — separate and integrate.
\[ \frac{dv}{v^2} = \frac{dx}{x} \;\Rightarrow\; -\frac{1}{v} = \log|x| + C \]
Step 4 — convert back to \( x \) and \( y \).
Replace \( v \) by \( y/x \) — the step students forget:
\[ -\frac{x}{y} = \log|x| + C \]
Step 5 — fix the constant.
With \( y(1) = 1 \): \( -1 = 0 + C \), so \( C = -1 \).
\[ -\frac{x}{y} = \log|x| – 1 \;\Rightarrow\; \frac{x}{y} = 1 – \log|x| \]
Final answer: \( y = \frac{x}{1 – \log|x|} \). Here \( \log \) is the natural logarithm, the convention used throughout the book.
Worked example 3 — linear. Solve \( \frac{dy}{dx} + 2y = e^x \), given \( y(0) = 1 \).
Step 1 — check the standard form.
The equation is already \( \frac{dy}{dx} + Py = Q \) with \( P = 2 \) and \( Q = e^x \), both functions of \( x \) only, so no division is needed (NCERT, §9.4.3).
Step 2 — write the integrating factor.
\[ \text{I.F.} = e^{\int 2\,dx} = e^{2x} \]
Step 3 — multiply and collapse the left side.
\[ e^{2x}\frac{dy}{dx} + 2e^{2x}y = e^{3x} \;\Rightarrow\; \frac{d}{dx}\left(y e^{2x}\right) = e^{3x} \]
Step 4 — integrate both sides.
\[ y e^{2x} = \int e^{3x}\,dx + C = \frac{e^{3x}}{3} + C \;\Rightarrow\; y = \frac{e^x}{3} + Ce^{-2x} \]
Step 5 — fix the constant.
With \( y(0) = 1 \): \( 1 = \frac{1}{3} + C \), so \( C = \frac{2}{3} \).
Final answer: \( y = \frac{e^x}{3} + \frac{2}{3}e^{-2x} \).
Notice the two final acts in every example: solve for \( y \) explicitly, then use the given point to eliminate the constant. A particular solution must contain no arbitrary constant at the end.
Common mistakes and how to fix them
The errors below are the ones students make most often in this chapter, each named before you make it, with the correction and a way to check your work.
| The mistake | The correct rule | How to check your answer |
|---|---|---|
| Quoting a degree for equations like \( \sin\left(\frac{dy}{dx}\right) = 0 \) or \( e^{y’} \) present | The degree exists only when the equation is a polynomial in derivatives (NCERT, §9.2.2, equations (9)–(11)). | If any derivative sits inside \( \sin, \cos, e, \log \), write degree not defined — never a number. |
| Reading the degree from a lower-order derivative’s power | The degree is the highest power of the highest order derivative (Example 1(ii)). | Locate the highest derivative first; only its power counts. |
| Reading \( P \) and \( Q \) before the coefficient of \( \frac{dy}{dx} \) is 1 | Divide through first. Example 15 rewrites \( x\frac{dy}{dx} + 2y = x^2 \) as \( \frac{dy}{dx} + \frac{2}{x}y = x \). | The form must be exactly \( \frac{dy}{dx} + Py = Q \) with nothing in front of \( \frac{dy}{dx} \). |
| Forgetting to replace \( v \) by \( y/x \) after integrating a homogeneous equation | The final answer must be in \( x \) and \( y \); \( v \) is only a working variable. | If your answer still contains \( v \), it is incomplete. |
| Wrong variable in the integrating factor for the \( x \)-form | For \( \frac{dx}{dy} + P_1x = Q_1 \), the factor is \( e^{\int P_1\,dy} \) — integrated with respect to \( y \). | The factor’s integration variable must match the derivative variable on the left. |
| Miscounting arbitrary constants | The general solution of an \( n \)th order equation has exactly \( n \) arbitrary constants; a particular solution has none (Exercise 9.2 Q11–12). | Count the constants left in the final answer — zero after a particular-solution condition is applied. |
| Losing a solution when separating variables | Separation divides by \( h(y) \), so the case \( h(y) = 0 \) must be checked separately (Example 6 discards \( y = 0 \) as a possible solution). | Set the denominator of the separated form to zero and test whether that constant function satisfies the original equation. |
Exam notes: how this chapter is asked
This section maps the question shapes the chapter produces, in the order worth revising them. It makes no promises about specific questions appearing — these are the shapes the book itself builds, in the chapter’s own order: classify, verify, solve, apply.
- Order and degree open the chapter: Exercise 9.1 Q1–10 and the objective questions Q11–12, plus Miscellaneous Q1. These are the quickest marks once the polynomial-in-derivatives test is automatic.
- Verification is pure differentiation practice: Exercise 9.2 Q1–10 and Examples 2–3. Differentiate the given function, substitute into the left side, and show it reduces to the right side.
- Solving is drilled one method per exercise, and each exercise ends in an objective-type question: Exercise 9.3 (separable; Q1–10 general, Q11–14 particular, Q23 objective), Exercise 9.4 (homogeneous; Q1–10 general, Q11–15 particular, Q16–17 objective), Exercise 9.5 (linear; Q1–12 general, Q13–15 particular, Q18–19 objective). Use the closing objective questions as self-tests.
- Applications follow two scripts: curves from a given tangent slope, and growth or interest models where the rate is proportional to the quantity. The Miscellaneous Exercise then mixes every method, including the hardest recognition cases.
A complete answer should show the separated or substituted equation before integrating, the arbitrary constant written explicitly, the substitution \( v \to y/x \) performed for homogeneous equations, and the constant evaluated whenever a condition is given. For order-and-degree questions, state degree not defined when it is not, instead of leaving the answer blank.
The application questions in Exercise 9.3 map to their scripts like this:
| Application | The script behind it | Where it appears |
|---|---|---|
| Equation of a curve whose tangent slope is given | Slope means \( \frac{dy}{dx} \); write \( \frac{dy}{dx} = \) (the given expression), solve, then fix \( C \) with the point the curve passes through. | Exercise 9.3 Q15–18 (also Examples 7, 8, 13, 18) |
| Bank interest, principal growing continuously | \( \frac{dP}{dt} = \frac{r}{100}P \) — the rate is proportional to the principal itself. | Exercise 9.3 Q20–21 (also Example 9) |
| Bacteria growth | \( \frac{dN}{dt} = kN \) — growth proportional to the number present. | Exercise 9.3 Q22 |
| Balloon being inflated | The constant-rate variant: \( \frac{dV}{dt} = \) constant, with volume linked to radius by the sphere formula \( V = \frac{4}{3}\pi r^3 \). | Exercise 9.3 Q19 |
Textbook contents and the examinable syllabus are not always identical — check the current official CBSE syllabus before the exam.
Revision summary: the chapter in ten lines
This condenses the NCERT Summary into the ten facts that carry the whole chapter — read these last, and they should trigger every formula above.
- A differential equation is an equation involving derivatives of the dependent variable with respect to the independent variable.
- The order is the order of the highest order derivative present.
- The degree is defined only when the equation is a polynomial in its derivatives.
- The degree, when defined, is the highest power (a positive integer) of the highest order derivative.
- A solution is any function that satisfies the equation.
- The general solution carries as many arbitrary constants as the order; the particular solution carries none.
- The variable separable method applies when every \( y \)-term can stay with \( dy \) and every \( x \)-term with \( dx \), then both sides integrate.
- A homogeneous differential equation has \( \frac{dy}{dx} = f(x, y) \) where \( f \) is a homogeneous function of degree zero, and is solved by the substitution \( y = vx \).
- A first order linear differential equation has the form \( \frac{dy}{dx} + Py = Q \) with \( P, Q \) of \( x \) only, and is solved by multiplying by the integrating factor \( e^{\int P\,dx} \).
- For the \( x \)-dependent form \( \frac{dx}{dy} + P_1x = Q_1 \), the same recipe runs with \( e^{\int P_1\,dy} \) and \( x \cdot \text{I.F.} = \int (Q_1 \times \text{I.F.})\,dy + C \).
Related resources
This listing is maintained for the 2026-27 academic session using the NCERT textbook information available to us. NCERT remains the authority for confirming the latest edition.
If this chapter is settled, the chapters that use the same toolkit are the next places to go: the Application of Integrals chapter and the Determinants chapter feed the same problem-solving muscles, and the Class 12 Mathematics notes hub keeps every chapter in one place. For the wider picture, the Class 12 hub and the main CBSE notes index cover all subjects.
Reference: NCERT Class 12 Mathematics Part II textbook, chapter 3 (Differential Equations), official edition on ncert.nic.in.
Sources and data verification
The figures on this page describe the NCERT Class 12 Mathematics Part II textbook, Differential Equations chapter, in its current rationalised edition. This page covers only this chapter of this book, not other chapters or other subjects.
NCERT publishes the official textbook PDFs at ncert.nic.in and settles textbook content, editions and numbering. CBSE settles the curriculum, syllabus and examinations — textbook contents and the examinable syllabus are not always identical, so check the current official syllabus.
This page is maintained for the current academic session using the NCERT information available to us.
Frequently asked questions
When is the degree of a differential equation not defined?
When the equation is not a polynomial in its derivatives. If any derivative appears inside \( \sin, \cos, e \) or \( \log \) — as in \( \frac{dy}{dx} + \sin\left(\frac{dy}{dx}\right) = 0 \), equation (11) of §9.2.2, or \( \sin(y”) \) — the degree has no meaning, even though the order is still perfectly defined.
Exercise 9.1 Q1, Q4 and Q11 are built exactly for this: write degree not defined, never a guess.
How many arbitrary constants are there in the general solution of a fourth order differential equation?
Four. The general solution of an \( n \)th order equation carries exactly \( n \) arbitrary constants, because each integration introduces one (NCERT, §9.3 and Summary). Exercise 9.2 Q11 asks this directly and the correct option is (D) 4. The particular solution of a third order equation, asked in Exercise 9.2 Q12, carries none — option (D) 0.
How do I decide between the variable separable, homogeneous and linear methods?
By the shape of the right side of \( \frac{dy}{dx} = F(x, y) \), before you touch a pen. First test: does \( F \) factor as \( g(x) \cdot h(y) \)? Then it is separable. Second test: does \( F \) depend only on \( y/x \) — is it unchanged when \( x, y \) are replaced by \( \lambda x, \lambda y \)?
Then it is homogeneous; substitute \( y = vx \). Third test: can it be rearranged to \( \frac{dy}{dx} + Py = Q \) with \( P, Q \) of \( x \) only? Then it is linear; use the integrating factor. The decision table above runs exactly these tests in order.
Why do we multiply a linear differential equation by an integrating factor?
To turn the left side into a single derivative so that one integration solves the equation. Multiplying \( \frac{dy}{dx} + Py = Q \) by an unknown function \( g(x) \) and comparing the left side with the product rule for \( \frac{d}{dx}(y \cdot g(x)) \) forces \( g’ = Pg \), which integrates to \( g = e^{\int P\,dx} \) (NCERT, §9.4.3).
After multiplying, the left side is exactly \( \frac{d}{dx}\left(y \cdot e^{\int P\,dx}\right) \), and both sides integrate directly.
In a homogeneous differential equation, when do I substitute x = vy instead of y = vx?
When the equation is naturally written with \( x \) as the dependent variable — the form \( \frac{dx}{dy} = h\left(\frac{x}{y}\right) \) (NCERT, §9.4.2, note).
If the ratios that appear are \( x/y \), for example a term like \( e^{x/y} \), then \( x = vy \) is the cleaner substitution, and you differentiate with respect to \( y \): \( \frac{dx}{dy} = v + y\frac{dv}{dy} \). At the end, replace \( v \) by \( x/y \).
Example 12 and Exercise 9.4 Q10 run this script, and Exercise 9.4 Q16 checks that you know which substitution to pick.
How do I find the equation of a curve when the slope of its tangent is given?
Translate the sentence into mathematics first: the slope of the tangent at \( (x, y) \) is \( \frac{dy}{dx} \), so slope at any point is equal to that expression becomes \( \frac{dy}{dx} = \) that expression (Examples 7, 8, 13 and 18 all follow this). Solve the differential equation by the method its shape dictates, and the answer is a family of curves.
The point the curve passes through fixes the arbitrary constant, leaving the equation of that single curve.
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