Differential Equations Class 12 Notes

These differential equations class 12 notes condense Chapter 9 of the NCERT Mathematics Part II textbook into one revision page for the 2026-27 session. You get order and degree, general and particular solutions, the three solving methods with formulas, worked examples with fresh numbers, and exam cautions that cost marks.

Every claim carries its textbook page number so you can cross-check anything you doubt, including in the Class 12 Mathematics notes hub.

What is a differential equation? The definition in plain words

An algebraic equation such as \(x + y = 7\) ties variables together. A differential equation goes further — it contains a derivative of the dependent variable with respect to the independent variable (NCERT, p. 301).

For instance, \(x\frac{dy}{dx} + y = 0\) involves the derivative \(\frac{dy}{dx}\), so it is a differential equation. The equation \(x + y = 7\) has no derivative and is merely algebraic.

When all derivatives are taken with respect to only one independent variable, the equation is an ordinary differential equation (NCERT, p. 302). This chapter restricts itself to ordinary differential equations; equations with several independent variables belong to the partial type and are left aside.

The chapter uses a compact derivative notation that you must recognise on sight:

Each dash adds one order, and the subscript form \(y_n\) is used for high-order derivatives where dashes become clumsy (NCERT note, p. 302).

Order and degree of a differential equation

Order is the order of the highest-order derivative present in the equation (NCERT, p. 302). So an equation containing \(\frac{d^2y}{dx^2}\) but no higher derivative has order 2.

Degree is the highest power of that highest-order derivative, and it is defined only when the equation is a polynomial in the derivatives (NCERT, p. 303). Three quick examples fix the idea:

• \(\left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^3 + y = 0\) → order 2, degree 2
• \(y”’ + \sin\left(\frac{dy}{dx}\right) = 0\) → order 3, degree not defined
• \(\left(\frac{dy}{dx}\right)^3 + y = 0\) → order 1, degree 3

Memory device: the degree is the power sitting on the highest-order derivative — “the exponent on the top step”. Degree is left undefined whenever a derivative hides inside a function such as sine or \(e\), because you cannot take a “power of a sin”.

Both order and degree, when degree is defined, are always positive integers (NCERT, p. 303).

The last two items of Exercise 9.1 (p. 304) are one-mark standbys: Q11 gives an equation containing \(\sin\left(\frac{dy}{dx}\right)\), so its degree is not defined; Q12 asks for the order of \(2x^2\frac{d^2y}{dx^2} – 3\frac{dy}{dx} + y = 0\), which is 2.

General solution vs particular solution: why the constants appear

A solution of a differential equation is a function that satisfies the equation when it and its derivatives are substituted in; its graph is the solution curve or integral curve (NCERT, p. 305).

The general solution contains arbitrary constants — an equation of order \(n\) yields \(n\) of them. The particular solution comes from fixing those constants to specific values, so it carries none (NCERT, p. 305).

Take \(\frac{d^2y}{dx^2} + y = 0\). The function \(y = a\sin(x + b)\) satisfies it because \(y” = -a\sin(x+b)\), so \(y” + y = 0\) — two constants for order two. Setting \(a = 2\) and \(b = \frac{\pi}{4}\) yields the particular solution \(y = 2\sin\left(x + \frac{\pi}{4}\right)\) (NCERT, p. 305).

To verify that a given function is a solution — the task throughout Exercise 9.2 (p. 307) — differentiate it as many times as the order demands, substitute the derivatives into the left side, and check that it equals the right side. Examples 2 and 3 (p. 306) do exactly this for \(y = e^{-3x}\) and \(y = a\cos x + b\sin x\).

Exercise 9.2 Q11-12 quiz the constant count: the general solution of a fourth-order equation has 4 constants, and a particular solution of a third-order equation has 0.

Key terms at a glance: definitions table

Method 1: Variables separable — split, integrate, done

A first-order equation \(\frac{dy}{dx} = F(x,y)\) is of the variable separable type when \(F(x,y)\) splits into a product \(g(x)h(y)\) — an x-part times a y-part (NCERT, p. 307).

1. Provided \(h(y) \neq 0\), rewrite as \(\frac{1}{h(y)}\,dy = g(x)\,dx\).
2. Integrate both sides: \(\int \frac{1}{h(y)}\,dy = \int g(x)\,dx\).
3. Write the result as \(H(y) = G(x) + C\) with one arbitrary constant (NCERT, p. 307).

The condition \(h(y) \neq 0\) is the algebraic guard: you divided by \(h(y)\), so a zero in the denominator is not allowed. In Example 4, \(\frac{dy}{dx} = \frac{x+1}{2-y}\) separates to \((2-y)\,dy = (x+1)\,dx\) and the stated restriction is \(y \neq 2\) (NCERT, p. 308).

The same idea carries the chapter’s first applications: Example 8 finds a curve from its slope \(\frac{dy}{dx} = \frac{2x}{y^2}\) (p. 310), and Example 9 models bank principal growing at 5% per year, \(\frac{dP}{dt} = \frac{P}{20}\), separating to \(P = 1000e^{t/20}\) (pp. 310–311).

Exercise 9.3 (pp. 311–312) gives 23 items — general solutions, particular solutions, curve questions, and bank and bacteria problems. The integration step leans on standard antiderivatives, so refresh them from our integrals notes if needed.

Constant habit: fold every constant into a single \(C\) at the end instead of carrying two. Example 4 integrates to \(2y – \frac{y^2}{2} = \frac{x^2}{2} + x + C_1\) and then renames \(2C_1\) as one \(C\) (NCERT, p. 308).

Method 2: Homogeneous equations — substitute y = vx

A function \(F(x,y)\) is homogeneous of degree \(n\) when \(F(\lambda x, \lambda y) = \lambda^n F(x,y)\) for any nonzero constant \(\lambda\). Every such function can be rewritten as \(x^n g\left(\frac{y}{x}\right)\) or \(y^n h\left(\frac{x}{y}\right)\) (NCERT, p. 313).

A differential equation \(\frac{dy}{dx} = F(x,y)\) is homogeneous when \(F\) has degree zero — that is, the right-hand side is a function of \(\frac{y}{x}\) only (NCERT, p. 314).

Recognition shortcut: divide every term by the highest power of \(x\). If the result still contains only the ratio \(\frac{y}{x}\), the equation is homogeneous. Example 13 rewrites \(\frac{x^2 + y^2}{2xy}\) as \(\frac{1 + (y/x)^2}{2(y/x)}\), a function of \(\frac{y}{x}\) alone (NCERT, pp. 320–321).

Solving routine:

1. Put \(y = vx\).
2. Then \(\frac{dy}{dx} = v + x\frac{dv}{dx}\) — both terms, by the product rule.
3. Substitute both into the equation; it becomes separable in \(v\) and \(x\).
4. Integrate: \(\int \frac{dv}{g(v) – v} = \int \frac{dx}{x} + C\) (NCERT, p. 314).
5. Replace \(v\) by \(\frac{y}{x}\) to write the general solution.

Textbook patterns worth practising: the \(\cos(y/x)\) form in Example 11 (p. 317) and the \(e^{x/y}\) form in Example 12 (pp. 318–320). Example 12 also shows the swap to \(x = vy\) when the equation is written as \(\frac{dx}{dy}\) (NCERT, p. 314).

Exercise 9.4 (p. 322) contains 17 items: the first ten ask you to show the equation is homogeneous and solve, the next five demand particular solutions, and Q16–17 are one-mark checks — Q16 wants the correct substitution (\(x = vy\)) and Q17 which of four equations is homogeneous.

Method 3: Linear equations — the integrating factor

Standard form: \(\frac{dy}{dx} + Py = Q\), where \(P\) and \(Q\) are functions of \(x\) (or constants). If the coefficient of \(\frac{dy}{dx}\) is not 1, divide through first — \(x\frac{dy}{dx} + 2y = x^2\log x\) becomes \(\frac{dy}{dx} + \frac{2}{x}y = x\log x\), so \(P = \frac{2}{x}\).

The recipe: integrating factor \(IF = e^{\int P\,dx}\), then \(y\cdot IF = \int Q \cdot IF \, dx + C\). Exercise 9.5 Q1–17 (p. 329) drill the full pattern, and Q18–19 ask for the IF alone.

Why it works: \(\frac{d}{dx}(IF) = IF\cdot P\), so multiplying the standard form by IF turns the left side into one derivative: \(IF\frac{dy}{dx} + IF\cdot Py = \frac{d}{dx}(y\cdot IF) = IF\cdot Q\). The left side integrates directly — that is the whole trick.

In Q18, \(x\frac{dy}{dx} – y = 2x^2\) divides to \(\frac{dy}{dx} – \frac{1}{x}y = 2x\), so \(P = -\frac{1}{x}\) and \(IF = e^{-\int \frac{1}{x}dx} = \frac{1}{x}\). In Q19 the roles are swapped: the equation is \(\frac{dx}{dy} + \frac{y}{1-y^2}x = \frac{ay}{1-y^2}\), so the IF follows from \(P = \frac{y}{1-y^2}\).

The textbook derives this method on pp. 323–328; if you need the full argument, open the official chapter PDF (lemh203.pdf) hosted at ncert.nic.in and read through the build-up before the exercises.

Differential equations class 12 notes: the three methods side by side

Swapped case: for a homogeneous equation given as \(\frac{dx}{dy} = h\left(\frac{x}{y}\right)\), swap the roles and put \(x = vy\) (NCERT, p. 314).

Worked examples with fresh numbers: copy these steps

Worked example 1: Variables separable with a particular solution

Worked example 2: Homogeneous — substitute y = vx

Worked example 3: Linear with the integrating factor

Where differential equations show up: the three NCERT problem families

NCERT wraps the chapter’s word problems into three families. Knowing the family tells you the model, and the model tells you the method.

• Geometry of curves: the slope of the tangent is given as a function of \(x\) and \(y\), and you find the curve. Example 8 (p. 310), Example 13 (pp. 320–321), Exercise 9.3 Q15–18.
• Growth and decay: a quantity changes at a rate proportional to itself — bank principal at \(\frac{dP}{dt} = \frac{P}{20}\) (Example 9, pp. 310–311) and the bacteria count doubling (Exercise 9.3 Q22). These are separable.
• Rates tied to a known formula: a spherical balloon inflating with \(V = \frac{4}{3}\pi r^3\) and constant \(\frac{dV}{dt}\) (Exercise 9.3 Q19).

Fresh application — Newton’s law of cooling. A hot drink cools at a rate proportional to its temperature gap with the room: \(\frac{dT}{dt} = -k(T – T_{\text{room}})\). This is exactly the separable shape of the bank problem — gather the \(T\)-terms on one side and integrate.

Problem setups like this lead naturally into the area-and-curve work in our application of integrals notes.

Common mistakes that cost marks

Exam notes: what earns the mark

• Eight one-mark items decide the chapter’s vocabulary marks: order (Exercise 9.1 Q12), degree (Q11), constants in the general and particular solution (Exercise 9.2 Q11–12), the right substitution (Exercise 9.4 Q16), which equation is homogeneous (Q17), and the integrating factor (Exercise 9.5 Q18–19, p. 329). Practise exactly these eight.
• In show-it-is-homogeneous questions, writing \(F(\lambda x, \lambda y) = \lambda^0 F(x,y)\) or rewriting the RHS as \(g(y/x)\) earns a visible step; skipping it forfeits a mark (Examples 10–12, pp. 315–320).
• Verification questions (Exercise 9.2) are marked on the substitution: compute the derivatives, replace them in the LHS, and write “LHS = RHS” explicitly — not just the function.
• For curve and application problems, close with the units (rupees, years, hours) and give \(y\) explicitly when the question asks for the curve (Examples 7–9, Exercise 9.3 Q15–22).

Revision summary: the whole chapter on one page

• Definition and notation: \(\frac{dy}{dx} = y’\), \(\frac{d^2y}{dx^2} = y”\), and \(y_n = \frac{d^n y}{dx^n}\) (NCERT, p. 302).
• Order: order of the highest derivative present. Degree: power of that derivative, only when the equation is a polynomial in derivatives (NCERT, pp. 302–303).
• Solutions: general solution carries \(n\) arbitrary constants for order \(n\); particular solution carries none (NCERT, p. 305).
• Three methods: use the formula-sheet table above — separable (split and integrate), homogeneous (\(y = vx\)), linear (\(IF = e^{\int P\,dx}\)).

Reference: NCERT Class 12 Mathematics textbook, chapter Differential Equations.

Frequently asked questions about differential equations

When is the degree of a differential equation not defined?

Degree is undefined whenever the equation is not a polynomial in the derivatives — the classic case is a derivative appearing inside a function, like \(\sin\left(\frac{dy}{dx}\right)\) or \(e^{y’}\) (NCERT, p. 303). For example, \(\frac{dy}{dx} + \sin\left(\frac{dy}{dx}\right) = 0\) has order 1 but its degree is not defined.

Note that \(\sin y\) alone does not destroy the degree, because \(y\) is not a derivative.

How many arbitrary constants does the general solution of an nth order differential equation contain?

Exactly \(n\). The second-order equation \(\frac{d^2y}{dx^2} + y = 0\) gives the general solution \(y = a\sin(x+b)\) with two constants (NCERT, p. 305), and Exercise 9.2 Q11 answers 4 for a fourth-order equation. A particular solution, obtained by fixing those constants, contains 0 (Q12).

How do I decide between y = vx and x = vy in a homogeneous differential equation?

Use \(y = vx\) when the equation is written as \(\frac{dy}{dx} = g\left(\frac{y}{x}\right)\) — the RHS is a function of \(y/x\). Use \(x = vy\) when it is \(\frac{dx}{dy} = h\left(\frac{x}{y}\right)\), the RHS a function of \(x/y\) (NCERT, p. 314).

Exercise 9.4 Q16 asks exactly this, and the correct substitution is \(x = vy\).

What is the integrating factor and when do I use it?

It is \(IF = e^{\int P\,dx}\), used to solve first-order linear equations written as \(\frac{dy}{dx} + Py = Q\). Multiply both sides by IF, and the left side collapses into \(\frac{d}{dx}(y\cdot IF)\), so \(y\cdot IF = \int Q\cdot IF\, dx + C\). Exercise 9.5 Q18–19 (p. 329) test the IF directly.

How do I find a particular solution from a general solution?

Solve for the general solution with its arbitrary constant \(C\), substitute the given point into that general form to find \(C\), then write the answer with that \(C\). The substitution with \(C\) still present is the mark-bearing step — that is how Examples 6 and 7 (p. 309) proceed.

How do I verify that a given function is a solution of a differential equation?

Differentiate the function as many times as the order of the equation, substitute the derivatives into the LHS, and show it equals the RHS. Example: \(y = e^{-3x}\) satisfies \(y” + y’ – 6y = 0\) because \(9e^{-3x} – 3e^{-3x} – 6e^{-3x} = 0\) (Example 2, NCERT p. 306).