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Differential Equations

Chapter 9Notes + practice

CBSE Class 12 Mathematics · NCERT Mathematics Part-II

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Shishya's notes

What this chapter is about

A differential equation is an equation that involves an unknown function and one or more of its derivatives. In Class 12, you learn to identify, form and solve such equations. This chapter connects your knowledge of differentiation and integration: differentiation creates the equation from a family of curves, and integration recovers the solution.

You study the order and degree of a differential equation, how to form a differential equation from a given general solution, and several methods to solve first-order equations. The methods include variable-separable form, homogeneous equations, and linear differential equations of the type dy/dx + Py = Q. Each method converts the equation into a form where integration can be applied directly.

After this chapter you should be able to classify a differential equation by order and degree, write the differential equation whose general solution is a given family of curves, recognise which solving method applies, and carry out the solution to express y in terms of x (or vice versa).

Key ideas

  • A differential equation contains derivatives such as dy/dx, d²y/dx², etc., along with x, y and constants.
  • The order is the highest derivative that appears; the degree is the power of that highest derivative after the equation is made free of radicals and fractions in derivatives.
  • The number of arbitrary constants in the general solution equals the order of the equation.
  • Variable-separable equations can be written as f(y) dy = g(x) dx; integrate both sides directly.
  • A homogeneous equation of degree zero in x and y is solved by the substitution y = vx (or x = vy), which converts it to variable-separable form.
  • A linear first-order equation dy/dx + P(x)y = Q(x) is solved using the integrating factor IF = e^(∫P dx); multiply through and integrate.
  • A particular solution is obtained by assigning specific values to the arbitrary constants using initial or boundary conditions.

Formulas and facts to remember

  • Item: Order · Formula / Rule: Highest derivative in the equation · Meaning: Tells how many integrations are needed in principle
  • Item: Degree · Formula / Rule: Exponent of highest-order derivative (equation polynomial in derivatives) · Meaning: Must remove radicals/fractions first
  • Item: Variable-separable · Formula / Rule: ∫f(y) dy = ∫g(x) dx + C · Meaning: Separate y-terms to one side, x-terms to the other
  • Item: Homogeneous substitution · Formula / Rule: y = vx ⇒ dy/dx = v + x dv/dx · Meaning: Converts dy/dx = F(y/x) to separable form
  • Item: Linear equation form · Formula / Rule: dy/dx + Py = Q · Meaning: P and Q are functions of x only
  • Item: Integrating factor (IF) · Formula / Rule: IF = e^(∫P dx) · Meaning: Multiply entire equation by IF
  • Item: Solution of linear equation · Formula / Rule: y × IF = ∫(Q × IF) dx + C · Meaning: After multiplying by IF, left side is d/dx (y × IF)

Worked examples

Example 1 – Finding order and degree

Equation: (d²y/dx²)³ + 2(dy/dx)² − y = 0

Step 1: Identify the highest derivative. Here it is d²y/dx², so the order is 2.

Step 2: The equation is already polynomial in the derivatives. The highest-order derivative d²y/dx² is raised to the power 3, so the degree is 3.

Answer: Order = 2, Degree = 3.


Example 2 – Solving a variable-separable equation

Solve dy/dx = (1 + y²)/(1 + x²).

Step 1: Separate variables. dy/(1 + y²) = dx/(1 + x²)

Step 2: Integrate both sides. ∫dy/(1 + y²) = ∫dx/(1 + x²) tan⁻¹ y = tan⁻¹ x + C

Step 3: Write the general solution. tan⁻¹ y − tan⁻¹ x = C, or equivalently y = tan(tan⁻¹ x + C).


Example 3 – Solving a linear first-order equation

Solve dy/dx + (2/x)y = x², given y(1) = 1.

Step 1: Identify P = 2/x and Q = x².

Step 2: Find the integrating factor. IF = e^(∫2/x dx) = e^(2 ln|x|) = x².

Step 3: Multiply the equation by IF. x² dy/dx + 2xy = x⁴ The left side is d/dx (x² y).

Step 4: Integrate. x² y = ∫x⁴ dx = x⁵/5 + C y = x³/5 + C/x².

Step 5: Use the initial condition y(1) = 1. 1 = 1/5 + C ⇒ C = 4/5.

Particular solution: y = x³/5 + 4/(5x²).

Common mistakes

  • Forgetting to remove a square root before finding the degree → clear radicals first so the equation is polynomial in derivatives.
  • Dropping the arbitrary constant after integration → always include + C; its value is fixed only when an initial condition is given.
  • Substituting y = vx but leaving dy/dx unchanged → replace dy/dx by v + x dv/dx.
  • Using IF = ∫P dx instead of IF = e^(∫P dx) → the integrating factor is an exponential, not the integral itself.
  • Mixing up order and degree → order counts which derivative, degree counts its power.

Quick revision

  • Order = highest derivative present; Degree = power of that derivative (after rationalising).
  • Variable-separable: move all y and dy to one side, all x and dx to the other, then integrate.
  • Homogeneous: substitute y = vx, simplify, separate, integrate, replace v by y/x.
  • Linear dy/dx + Py = Q: multiply by IF = e^(∫P dx), then integrate to get y × IF = ∫Q × IF dx + C.
  • Number of arbitrary constants in general solution = order of the equation.
  • Apply initial conditions only after finding the general solution to obtain the particular solution.

Written by Shishya's AI on 26 Sept 2026 from the chapter's title and class level, in Shishya's own words — not a copy or summary of the textbook. Read the official chapter for the book's own text, activities and exercises.

Practice: 5 questions on Differential Equations

One question at a time, with the answer and a short explanation after each. No account needed, and no result is saved to any account or profile: Shishya records only an anonymous usage event (which chapter was practised and the score).

These practice questions are Shishya's own, written by AI and answer-checked before they are shown. They are not taken from the NCERT book or any board paper.