Question 111 markMCQSection A

The integrating factor of differential equation R(dxdy)+Px=QR\left(\frac{dx}{dy}\right) + Px = Q, where P, Q, R are functions of y, is

Differential Equations
Integrating Factor of a Linear Differential Equation (x as function of y)

Options

(A)e∫(P/Q) dye^{\int (P/Q)\,dy}
(B)e∫P dye^{\int P\,dy}
(C)e∫(P/R) dye^{\int (P/R)\,dy}
(D)e∫(P/R) dxe^{\int (P/R)\,dx}

Official answer

Correct option: (C) e∫(P/R) dye^{\int (P/R)\,dy}


Dividing throughout by R converts the equation to the standard linear form dxdy+(PR)x=QR\frac{dx}{dy} + \left(\frac{P}{R}\right)x = \frac{Q}{R}, whose integrating factor is e raised to the integral of (coefficient of x) with respect to y, i.e. e∫(P/R) dye^{\int (P/R)\,dy}.

Answer keywords:integrating factorlinear differential equationdx/dy formstandard formfirst order first degreedifferential equations class 12

Marking scheme

  1. 1.1 mark: awarded only for selecting option (C); no partial credit.

Hint

First divide the whole equation by R to get standard linear form dxdy+(PR)x=QR\frac{dx}{dy} + \left(\frac{P}{R}\right)x = \frac{Q}{R}, then apply I.F. = e^(integral of the coefficient of x, with respect to y).

Quick oral answer

Divide by R to get dxdy+(PR)x=QR\frac{dx}{dy} + \left(\frac{P}{R}\right)x = \frac{Q}{R} in standard linear form, so the integrating factor is e to the integral of (P/R) with respect to y.

Common mistakes

  1. 1.Not dividing by R before identifying the coefficient of x, leading to option (B).
  2. 2.Confusing which side (P or Q) forms the numerator of the coefficient, leading to option (A).
  3. 3.Integrating with respect to x instead of y, since the equation here is linear in x with y as the independent variable.

Analysis and explanation

Concept: Standard form of a linear differential equation in x.


A linear differential equation in x (as a function of y) has the standard form dxdy+P1x=Q1\frac{dx}{dy} + P_1 x = Q_1, and its integrating factor (I.F.) is e∫P1 dye^{\int P_1\,dy}. Here, dividing R(dxdy)+Px=QR\left(\frac{dx}{dy}\right) + Px = Q throughout by R gives dxdy+(PR)x=QR, so P1=PR\frac{dx}{dy} + \left(\frac{P}{R}\right)x = \frac{Q}{R}, \text{ so } P_1 = \frac{P}{R}. Hence I.F. = e∫(P/R) dye^{\int (P/R)\,dy}.


Why the distractors fail:

  • (A) e∫(P/Q) dye^{\int (P/Q)\,dy} wrongly uses Q, the term on the right-hand side, in place of R, the coefficient of dx/dy — a mismatch of roles.
  • (B) e∫P dye^{\int P\,dy} forgets to divide by R first, treating the equation as already in standard form when it is not (R is not necessarily 1).
  • (D) e∫(P/R) dxe^{\int (P/R)\,dx} uses the wrong variable of integration; since the equation is linear in x with y as the independent variable, the integrating factor must be integrated with respect to y, not x.

Exam tip: Always reduce a linear differential equation to standard form (coefficient of the derivative = 1) before reading off P1 and computing the I.F.; also check which variable is independent.

Interesting facts

Linear differential equations of the form dxdy+P1x=Q1\frac{dx}{dy} + P_1 x = Q_1 (x as dependent variable) are the mirror-image of the more commonly taught dydx+Py=Q\frac{dy}{dx} + Py = Q, but use exactly the same integrating factor method.

The method of integrating factors for linear ODEs traces back to Leibniz and the Bernoulli brothers in the late 17th century, over 300 years before it entered the CBSE syllabus.

Question details

Q11 of 44 questions

Board
CBSE
Class
Class 12
Subject
Mathematics
Year
2026
Section
Section A
Marks
1
Type
MCQ
Chapter
Differential Equations
Topic
Integrating Factor of a Linear Differential Equation (x as function of y)
Bloom's level
understand
Syllabus topic
Unit 4: Differential Equations — Linear Differential Equations: Integrating Factor of dxdy+(PR)x=QR\frac{dx}{dy} + \left(\frac{P}{R}\right)x = \frac{Q}{R}
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why do we divide by R first?

Because the integrating factor formula e∫P1 dye^{\int P_1\,dy} only applies when the equation is in standard form dxdy+P1x=Q1\frac{dx}{dy} + P_1 x = Q_1, i.e., when the coefficient of dx/dy is exactly 1; dividing by R achieves this.

Why is the integration with respect to y and not x?

Because here x is treated as the dependent variable and y as the independent variable (the equation is written with dxdy\frac{dx}{dy}), so the integrating factor must be a function of y, integrated with respect to y.