Board 2026 Class 12 Maths Q11 Solution: Integrating factor of R dx/dy + Px = Q

The question

The integrating factor of differential equation $R\frac{dx}{dy} + Px = Q$ where P, Q, R are functions of y is

(A) $e^{\int \frac{P}{Q}\,dy}$

(B) $e^{\int P\,dy}$

(C) $e^{\int \frac{P}{R}\,dy}$

(D) $e^{\int \frac{P}{R}\,dx}$

Exam answer

Dividing by $R$ ⁠:

\frac{dx}{dy} + \frac{P}{R}\,x = \frac{Q}{R}

Why?

The standard linear form needs the coefficient of $\frac{dx}{dy}$ to be 1, so every term is divided by $R$ (with $R \ne 0$ ⁠).

This is linear in $x$ ⁠, of the form $\frac{dx}{dy} + P_1x = Q_1$ with $P_1 = \frac{P}{R}$ ⁠.

Why?

Here $x$ is the dependent variable and $y$ the independent one. So the linear form is

\frac{dx}{dy} + P_1x = Q_1

with $P_1$ and $Q_1$ functions of $y$ only.

$\therefore\ \text{I.F.} = e^{\int P_1\,dy} = e^{\int \frac{P}{R}\,dy}$

Why?

For this form the integrating factor is $e^{\int P_1\,dy}$ ⁠. Multiplying by it turns the left side into one derivative:

\begin{aligned} &\frac{d}{dy}\left(x\,e^{\int P_1\,dy}\right) \\ &= e^{\int P_1\,dy}\left(\frac{dx}{dy} + P_1x\right) \end{aligned}

Watch out: integrate with respect to $y$ ⁠, because $P_1$ is a function of $y$ ⁠. (D) uses $dx$ ⁠, and (B) forgets to divide by $R$ ⁠.

Answer: (C) $e^{\int \dfrac{P}{R}\,dy}$

Where this comes from

Chapter 9, Differential Equations: Linear differential equations: integrating factor.

Question type: Linear in $x$

How to spot it: click to see

Type 8. Linear in $x$ ⁠: when $y$ will not behave, solve for $x$

Spot it: there is a $y^2$ or $\tan^{-1}y$ next to $dy$ ⁠, but $x$ appears only to power 1.

Here: The equation already has $\frac{dx}{dy}$ ⁠, with $x$ to power 1 and $P, Q, R$ in $y$ ⁠. Divide by $R$ so that $\frac{dx}{dy}$ stands alone: $P_1 = \frac{P}{R}$ ⁠, and the I.F. is $e^{\int\frac{P}{R}\,dy}$ ⁠.

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Q11: Integrating factor of Rdxdy+Px=QR\frac{dx}{dy} + Px = QRdydx​+Px=Q

Q11: Integrating factor of $R\frac{dx}{dy} + Px = Q$