The integrating factor of the differential equation $(1 - y^2)\frac{dx}{dy} + yx = ay$, $(-1 < y < 1)$ is: (a) $\frac{1}{y^2 - 1}$ (b) $\frac{1}{\sqrt{y^2 - 1}}$ (c) $\frac{1}{1 - y^2}$ (d) $\frac{1}{\sqrt{1 - y^2}}$

The integrating factor of the differential equation (1 — CBSE Class 12 Mathematics 2023 PYQ

Question

(1 - y^2)\frac{dx}{dy} + yx = ay

(a)

\frac{1}{y^2 - 1}

(b)

\frac{1}{\sqrt{y^2 - 1}}

(c)

\frac{1}{1 - y^2}

(d)

\frac{1}{\sqrt{1 - y^2}}

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Concept understanding — Integrating Factor Method

Integrating Factor Method

Some first-order differential equations refuse to separate — you cannot get all the $y$ 's on one side and all the $x$ 's on the other. The integrating factor method is the standard trick for a special (and very common) family of these: the linear first-order equation. The idea is beautifully simple: multiply the whole equation by one cleverly chosen function, and the messy left-hand side collapses into a single derivative that we can integrate directly.

The Standard Form

An equation is linear of first order if it can be written as

\frac{dy}{dx} + P\,y = Q

where $P$ and $Q$ are functions of $x$ alone (or constants). Notice $y$ and $\frac{dy}{dx}$ appear only to the first power, and never multiplied together — that is what "linear" means here.

Tip

Always rearrange into this exact shape first. The coefficient of $\frac{dy}{dx}$ must be $1$ before you read off $P$ and $Q$ .

The Integrating Factor

The magic multiplier is

\text{I.F.} = e^{\int P\,dx}.

Why this one? Multiply the equation by $e^{\int P\,dx}$ :

e^{\int P\,dx}\frac{dy}{dx} + P\,e^{\int P\,dx}\,y = Q\,e^{\int P\,dx}.

By the product rule, the left-hand side is exactly $\dfrac{d}{dx}\!\left(y\cdot e^{\int P\,dx}\right)$ , because the derivative of $e^{\int P\,dx}$ is $P\,e^{\int P\,dx}$ . So the equation becomes

\frac{d}{dx}\big(y\cdot \text{I.F.}\big) = Q\cdot \text{I.F.}

The left side is now a single derivative — that is the whole point of choosing this factor.

The Solution

Integrate both sides with respect to $x$ :

y\cdot \text{I.F.} = \int \big(Q\cdot \text{I.F.}\big)\,dx + C.

This is the general solution. In words: (solution) × (integrating factor) = integral of (Q × integrating factor), plus a constant.

A Quick Illustration

For $\dfrac{dy}{dx} + \dfrac{1}{x}\,y = x$ , we read $P = \tfrac1x$ , $Q = x$ . Then $\int P\,dx = \log x$ , so $\text{I.F.} = e^{\log x} = x$ . The solution is

y\cdot x = \int x\cdot x\,dx + C = \frac{x^3}{3} + C.

--- …

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The key idea is to rewrite the equation in the standard linear form $\frac{dx}{dy} + P(y)x = Q(y)$ and then compute the integrating factor as $\mu(y) = e^{\int P(y)\,dy}$ . For this problem, the integrating factor is $\frac{1}{\sqrt{1 - y^2}}$ , which corresponds to option (d).

The Integrating Factor (IF) method is the standard tool for solving first-order linear differential equations. The equation is linear in $x$ as a function of $y$ , so we aim to write it in the form $\frac{dx}{dy} + P(y)x = Q(y)$ . Once in that form, the IF is $e^{\int P\,dy}$ — it works because multiplying through by the IF turns the left side into the exact derivative of $(x \cdot \text{IF})$ , making integration straightforward.

Let’s go step by step.

Rewrite the given equation in standard linear form. The equation is $(1 - y^2)\frac{dx}{dy} + yx = ay$ . Divide through by $(1 - y^2)$ (valid since $-1 < y < 1$ , so $1 - y^2 > 0$ ):

\frac{dx}{dy} + \frac{y}{1 - y^2}\,x = \frac{ay}{1 - y^2}.

Now it matches $\frac{dx}{dy} + P(y)x = Q(y)$ with $P(y) = \frac{y}{1 - y^2}$ .

Find the integrating factor. The integrating factor is $\mu(y) = e^{\int P(y)\,dy} = e^{\int \frac{y}{1 - y^2}\,dy}$ . Compute the integral: let $u = 1 - y^2$ , so $du = -2y\,dy$ , hence $y\,dy = -\frac{1}{2}du$ . Then

\int \frac{y}{1 - y^2}\,dy = \int \frac{1}{u}\left(-\frac{1}{2}\right)du = -\frac{1}{2}\log|u| + C = -\frac{1}{2}\log(1 - y^2) + C,

since $1 - y^2 > 0$ in the given interval, so absolute value is unnecessary.

Therefore,

$\mu(y) = e^{-\frac{1}{2}\log(1 - y^2)} = e^{\log\left((1 - y^2)^{-1/2}\right)} = \frac{1}{\sqrt{1 - y^2}}.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 36 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$R\dfrac{dx}{dy} + Px = Q$
›Reveal solutionSolution
The key idea is to rewrite the given equation in the standard linear form $\frac{dx}{dy} + P_1 x = Q_1$ by dividing through by $R$ , then the integrating factor is $e^{\int P_1\, dy} = e^{\int \frac{P}{R}\, dy}$ . The correct option is (C).

The Integrating Factor (IF) method is a systematic way to solve first-order linear differential equations. The core insight is that we want to multiply the entire equation by a function that turns the left-hand side into the exact derivative of a product — specifically, the derivative of $x$ times some function. This works because if you have an equation of the form $\frac{dx}{dy} + f(y) x = g(y)$ , multiplying both sides by $e^{\int f(y)\, dy}$ makes the left side become $\frac{d}{dy}\left( x \cdot e^{\int f(y)\, dy} \right)$ , which is then easy to integrate.

Here, the given equation is $R\frac{dx}{dy} + Px = Q$ , where $P$ , $Q$ , $R$ are functions of $y$ alone. Notice that the coefficient of $\frac{dx}{dy}$ is $R$ , not 1. So our first job is to put it into the standard form.

Rewrite in standard linear form. Divide every term by $R$ (assuming $R \neq 0$ ):

$\frac{dx}{dy} + \frac{P}{R}\, x = \frac{Q}{R}.$
Now it matches the pattern $\frac{dx}{dy} + P_1(y)\, x = Q_1(y)$ , where $P_1(y) = \frac{P}{R}$ and $Q_1(y) = \frac{Q}{R}$ .

Recall the formula for the integrating factor. For a first-order linear ODE in the form $\frac{dx}{dy} + P_1(y)\, x = Q_1(y)$ , the integrating factor is

$\text{IF} = e^{\int P_1(y)\, dy}.$
This is a standard result — the exponential of the integral of the coefficient of $x$ .

Substitute $P_1$ into the formula. Here $P_1(y) = \frac{P}{R}$ , so $\text{IF} = e^{\int \frac{P}{R}\, dy}.$ …
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CBSE 2026Set 65/2/11 markMCQ
$2x \frac{dy}{dx} - y = 3$
›Reveal solutionSolution
To find the integrating factor, we first convert the given differential equation into the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$ . From this, we identify $P(x) = -\frac{1}{2x}$ , and the integrating factor is calculated as $e^{\int P(x) dx}$ , which evaluates to $\frac{1}{\sqrt{x}}$ .

The integrating factor method is a powerful technique used to solve first-order linear differential equations. A first-order linear differential equation has the general form:
$\frac{dy}{dx} + P(x)y = Q(x)$
where $P(x)$ and $Q(x)$ are functions of $x$ (or constants).

Why the Integrating Factor?

The core idea is to transform the left-hand side (LHS) of this equation into the derivative of a product. Specifically, we want to make the LHS look like $\frac{d}{dx} (y \cdot \text{some function})$ .

Let's say we multiply the entire equation by a function, $\mu(x)$ , which we call the integrating factor:
$\mu(x) \frac{dy}{dx} + \mu(x) P(x)y = \mu(x) Q(x)$
Now, consider the product rule for differentiation: $\frac{d}{dx} (\mu(x) y) = \mu(x) \frac{dy}{dx} + y \frac{d\mu}{dx}$ .

For our modified LHS to be exactly $\frac{d}{dx} (\mu(x) y)$ , we need the term $\mu(x) P(x)y$ to be equal to $y \frac{d\mu}{dx}$ .

This means:
$\mu(x) P(x) = \frac{d\mu}{dx}$
This is a separable differential equation for $\mu(x)$ . We can rewrite it as:
$\frac{d\mu}{\mu} = P(x) dx$
Integrating both sides:
$\int \frac{d\mu}{\mu} = \int P(x) dx$ $\log|\mu| = \int P(x) dx$
Exponentiating both sides (and typically taking the positive value for $\mu(x)$ as a convention, and omitting the constant of integration since any constant factor in $\mu(x)$ would cancel out later):
$\mu(x) = e^{\int P(x) dx}$
This $\mu(x)$ is the integrating factor. Once we multiply the original equation by this $\mu(x)$ , the LHS becomes $\frac{d}{dx} (\mu(x) y)$ , which can then be easily integrated to solve for $y$ .

Let's apply this method to the given problem.

Convert to Standard Form The given differential equation is $2x \frac{dy}{dx} - y = 3$ . The standard form for a first-order linear differential equation is $\frac{dy}{dx} + P(x)y = Q(x)$ . To achieve this, we need the coefficient of $\frac{dy}{dx}$ to be 1. We can do this by dividing the entire equation by $2x$ :

$\frac{2x \frac{dy}{dx}}{2x} - \frac{y}{2x} = \frac{3}{2x}$ $\frac{dy}{dx} - \frac{1}{2x} y = \frac{3}{2x}$

Identify $P(x)$ Now, comparing our equation $\frac{dy}{dx} - \frac{1}{2x} y = \frac{3}{2x}$ with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$ , we can identify $P(x)$ and $Q(x)$ :

$P(x) = -\frac{1}{2x}$
$$ Q(x) = \frac{3}{2x} $$ …
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CBSE 2026Set A1 markMCQ
$(1 + x^2)\frac{dy}{dx} + y = e^{\tan^{-1}x}$
(a) $e^{\tan^{-1}x}$
(b) $e^{\sin^{-1}x}$
(c) $\tan^{-1}x$
(d) $\sin^{-1}x$
›Reveal solutionSolution

Standard form gives $P=\tfrac{1}{1+x^2}$ , so I.F. $=e^{\int\frac{dx}{1+x^2}}=e^{\tan^{-1}x}$ .

Divide the equation by $(1+x^2)$ to get the linear form $\dfrac{dy}{dx}+Py=Q$ :

$\dfrac{dy}{dx}+\dfrac{1}{1+x^2}\,y=\dfrac{e^{\tan^{-1}x}}{1+x^2}$ , so $P=\dfrac{1}{1+x^2}$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The Integrating factor of differential equation (1 - y²) dy/dx + y\cdotx = ay is$
(a) 1/(y²−1)
(b) 1/√(y²−1)
(c) 1/(1−y²)
(d) 1/√(1−y²)
›Reveal solutionSolution

Written correctly this is a linear equation in $x$ as a function of $y$ : $(1-y^2)\dfrac{dx}{dy} + xy = ay$ . Its integrating factor works out to $1/\sqrt{1-y^2}$ .

Divide through by $(1-y^2)$ to get the standard linear form $\dfrac{dx}{dy} + P(y)\,x = Q(y)$ :

$\frac{dx}{dy} + \frac{y}{1-y^2}\,x = \frac{ay}{1-y^2}, \qquad P(y) = \frac{y}{1-y^2}.$

Integrating factor $\text{I.F.} = e^{\int P(y)\,dy} = e^{\int \frac{y}{1-y^2}dy}$ .

Let $u = 1-y^2$ , $du = -2y\,dy$ : …

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CBSE 2026Set ANNUAL1 mark
$Q. Find the integrating factor of the differential equation x dy/dx + y = 10.$
›Reveal solutionSolution

Write the equation in the standard linear form $dy/dx+Py=Q$ first, then use $I.F.=e^{\int P\,dx}$ .

$x\dfrac{dy}{dx}+y=10 \;\Rightarrow\; \dfrac{dy}{dx}+\dfrac1x y = \dfrac{10}{x}$

Here $P=1/x$ .

…

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CBSE 2026Set ANNUAL1 mark
$\frac{dx}{dy} + P_1 x = Q_1$
›Reveal solutionSolution

Linear DE in $x$ ; multiply by integrating factor $e^{\int P_1\,dy}$ .

For $\dfrac{dx}{dy}+P_1x=Q_1$ , the integrating factor is $\text{IF}=e^{\int P_1\,dy}$ , and the solution is

…

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CBSE 2025Set 65/1/11 markMCQ
$(x + 2y^3) \frac{dy}{dx} = 2y$
›Reveal solutionSolution
The given equation is not linear in $y$ , but it is linear in $x$ when rewritten as $\frac{dx}{dy} - \frac{x}{2y} = y^2$ . The integrating factor is $e^{\int -\frac{1}{2y}\,dy} = \frac{1}{\sqrt{y}}$ , so the correct option is (B).

We often learn the integrating factor method for first-order linear ODEs of the form $\frac{dy}{dx} + P(x)y = Q(x)$ . But here, the equation is $(x + 2y^3) \frac{dy}{dx} = 2y$ . If we try to write it as $\frac{dy}{dx} + P(x)y = Q(x)$ , we run into trouble because the coefficient of $\frac{dy}{dx}$ involves both $x$ and $y$ , and the right-hand side is $2y$ — not a function of $x$ alone. That path leads nowhere.

The trick is to swap the roles of $x$ and $y$ . Notice that the equation contains $x$ and $y$ in a way that suggests $x$ might be the dependent variable. If we rewrite it as $\frac{dx}{dy}$ , we get a linear equation in $x$ — and then the integrating factor method works cleanly.

Let’s do it step by step.

Rewrite the equation in terms of $\frac{dx}{dy}$ . Start with

$(x + 2y^3) \frac{dy}{dx} = 2y.$
Divide both sides by $\frac{dy}{dx}$ (which is fine as long as $y$ is not constant):
$x + 2y^3 = 2y \frac{dx}{dy}.$
Now isolate $\frac{dx}{dy}$ :
$\frac{dx}{dy} = \frac{x + 2y^3}{2y} = \frac{x}{2y} + y^2.$

Bring it to the standard linear form. A first-order linear ODE in $x$ (as a function of $y$ ) looks like

$\frac{dx}{dy} + P(y)x = Q(y).$
From $\frac{dx}{dy} = \frac{x}{2y} + y^2$ , subtract $\frac{x}{2y}$ from both sides:
$\frac{dx}{dy} - \frac{x}{2y} = y^2.$
So here $P(y) = -\frac{1}{2y}$ and $Q(y) = y^2$ .

Find the integrating factor. The integrating factor for $\frac{dx}{dy} + P(y)x = Q(y)$ is

$\mu(y) = e^{\int P(y)\,dy}.$
With $P(y) = -\frac{1}{2y}$ , we have

$\int -\frac{1}{2y}\,dy = -\frac{1}{2} \log|y| = \log\left(y^{-1/2}\right).$ …
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CBSE 2025Set ANNUAL1 mark
$\dfrac{dy}{dx}+y=x$
›Reveal solutionSolution

For a linear equation $\dfrac{dy}{dx}+Py=Q$ , the integrating factor is $e^{\int P\,dx}$ .

Here $\dfrac{dy}{dx}+y=x$ is linear with $P=1$ , $Q=x$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$\dfrac{dy}{dx} - \dfrac{y}{x} = x^4 - 3x$
(a) $x$
(b) $\log x$
(c) $\dfrac{1}{x}$
(d) $-x$
›Reveal solutionSolution

The integrating factor of a linear DE depends only on the coefficient of y, not on the right-hand-side function of x — so it is unaffected by the exact power in the RHS term.

The equation $\dfrac{dy}{dx} - \dfrac{y}{x} = x^4 - 3x$ is a first-order linear differential equation of the standard form $\dfrac{dy}{dx} + P(x)y = Q(x)$ , with:

$P(x) = -\dfrac{1}{x}$ , and $Q(x) = x^4-3x$ (the right-hand side, whatever its exact power).

The integrating factor is:

$\text{I.F.} = e^{\int P(x)\,dx} = e^{\int -\frac{1}{x}dx} = e^{-\ln x} = \dfrac{1}{x}$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The integrating factor of ydx - (x + 2y²)dy = 0 is :$
(a) 1/y²
(b) −1/y
(c) 1/y
(d) y
›Reveal solutionSolution

Rewritten as a linear equation in $x$ (treating $y$ as the independent variable), the integrating factor is $e^{\int -\frac1y\,dy}=\dfrac1y$ .

Given: $y\,dx-(x+2y^2)\,dy=0$ , i.e. $y\,dx = (x+2y^2)\,dy$ , so

$\frac{dx}{dy} = \frac{x}{y}+2y \quad\Longrightarrow\quad \frac{dx}{dy} - \frac1y\,x = 2y.$

…

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CBSE 2025Set ANNUAL1 markMCQ
$x\dfrac{dy}{dx} - y = 2x^2$
(a) $e^{-x}$
(b) $e^{-y}$
(c) $\dfrac{1}{x}$
(d) $x$
›Reveal solutionSolution

Put the equation in standard linear form $\frac{dy}{dx}+Py=Q$ and compute $e^{\int P\,dx}$ .

Starting from $x\dfrac{dy}{dx}-y=2x^2$ , divide throughout by $x$ :

$\frac{dy}{dx}-\frac{1}{x}y=2x.$

This is linear in $y$ with $P(x)=-\dfrac1x$ and $Q(x)=2x$ .

The integrating factor is

$\text{IF}=e^{\int P\,dx}=e^{\int-\frac1x\,dx}=e^{-\log x}=e^{\log x^{-1}}=\frac1x.$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The integrating factor of differential equation cos x (dy/dx) + y sin x = 1 is .......$
(a) sec x + tan x
(b) Log(sec x + tan x)
(c) e^(sec x)
(d) sec x
›Reveal solutionSolution

Writing the equation in standard linear form $\dfrac{dy}{dx} + Py = Q$ gives $P = \tan x$ , and the integrating factor $e^{\int P\,dx} = e^{\ln|\sec x|} = \sec x$ .

Given: $\cos x \dfrac{dy}{dx} + y\sin x = 1$

Step 1 — standard form: Divide throughout by $\cos x$ :

$\frac{dy}{dx} + y\tan x = \sec x$ …

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Q.The integrating factor of the differential equation (1−y2)dxdy+yx=ay(1 - y^2)\frac{dx}{dy} + yx = ay(1−y2)dydx​+yx=ay, (−1<y<1)(-1 < y < 1)(−1<y<1) is:

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Q.The integrating factor of the differential equation $(1 - y^2)\frac{dx}{dy} + yx = ay$ , $(-1 < y < 1)$ is:

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