(iv) $\frac{dy}{dx}+\frac{y}{x\log x}=\frac{1}{x}$ is an equation of the type ______.

(iv) (dy)/(dx)+(y)/(xlog x)=(1)/(x) is an equation of the type ___.

NCERT Exemplar · Q37

\frac{dy}{dx}+\frac{y}{x\log x}=\frac{1}{x}

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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 equation fits $\frac{dy}{dx}+P(x)\,y=Q(x)$ , so it is a first-order linear differential equation.

We are asked to classify

\frac{dy}{dx}+\frac{y}{x\log x}=\frac{1}{x}.

What makes an equation "linear"

A first-order equation is linear when it can be written as

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

where $P$ and $Q$ depend on $x$ only, and $y$ together with $\frac{dy}{dx}$ appear to the first power and are never multiplied by each other.

Match the pattern

Read off the coefficients directly:

P(x)=\frac{1}{x\log x},\qquad Q(x)=\frac{1}{x}.

Both are functions of $x$ alone, and $y$ occurs only linearly. So the equation is exactly of the linear type.

How such an equation is solved

The integrating factor is …

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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 15 on this concept.

KEAM 2024Set eng-2024-06064 marksMCQ
$x\frac{dy}{dx}+2y=xe^{x}$
›Reveal solutionSolution

Dividing by $x$ gives $P(x)=\frac{2}{x}$ , so the integrating factor is $x^2$ .

Divide $x\frac{dy}{dx}+2y=xe^x$ by $x$ : $\frac{dy}{dx}+\frac{2}{x}y=e^x$ . …
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KEAM 2024Set eng-2024-06054 marksMCQ
$\left(1 + 2e^{-x}\right)\frac{dy}{dx} - 2e^{-x}y = 1 + e^{-x}$
›Reveal solutionSolution

$\int P\,dx=\log(1+2e^{-x})$ , so the integrating factor is $1+2e^{-x}$ .

Divide by $(1+2e^{-x})$ :
$\frac{dy}{dx}-\frac{2e^{-x}}{1+2e^{-x}}\,y=\frac{1+e^{-x}}{1+2e^{-x}},$
so $P(x)=-\dfrac{2e^{-x}}{1+2e^{-x}}$ . With $w=1+2e^{-x}$ , $dw=-2e^{-x}dx$ : …
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KEAM 2025Set eng-2025-04254 marksMCQ
$(1+x^2)dy=(1-2xy)dx$
›Reveal solutionSolution

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

Rewrite $(1+x^{2})dy=(1-2xy)dx$ as
$\dfrac{dy}{dx}+\dfrac{2x}{1+x^{2}}\,y=\dfrac{1}{1+x^{2}}.$
Here $P=\dfrac{2x}{1+x^{2}}$ , so …
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KEAM 2024Set eng-2024-06094 marksMCQ
$\dfrac{dy}{dx} + y = 5$
›Reveal solutionSolution

The equation is separable: $\frac{dy}{5-y}=dx$ , which integrates to $-\log|5-y|=x+C$ .

Write $\frac{dy}{dx}=5-y$ , so $\frac{dy}{5-y}=dx$ . …
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KEAM 2021Set eng-2021-P2-B14 marksMCQ
$xy' + 2y - 7x^3 = 0$
›Reveal solutionSolution
The integrating factor is $x^2$ .

Concept and Intuition

A linear ODE $y'+P(x)y=Q(x)$ has integrating factor $e^{\int P\,dx}$ . First put the equation in standard form by dividing through by the coefficient of $y'$ .

Step-by-Step Solution

$xy'+2y-7x^3=0\Rightarrow y'+\dfrac{2}{x}y=7x^2$ .

Here $P(x)=\dfrac{2}{x}$ , so $\int P\,dx=2\log|x|$ . …
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KEAM 2022Set eng-2022-P2-B14 marksMCQ
$4x\,dy - e^{-2y}dy + dx = 0$
›Reveal solutionSolution
The integrating factor is $e^{4y}$ .

Concept and Intuition

Rewrite the equation as linear in $x$ with respect to $y$ ; the integrating factor is $e^{\int P\,dy}$ .

Step-by-Step Solution

$4x\,dy-e^{-2y}\,dy+dx=0\Rightarrow (4x-e^{-2y})\,dy+dx=0$ .

Divide by $dy$ : $\frac{dx}{dy}+4x=e^{-2y}$ .

This is linear with $P(y)=4$ .

Integrating factor $=e^{\int4\,dy}=e^{4y}$ .

Common Mistakes …
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KEAM 2022Set eng-2022-P2-B14 marksMCQ
$(x^2 y^2 + y)dx - (x - 2x^3 y)dy = 0$
›Reveal solutionSolution
The general solution is $xy^2-\frac{y}{x}=C$ .

Concept and Intuition

Regroup terms into recognizable exact differentials $d(xy^2)$ and $d(y/x)$ .

Step-by-Step Solution

Expand: $(x^2y^2+y)dx-(x-2x^3y)dy=0\Rightarrow x^2y^2\,dx+y\,dx-x\,dy+2x^3y\,dy=0$ .

Note $d(xy^2)=y^2\,dx+2xy\,dy$ , so $x^2y^2\,dx+2x^3y\,dy=x^2\,d(xy^2)$ .

Note $d(y/x)=\frac{x\,dy-y\,dx}{x^2}$ , so $y\,dx-x\,dy=-x^2\,d(y/x)$ .

Equation: $x^2\,d(xy^2)-x^2\,d(y/x)=0\Rightarrow d(xy^2)-d(y/x)=0$ . …
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KEAM 2026Set eng-2026-04224 marksMCQ
$(y + x^2)\, dx = x\, dy, x > 0$
›Reveal solutionSolution

Rewrite as a first-order linear ODE $y'-y/x=x$ , use integrating factor $1/x$ , then apply the point $(1,0)$ .

From $(y+x^2)dx=x\,dy$ : $x\frac{dy}{dx}=y+x^2\Rightarrow \frac{dy}{dx}-\frac{y}{x}=x$ .

Integrating factor $=e^{-\int dx/x}=\frac1x$ , so $\frac{d}{dx}\Big(\frac{y}{x}\Big)=1$ . …
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KEAM 2026Set eng-2026-04184 marksMCQ
$\dfrac{dy}{dx} + \dfrac{2y}{20 - x} = 10$
›Reveal solutionSolution

The integrating factor is $e^{\int P\,dx}$ with $P=\dfrac{2}{20-x}$ , which evaluates to $\dfrac{1}{(20-x)^2}$ .

For $\dfrac{dy}{dx}+P\,y=Q$ with $P=\dfrac{2}{20-x}$ ,

$\int P\,dx=\int\dfrac{2}{20-x}\,dx=-2\log|20-x|.$ …
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KEAM 2026Set eng-2026-04174 marksMCQ
$x^3 \dfrac{dy}{dx} + 3x^2 y = \cos x$
›Reveal solutionSolution

Recognize the exact derivative $\dfrac{d}{dx}(x^3y)$ .

Since $\dfrac{d}{dx}(x^3y)=x^3\dfrac{dy}{dx}+3x^2y$ , the equation $x^3\dfrac{dy}{dx}+3x^2y=\cos x$ becomes

$\frac{d}{dx}(x^3y)=\cos x.$ …
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KEAM 2024Set eng-2024-06074 marksMCQ
$2y\tan x+\dfrac{dy}{dx}=5\sin x$
›Reveal solutionSolution

Solve the linear ODE with integrating factor $\sec^2 x$ .

Rewrite as $\frac{dy}{dx}+2\tan x\,y=5\sin x$ . Integrating factor:
$\mu=e^{\int 2\tan x\,dx}=e^{2\log\sec x}=\sec^2 x.$
Then

$\frac{d}{dx}\left(y\sec^2 x\right)=5\sin x\sec^2 x=5\frac{\sin x}{\cos^2 x}.$ …
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KEAM 2025Set eng-2025-04234 marksMCQ
$\sin x\, dy = \dfrac{1}{2}(\sin 2x + 2y\cos x)\, dx$
›Reveal solutionSolution
The equation reduces to y' - (cot x) y = cos x, giving integrating factor e^{-integral cot x dx} = csc x.

Concept and Intuition

Rewrite the equation in standard linear form dy/dx + P(x)y = Q(x); the integrating factor is e^{integral P dx}.

Step-by-Step Solution

sin x dy = (1/2)(sin2x + 2y cos x) dx; using sin2x = 2 sin x cos x, the RHS = (sin x cos x + y cos x) dx.

So sin x (dy/dx) = sin x cos x + y cos x, i.e. dy/dx = cos x + y cot x. …
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←Previous questionQ36 — (iii) The number of arbitrary constants in the general solution of a d…→Next questionQ38 — (v) General solution of the differential equation of the type (dx)/(dy…

Q.(iv) $\frac{dy}{dx}+\frac{y}{x\log x}=\frac{1}{x}$ is an equation of the type ______.

Concept understanding — Integrating Factor Method

Integrating Factor Method

The Standard Form

The Integrating Factor

The Solution

A Quick Illustration

What makes an equation "linear"

Match the pattern

How such an equation is solved

Method: Identifying the Type of a First-Order Differential Equation

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.(iv) dydx+yxlog⁡x=1x\frac{dy}{dx}+\frac{y}{x\log x}=\frac{1}{x}dxdy​+xlogxy​=x1​ is an equation of the type ______.

What makes an equation "linear"

Match the pattern

How such an equation is solved

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Q.(iv) $\frac{dy}{dx}+\frac{y}{x\log x}=\frac{1}{x}$ is an equation of the type ______.

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