General solution of $\frac{dy}{dx}+y\tan x=\sec x$ is: (A) $y\sec x=\tan x+c$ (B) $y\tan x=\sec x+c$ (C) $\tan x=y\tan x+c$ (D) $x\sec x=\tan y+c$

General solution of (dy)/(dx)+ytan x=sec x is: (A) ysec x=tan x+c (B) ytan x=sec x+c (C) tan x=ytan x+c (D) xs

NCERT Exemplar · Q93

\frac{dy}{dx}+y\tan x=\sec x

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Appeared in past exams:KCET 2025· Set A-1· 1mexact

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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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With integrating factor $\sec x$ , the equation gives $y\sec x=\tan x+c$ — option (A).

The equation

\frac{dy}{dx}+y\tan x=\sec x

is first-order linear, of the form $\frac{dy}{dx}+P(x)y=Q(x)$ with $P(x)=\tan x$ and $Q(x)=\sec x$ .

1. Integrating factor

\int\tan x\,dx=\log|\sec x|,\qquad \text{I.F.}=e^{\log|\sec x|}=\sec x.

2. Multiply through

\sec x\frac{dy}{dx}+y\sec x\tan x=\sec^2 x.

The left side is exactly $\frac{d}{dx}(y\sec x)$ , since $\frac{d}{dx}(\sec x)=\sec x\tan x$ . So …

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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-2024-06074 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 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 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 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 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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KEAM 2024Set eng-2024-06074 marksMCQ
$(3\sin x\cos x)\,dy=(1+3y\sin^2 x)\,dx$
›Reveal solutionSolution

Put in linear form; the coefficient of $y$ gives IF $=\cos x$ .

From $(3\sin x\cos x)dy=(1+3y\sin^2 x)dx$ :
$\frac{dy}{dx}=\frac{1+3y\sin^2 x}{3\sin x\cos x}=\frac{1}{3\sin x\cos x}+\frac{3\sin^2 x}{3\sin x\cos x}y.$
The $y$ -coefficient is $\frac{\sin x}{\cos x}=\tan x$ , so …
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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 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 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 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 2026Set eng-2026-04174 marksMCQ
$2\, dy = (y + \cos x)\, dx$
›Reveal solutionSolution

Put in standard linear form and read off the integrating factor.

From $2\,dy=(y+\cos x)\,dx$ :

$2\frac{dy}{dx}=y+\cos x\ \Rightarrow\ \frac{dy}{dx}-\frac12 y=\frac12\cos x.$ …
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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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←Previous questionQ92 — Which of the following is the general solution of (d²y)/(dx²)-2(dy)/(d…→Next questionQ94 — Solution of the differential equation (dy)/(dx)+(y)/(x)=sin x is: (A) …

Q.General solution of $\frac{dy}{dx}+y\tan x=\sec x$ is:
(A) $y\sec x=\tan x+c$
(B) $y\tan x=\sec x+c$
(C) $\tan x=y\tan x+c$
(D) $x\sec x=\tan y+c$

Concept understanding — Integrating Factor Method

Integrating Factor Method

The Standard Form

The Integrating Factor

The Solution

A Quick Illustration

1. Integrating factor

2. Multiply through

Method: Integrating factor with a trigonometric coefficient

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Q.General solution of dydx+ytan⁡x=sec⁡x\frac{dy}{dx}+y\tan x=\sec xdxdy​+ytanx=secx is: (A) ysec⁡x=tan⁡x+cy\sec x=\tan x+cysecx=tanx+c (B) ytan⁡x=sec⁡x+cy\tan x=\sec x+cytanx=secx+c (C) tan⁡x=ytan⁡x+c\tan x=y\tan x+ctanx=ytanx+c (D) xsec⁡x=tan⁡y+cx\sec x=\tan y+cxsecx=tany+c

1. Integrating factor

2. Multiply through

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Q.General solution of $\frac{dy}{dx}+y\tan x=\sec x$ is:
(A) $y\sec x=\tan x+c$
(B) $y\tan x=\sec x+c$
(C) $\tan x=y\tan x+c$
(D) $x\sec x=\tan y+c$

Unlock everything free for 14 days