The solution of $\dfrac1x\cdot\dfrac{dy}{dx}=\tan^{-1}x$ is... (A) $\dfrac{x^2\tan^{-1}x}{2}+c=0$ (B) $x\tan^{-1}x+c=0$ (C) $x-\tan^{-1}x=c$ (D) $y=\dfrac{x^2\tan^{-1}x}{2}-\dfrac12(x-\tan^{-1}x)+c$

The solution of dfrac1x·(dy)/(dx)=tan⁻¹x is... (A) dfracx²tan⁻¹x2+c=0 (B) xtan⁻¹x+c=0 (C) x-tan⁻¹x=c (D) y=dfr

Miscellaneous Exercise 6(I) · Q101

\dfrac1x\cdot\dfrac{dy}{dx}=\tan^{-1}x

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$\dfrac1x\cdot\dfrac{dy}{dx}=\tan^{-1}x$ gives $\dfrac{dy}{dx}=x\tan^{-1}x$ , so $y=\int x\tan^{-1}x\,dx$ . By parts ( $u=\tan^{-1}x,\ dv=x\,dx$ ): $y=\dfrac{x^2}{2}\tan^{-1}x-\int\dfrac{x^2}{2(1+x^2)}dx=\dfrac{x^2}{2}\tan^{-1}x-\dfrac12\int\lef …

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

CBSE 2024Set ANNUAL2 marks
$1+\dfrac{dy}{dx}=\csc(x+y)$
›Reveal solutionSolution

Put $u=x+y$ ; the equation becomes $\dfrac{du}{dx}=\csc u$ , separable.

$u=x+y \Rightarrow \dfrac{du}{dx}=1+\dfrac{dy}{dx}=\csc u$

$\sin u\,du=dx$

…

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CBSE 2023Set ANNUAL2 marksMCQ
$\dfrac{dx}{dt} = \dfrac{x\log x}{t}$
(a) $x = e^{ct}$
(b) $x + e^{ct} = 0$
(c) $x = e^t + t$
(d) $xe^{ct} = 0$
›Reveal solutionSolution

Separate variables and integrate.

$\dfrac{dx}{x\log x}=\dfrac{dt}{t}$

Integrating: $\log(\log x)=\log t+\log k=\log(kt)$

…

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CBSE 2023Set ANNUAL2 marks
$\cos x \cos y\,dy - \sin x \sin y\,dx = 0$
›Reveal solutionSolution

Separate variables: $\cot y\,dy=\tan x\,dx$ .

$\cos x\cos y\,dy=\sin x\sin y\,dx \Rightarrow \cot y\,dy=\tan x\,dx$

Integrating: $\log|\sin y|=-\log|\cos x|+\log k$

…

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CBSE 2022Set ANNUAL2 marks
$y\dfrac{dy}{dx} + x = 0$
›Reveal solutionSolution

Separate variables and integrate.

$y\dfrac{dy}{dx}+x=0 \implies y\,dy = -x\,dx$

…

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CBSE 2020Set ANNUAL2 marks
$\dfrac{dy}{dx} = x^2y + y$
›Reveal solutionSolution

Separate variables and integrate.

$\dfrac{dy}{dx} = y(x^2+1)$

Separating variables:

$\dfrac{dy}{y} = (x^2+1)\,dx$

Integrating both sides: …

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←Previous questionQ100 — The differential equation y(dy)/(dx)+x=0 represents a family of... (A)…→Next questionQ102 — The solution of (x+y)²(dy)/(dx)=1 is... (A) x=tan⁻¹(x+y)+c (B) ytan⁻¹(…

Q.The solution of $\dfrac1x\cdot\dfrac{dy}{dx}=\tan^{-1}x$ is... (A) $\dfrac{x^2\tan^{-1}x}{2}+c=0$ (B) $x\tan^{-1}x+c=0$ (C) $x-\tan^{-1}x=c$ (D) $y=\dfrac{x^2\tan^{-1}x}{2}-\dfrac12(x-\tan^{-1}x)+c$

Concept understanding — Differential Equations with Variables Separable

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