Solve the D.E. 3extan ydx+(1+ex)sec² ydy=0. – Class 12 Mathematics | ConceptFirst

Question 176 of 177

3e^x\tan y\,dx+(1+e^x)\sec^2 y\,dy=0

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2026Subjective· 3mImportance★★★★★

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This is a variable-separable D.E.; separate the $x$ and $y$ terms and integrate.

$3e^x\tan y\,dx+(1+e^x)\sec^2y\,dy=0$

Separate variables:

$\frac{3e^x}{1+e^x}\,dx=-\frac{\sec^2y}{\tan y}\,dy$

Integrating both sides:

$\int\frac{3e^x}{1+e^x}\,dx=-\int\frac{\sec^2y}{\tan y}\,dy$

For the left side, let $u=1+e^x,\ du=e^x dx$ : $\displaystyle\int\frac{3\,du}{u}=3\ln|1+e^x|$ .

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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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