Find the particular solution: $y(1+\log x)\dfrac{dx}{dy}-x\log x=0,\ y=e^2,$ when $x=e$.

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$y(1+\log x)\dfrac{dx}{dy}-x\log x=0$ separates as $\dfrac{1+\log x}{x\log x}dx=\dfrac{dy}{y}$ . With $u=\log x$ , the left side is $\int\dfrac{1+u}{u}du=\log u+u+c_1=\log(\log x)+\log x+c_1$ . So $\log(\log x)+\log x=\log y+c$ . At $x=e,y=e^2$ : $\log(1)+1-2=-1=c$ . Rearranging this back to an explicit form: $\log(x\log x)-\log y=-1$ , i.e. $\dfrac{x\l …

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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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Q.Find the particular solution: $y(1+\log x)\dfrac{dx}{dy}-x\log x=0,\ y=e^2,$ when $x=e$ .

Concept understanding — Differential Equations with Variables Separable

Q.Find the particular solution: y(1+log⁡x)dxdy−xlog⁡x=0, y=e2,y(1+\log x)\dfrac{dx}{dy}-x\log x=0,\ y=e^2,y(1+logx)dydx​−xlogx=0, y=e2, when x=ex=ex=e.

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Q.Find the particular solution: $y(1+\log x)\dfrac{dx}{dy}-x\log x=0,\ y=e^2,$ when $x=e$ .

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