Consider the differential equation
$\frac{d y}{d x} = \frac{y}{2 y \log y + y - x}$
Statement-1: $x y = y^{2} \log y + C$ is a solution of the given
differential equation
Statement-2: The differential equation is a linear equation
in $y$ and $x .$

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STATEMENT-1 is False, STATEMENT-2 is True

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for
STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is False

answer is A.

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Detailed Solution

$\frac{d x}{d y} = \frac{2 y \log y + y - x}{y} = 2 \log y + 1 - \frac{x}{y}$

$\frac{d x}{d y} + \frac{x}{y} = 2 \log y + 1$ which is linear in $x$ and $y .$

I.F $= e^{\int \frac{1}{y} d y} = y$ .

so $\frac{d}{d y} (x y) = 2 y \log y + y$

$\begin{array}{l} \Rightarrow x y = 2 \int y \log y d y + \frac{y^{2}}{2} + C \\ = 2 [\frac{y^{2}}{2} \log y - \frac{y^{2}}{4}] + \frac{y^{2}}{2} + C \\ = y^{2} \log y + C \end{array}$

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