Match List-I with List-II

List-I	List-II
Differential equation	Integrating factor
(A) $x\frac{dy}{dx} - y = 2x^2$	(I) $e^{-y}$
(B) $\frac{dy}{dx} + \frac{y}{x} = 2x$	(II) $\frac{1}{x}$
(C) $x\frac{dy}{dx} + 2y = x^2logx$	(III) $x$
(D) $\frac{dx}{dy} - x = y$	(IV) $x^2$

Choose the correct answer from the options given below:

For a differential equation in the form $\frac{dy}{dx} + P(x) \cdot y = Q(x)$, the integrating factor is $IF = e^{\int P(x)dx}$.

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(A) $x\frac{dy}{dx} - y = 2x^2$

Dividing by $x$:

$\frac{dy}{dx} - \frac{y}{x} = 2x$

Here $P(x) = -\frac{1}{x}$

$IF = e^{\int(-\frac{1}{x})dx}$

$IF = e^{-\ln|x|}$

$IF = e^{\ln(\frac{1}{x})}$

$IF = \frac{1}{x}$

Match: (A) → (II)

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(B) $\frac{dy}{dx} + \frac{y}{x} = 2x$

The equation is already in standard form.

Here $P(x) = \frac{1}{x}$

$IF = e^{\int(\frac{1}{x})dx}$

$IF = e^{\ln|x|}$

$IF = x$

Match: (B) → (III)

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(C) $x\frac{dy}{dx} + 2y = x^2\log x$

Dividing by $x$:

$\frac{dy}{dx} + \frac{2y}{x} = x\log x$

Here $P(x) = \frac{2}{x}$

$IF = e^{\int(\frac{2}{x})dx}$

$IF = e^{2\ln|x|}$

$IF = e^{\ln(x^2)}$

$IF = x^2$

Match: (C) → (IV)

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(D) $\frac{dx}{dy} - x = y$

This equation has $\frac{dx}{dy}$ instead of $\frac{dy}{dx}$.

For equations of the form $\frac{dx}{dy} + P(y) \cdot x = Q(y)$, the integrating factor is $IF = e^{\int P(y)dy}$.

Rewriting:

$\frac{dx}{dy} + (-1) \cdot x = y$

Here $P(y) = -1$

$IF = e^{\int(-1)dy}$

$IF = e^{-y}$

Match: (D) → (I)

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The correct matching is:

(A) → (II), (B) → (III), (C) → (IV), (D) → (I)