Match List-I with List-II

List-I	List-II
(A) Integrating factor of $xdy - (y + x^2)dx = 0$	(I) $x^2$
(B) Integrating factor of $xdy + (2y + x^2)dx = 0$	(II) $x^3$
(C) Integrating factor of $(3y - x^2)dx + xdy = 0$	(III) $x$
(D) Integrating factor of $(y + 3x^2)dx + xdy = 0$	(IV) $\frac{1}{x}$

Choose the correct answer from the options given below:

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

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(A) $xdy - (y + x^2)dx = 0$

Rearranging:

$xdy = (y + x^2)dx$

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

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

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

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

The integrating factor:

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

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

$\mu = e^{\ln|x|^{-1}}$

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

(A) matches with (IV)

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(B) $xdy + (2y + x^2)dx = 0$

Rearranging:

$xdy = -(2y + x^2)dx$

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

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

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

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

The integrating factor:

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

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

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

$\mu = x^2$

(B) matches with (I)

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(C) $(3y - x^2)dx + xdy = 0$

Rearranging:

$xdy = -(3y - x^2)dx$

$\frac{dy}{dx} = \frac{-3y + x^2}{x}$

$\frac{dy}{dx} = -\frac{3y}{x} + x$

$\frac{dy}{dx} + \frac{3}{x} \cdot y = x$

Here, $P(x) = \frac{3}{x}$

The integrating factor:

$\mu = e^{\int (\frac{3}{x})dx}$

$\mu = e^{3\ln|x|}$

$\mu = e^{\ln|x^3|}$

$\mu = x^3$

(C) matches with (II)

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(D) $(y + 3x^2)dx + xdy = 0$

Rearranging:

$xdy = -(y + 3x^2)dx$

$\frac{dy}{dx} = -\frac{y + 3x^2}{x}$

$\frac{dy}{dx} = -\frac{y}{x} - 3x$

$\frac{dy}{dx} + \frac{1}{x} \cdot y = -3x$

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

The integrating factor:

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

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

$\mu = x$

(D) matches with (III)

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Therefore, (A) - (IV), (B) - (I), (C) - (II), (D) - (III)