Match List-I with List-II

List-I	List-II
Differential Equation	Integrating Factor
(A) $y dx + (x - y^3)dy = 0$	(I) $e^{-x}$
(B) $x\frac{dy}{dx} + y = x^2$	(II) $\frac{1}{x}$
(C) $\frac{dy}{dx} - y = e^x$	(III) $y$
(D) $x dy - y dx = x^3 dx$	(IV) $x$

Choose the correct answer from the options given below:

For $y dx + (x - y^3)dy = 0$

$y dx = -(x - y^3)dy$

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

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

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

This is linear in $x$ with $P(y) = \frac{1}{y}$

Integrating Factor $= e^{\int P(y)dy}$

$= e^{\int \frac{1}{y}dy}$

$= e^{\ln y}$

$= y$

(A) → (III)

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

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

This is linear in $y$ with $P(x) = \frac{1}{x}$

Integrating Factor $= e^{\int P(x)dx}$

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

$= e^{\ln x}$

$= x$

(B) → (IV)

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For $\frac{dy}{dx} - y = e^x$

Already in standard form with $P(x) = -1$

Integrating Factor $= e^{\int P(x)dx}$

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

$= e^{-x}$

(C) → (I)

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For $x dy - y dx = x^3 dx$

$x dy = y dx + x^3 dx$

$x dy = (y + x^3) dx$

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

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

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

This is linear in $y$ with $P(x) = -\frac{1}{x}$

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

$= e^{-\ln x}$

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

$= \frac{1}{x}$

(D) → (II)

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Final Matching:

(A) → (III) $y$

(B) → (IV) $x$

(C) → (I) $e^{-x}$

(D) → (II) $\frac{1}{x}$