Match List-I with List-II

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

Choose the correct answer from the options given below:

For any linear first-order differential equation in the form:

$\frac{dy}{dx} + P(x) \cdot y = Q(x)$

The Integrating Factor is:

$I.F. = e^{\int P(x)dx}$

If the equation isn't in this standard form, divide the entire equation by the coefficient of $\frac{dy}{dx}$ first.

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(A) $\frac{dy}{dx} + 2xy = 1$

The equation is already in standard form.

$P(x) = 2x$

$I.F. = e^{\int 2x \, dx}$

$I.F. = e^{x^2}$

Match: (A) → (IV)

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(B) $x\frac{dy}{dx} + 2xy = 1$

The coefficient of $\frac{dy}{dx}$ is $x$. Dividing the entire equation by $x$:

$\frac{dy}{dx} + 2y = \frac{1}{x}$

$P(x) = 2$

$I.F. = e^{\int 2 \, dx}$

$I.F. = e^{2x}$

Match: (B) → (II)

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

Dividing the entire equation by $x$:

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

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

$I.F. = e^{\int \frac{1}{x} \, dx}$

$I.F. = e^{\ln x}$

$I.F. = x$

Match: (C) → (I)

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

Dividing the entire equation by $x$:

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

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

$I.F. = e^{\int \frac{2}{x} \, dx}$

$I.F. = e^{2\ln x}$

$I.F. = e^{\ln x^2}$

$I.F. = x^2$

Match: (D) → (III)

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Equation	Integrating Factor
(A)	(IV) $e^{x^2}$
(B)	(II) $e^{2x}$
(C)	(I) $x$
(D)	(III) $x^2$