Choose the correct answer from the options given below:

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

To match each differential equation with its correct integrating factor, we'll analyze each equation systematically:

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(A) For $x \, dy - (y + 2x^2) \, dx = 0$

Rearranging to standard form: $-(y + 2x^2) \, dx + x \, dy = 0$

Identifying $M = -(y + 2x^2)$ and $N = x$

Checking for exactness: $\dfrac{\partial M}{\partial y} = -1$ and $\dfrac{\partial N}{\partial x} = 1$

Since these aren't equal, we need an integrating factor.

We can verify that $\mu = \dfrac{1}{x}$ works as an integrating factor.

Multiplying the equation by $\dfrac{1}{x}$: $-\dfrac{y}{x} \, dx - 2x \, dx + dy = 0$

This is now exact, confirming that $\dfrac{1}{x}$ is the integrating factor.

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(B) For $(2x^2 - 3y) \, dx = x \, dy$

Rearranging: $(2x^2 - 3y) \, dx - x \, dy = 0$

Identifying $M = 2x^2 - 3y$ and $N = -x$

Checking the exactness condition shows this isn't exact.

Testing $\mu = x^2$ as the integrating factor:

Multiplying through: $x^2(2x^2 - 3y) \, dx - x^3 \, dy = 0$

This becomes $2x^4 \, dx - 3x^2y \, dx - x^3 \, dy = 0$

Verifying exactness confirms $x^2$ is the correct integrating factor.

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(C) For $(2y + 3x^2) \, dx + x \, dy = 0$

Identifying $M = 2y + 3x^2$ and $N = x$

Checking exactness: $\dfrac{\partial M}{\partial y} = 2$ and $\dfrac{\partial N}{\partial x} = 1$

These aren't equal, so we need an integrating factor.

Verifying $\mu = x^3$ works by multiplying through:

$x^3(2y + 3x^2) \, dx + x^4 \, dy = 0$

This becomes exact, confirming $x^3$ is the correct integrating factor.

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(D) For $2x \, dy + (3x^3 + 2y) \, dx = 0$

Identifying $M = 3x^3 + 2y$ and $N = 2x$

Verifying $\mu = x$ as the integrating factor:

$x(3x^3 + 2y) \, dx + 2x^2 \, dy = 0$

Which gives $3x^4 \, dx + 2xy \, dx + 2x^2 \, dy = 0$

This is now exact, confirming $x$ is the correct integrating factor.

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Therefore, the correct matches are:

• (A) → (I) $\dfrac{1}{x}$
  • (B) → (III) $x^2$
  • (C) → (IV) $x^3$
  • (D) → (II) $x$