Match List-I with List-II:

The correct option is (A)- (I), (B)- (IV), (C)- (III), (D)- (II).

To find the matching pairs between List-I and List-II, we determine the integrating factor (I.F.) for each of the first-order linear differential equations given in List-I.

Recall that a standard first-order linear differential equation is written in the form:

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

The integrating factor for this form is given by the formula:

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

Step-by-step derivation for each option:

(A) Integrating factor of $xdy-(y+2x^2)dx=0$:
We rearrange the given equation into standard form:
$xdy=(y+2x^2)dx$
Dividing by $xdx$ (assuming $x\ne 0$):
$$\frac{dy}{dx}-\frac{1}{x}y=2x$$ Here, the coefficient of $y$ is $P\left(x\right)=-\frac{1}{x}$.
Calculating the Integrating Factor:
$$\text{I.F.}=e^{\int -\frac{1}{x}dx}=e^{-lnx}=e^{ln\left(x^{-1}\right)}=x^{-1}=\frac{1}{x}$$ Therefore, (A) matches with (I).

(B) Integrating factor of $(2x^2-3y)dx=xdy$:
We rearrange the equation:
$x\frac{dy}{dx}+3y=2x^2$
Dividing by $x$:
$$\frac{dy}{dx}+\frac{3}{x}y=2x$$ Here, the coefficient of $y$ is $P\left(x\right)=\frac{3}{x}$.
Calculating the Integrating Factor:
$$\text{I.F.}=e^{\int \frac{3}{x}dx}=e^{3lnx}=e^{ln\left(x^3\right)}=x^3$$ Therefore, (B) matches with (IV).

(C) Integrating factor of $(2y+3x^2)dx+xdy=0$:
We rearrange the equation:
$x\frac{dy}{dx}+2y=-3x^2$
Dividing by $x$:
$$\frac{dy}{dx}+\frac{2}{x}y=-3x$$ Here, the coefficient of $y$ is $P\left(x\right)=\frac{2}{x}$.
Calculating the Integrating Factor:
$$\text{I.F.}=e^{\int \frac{2}{x}dx}=e^{2lnx}=e^{ln\left(x^2\right)}=x^2$$ Therefore, (C) matches with (III).

(D) Integrating factor of $2xdy+(3x^3+2y)dx=0$:
We rearrange the equation:
$2x\frac{dy}{dx}+2y=-3x^3$
Dividing by $2x$:
$$\frac{dy}{dx}+\frac{1}{x}y=-\frac{3}{2}{x}^2$$ Here, the coefficient of $y$ is $P\left(x\right)=\frac{1}{x}$.
Calculating the Integrating Factor:
$$\text{I.F.}=e^{\int \frac{1}{x}dx}=e^{lnx}=x$$ Therefore, (D) matches with (II).

Combining all the correct matches, we get:
(A) - (I), (B) - (IV), (C) - (III), (D) - (II)