Match List-I with List-II

Choose the correct answer from the options given below:

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

The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Given equations and integrator checks (write each in standard linear or exact form and find I.F.):

(A) $y\,dx+(x-y^3)\,dy=0$. Write as $y\,\frac{dx}{dy}=y^3-x$, so $\frac{dx}{dy}+\frac{1}{y}x=y^2$. This is linear in $x$ with $P(y)=\frac{1}{y}$, I.F. $=e^{\int \frac{1}{y}\,dy}=e^{\ln y}=y$. ⇒ (A) → (III)

(B) $x\frac{dy}{dx}+y=x^2$. Rewrite $\frac{dy}{dx}+\frac{1}{x}y=x$. Here $P(x)=\frac{1}{x}$, I.F. $=e^{\int \frac{1}{x}\,dx}=e^{\ln x}=x$. ⇒ (B) → (IV)

(C) $\frac{dy}{dx}-y=e^{x}$. Here $P(x)=-1$, I.F. $=e^{\int -1\,dx}=e^{-x}$. ⇒ (C) → (I)

(D) $x\,dy- y\,dx = x^{3}\,dx$. Divide by $dx$: $x\frac{dy}{dx}-y=x^{3}\Rightarrow \frac{dy}{dx}-\frac{1}{x}y=x^{2}$. Here $P(x)=-\frac{1}{x}$, I.F. $=e^{\int -\frac{1}{x}\,dx}=e^{-\ln x}=\frac{1}{x}$. ⇒ (D) → (II)