Match List-I with List-II

(c is an arbitrary constant and $y'=\frac{dy}{dx}$)

Choose the correct answer from the options given below :

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

The correct answer is Option (4) → (A)-(IV), (B)-(II),(C)-(I),(D)-(III)

(A) $y=cx⇒\frac{y}{x}=c$

so $\frac{xy'-yx'}{x^2}=0⇒xy'=y$ (IV)

(B) $y=e^{-3x}+c$

$y'=-3e^{-3x}$

$y'=-3y⇒y'+3y=0$ (II)

(C) $x+y=\sqrt{x}+c$

$1+y'=\frac{1}{2\sqrt{x}}⇒y'=\frac{1}{2\sqrt{x}}-1$ (I)

(D) $xy=\log y + c$

$⇒y'x+y=\frac{1}{y}y'$ so $y'(\frac{1}{y}-x)=y$

so $y'=\frac{y^2}{1-xy}$ (III)