Match List-I with List-II
| List-I | List-II | ||
| General solution | Differential Equation | ||
| (A) | $y=cx$ | (I) | $y'=\frac{1}{2\sqrt{x}}-1$ |
| (B) | $y=e^{-3x}+c$ | (II) | $y'+3y=0$ |
| (C) | $x+y=\sqrt{x}+c$ | (III) | $y'=\frac{y^2}{1-xy}(xy≠1)$ |
| (D) | $xy=log\, y + c$ | (IV) | $xy'=y$ |
(c is an arbitrary constant and $y'=\frac{dy}{dx}$)
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 4
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)