Match List-I with List-II
| List-I | List-II |
| Differential equation | Curve satisfying the differential equation |
| (A) \( \displaystyle\frac{d^2y}{dx^2}+y=0 \) | (I) \( x^2+y^2=C \) : Where C is an arbitrary constant |
| (B) \( \displaystyle\frac{dy}{dx}+y=\displaystyle\frac{x^2+y}{x} \) | (II) \( 3e^{-4y}+4e^{3x}=C \) : Where C is an arbitrary constant |
| (C) \( \displaystyle\frac{dy}{dx}+\displaystyle\frac{x}{y}=0 \) | (III) \( y=sinx+cosx+C \) : Where C is an arbitrary constant |
| (D) \( \log(dy/dx)=3x+4y \) | (IV) \( y=\ x+Cxe^{-x} \) : Where C is an arbitrary constant |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
Each equation is solved (or each curve checked) and matched.
(C) \( \displaystyle\frac{dy}{dx}+\displaystyle\frac{x}{y}=0 \)
This matches (I). Separating the variables:
\( y\,dy=-x\,dx \)
\( \displaystyle \frac{y^2}{2}=-\frac{x^2}{2}+k \)
\( x^2+y^2=C \)
(D) \( \log(dy/dx)=3x+4y \)
This matches (II).
\( \displaystyle \frac{dy}{dx}=e^{3x}e^{4y} \)
\( e^{-4y}dy=e^{3x}dx \)
\( \displaystyle -\frac{e^{-4y}}{4}=\frac{e^{3x}}{3}+k \)
\( 3e^{-4y}+4e^{3x}=-12k=C \)
(B) \( \displaystyle\frac{dy}{dx}+y=\displaystyle\frac{x^2+y}{x} \)
This matches (IV). The equation is \( \frac{dy}{dx}+\left(1-\frac{1}{x}\right)y=x \), a linear equation with integrating factor \( e^{x-\log x}=\frac{e^x}{x} \):
\( \displaystyle y\cdot\frac{e^x}{x}=\int e^x\,dx \)
\( \displaystyle y\cdot\frac{e^x}{x}=e^x+C \)
\( y=x+Cxe^{-x} \)
(A) \( \displaystyle\frac{d^2y}{dx^2}+y=0 \)
This matches (III), by elimination. (B), (C) and (D) are matched uniquely with (IV), (I) and (II), so (III) is the only curve left for (A).
However, the curve as printed does not satisfy this equation. For \( y=\sin x+\cos x+C \):
\( y''=-\sin x-\cos x \)
\( y''+y=C \)
This is 0 only when C = 0. Without the constant, \( y=\sin x+\cos x \) does satisfy \( y''+y=0 \).
So the matching is (A) - (III), (B) - (IV), (C) - (I), (D) - (II).
Note: It seems that there is a mistake in the original NTA paper, which prints List-II (III) as \( y=\sin x+\cos x+C \) when it should read \( y=\sin x+\cos x \), without the added constant C. The question can still be solved, because the other three items are matched uniquely and (A) takes the remaining entry (III).