Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Differential Equation | General solution | ||
| (A) | \(\displaystyle\frac{dy}{dx}=\frac{y}{x};\ x \ne 0\) | (I) | \(y = cx\): c is an arbitrary constant |
| (B) | \(xdx - ydy= 0;\ y \ne 0,\ x \ne 0\) | (II) | \(x^2-y^2 = c\): c is an arbitrary constant |
| (C) | \(\displaystyle\frac{(x^2-1)}{y^2+1}\frac{dy}{dx}= 1\) | (III) | \(2x+3y= c\): c is an arbitrary constant |
| (D) | \(2dx+3dy = 0\) | (IV) | \((x^3-y^3) = c + 3(x + y)\): c is an arbitrary constant |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(I), (B)-(II), (C)-(IV), (D)-(III)
Each equation is solved by separating the variables and integrating both sides.
(A) \(\displaystyle\frac{dy}{dx}=\frac{y}{x};\ x \ne 0\)
This matches (I) \(y = cx\).
\( \displaystyle \int \frac{dy}{y} = \int \frac{dx}{x} \)
\(\log|y| = \log|x| + \log|c|\)
\(y = cx\)
(B) \(xdx - ydy= 0;\ y \ne 0,\ x \ne 0\)
This matches (II) \(x^2-y^2 = c\).
\( \displaystyle \int x\,dx = \int y\,dy \)
\( \displaystyle \frac{x^2}{2} = \frac{y^2}{2} + c_1 \)
\(x^2 - y^2 = c\), with \(c = 2c_1\)
(C) \(\displaystyle\frac{(x^2-1)}{y^2+1}\frac{dy}{dx}= 1\)
This matches (IV) \((x^3-y^3) = c + 3(x + y)\), by elimination. As printed, the equation gives
\( \displaystyle \frac{dy}{dx} = \frac{y^2+1}{x^2-1} \)
Differentiating (IV) instead gives \(3x^2 - 3y^2\frac{dy}{dx} = 3 + 3\frac{dy}{dx}\), that is,
\( \displaystyle \frac{dy}{dx} = \frac{x^2-1}{y^2+1} \)
This is the reciprocal of the printed slope, so the printed (C) does not satisfy (IV). (IV) is the solution of the intended equation \(\displaystyle\frac{(y^2+1)}{x^2-1}\frac{dy}{dx}= 1\):
\((y^2 + 1)\,dy = (x^2 - 1)\,dx\)
\( \displaystyle \frac{y^3}{3} + y = \frac{x^3}{3} - x + c_1 \)
Multiplying by 3 and putting \(c = -3c_1\):
\(y^3 + 3y = x^3 - 3x - c\)
\(x^3 - y^3 = c + 3(x + y)\)
(D) \(2dx+3dy = 0\)
This matches (III) \(2x+3y= c\). Integrating term by term gives \(2x + 3y = c\).
So the matching is (A)-(I), (B)-(II), (C)-(IV), (D)-(III).
Note: It seems that there is a mistake in the original NTA paper, which prints (C) as \(\displaystyle\frac{(x^2-1)}{y^2+1}\frac{dy}{dx}= 1\) when it should read \(\displaystyle\frac{(y^2+1)}{x^2-1}\frac{dy}{dx}= 1\). The question can still be solved because (A), (B) and (D) match (I), (II) and (III) exactly, and Option (4) is the only option containing these three matches.