Match List - I with List - II.
| List - I | List - II | ||
| (A) | solution of \((x^2+1)\displaystyle\frac{dy}{dx}=x\) | (I) | 3 |
| (B) | solution of \(\displaystyle\frac{dy}{dx}=e^x\) | (II) | 1 |
| (C) | order of \(\displaystyle\frac{d^3y}{dx^3}=0\) | (III) | \(y=e^x+C\) |
| (D) | degree of \(x^2\displaystyle\frac{d^2y}{dx^2}+\left(\displaystyle\frac{dy}{dx}\right)^2=0\) | (IV) | \(y=\displaystyle\frac{1}{2}\log(x^2+1)+C\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(A) solution of \((x^2+1)\frac{dy}{dx}=x\)
This matches (IV) \(y=\frac{1}{2}\log(x^2+1)+C\).
Separating the variables:
\( \displaystyle dy = \frac{x}{x^2+1}\,dx \)
The numerator is half the derivative of \(x^2+1\), so
\( \displaystyle y = \frac{1}{2}\log(x^2+1)+C \)
(B) solution of \(\frac{dy}{dx}=e^x\)
This matches (III) \(y=e^x+C\).
\(y=\int e^x\,dx=e^x+C\)
(C) order of \(\frac{d^3y}{dx^3}=0\)
This matches (I) 3: the highest derivative is the third derivative.
(D) degree of \(x^2\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2=0\)
This matches (II) 1: the highest-order derivative \(\frac{d^2y}{dx^2}\) appears to the power 1. The square is on \(\frac{dy}{dx}\), which does not decide the degree.
So the matching is (A)-(IV), (B)-(III), (C)-(I), (D)-(II).