Match List-I with List-II
Choose the correct answer from the options given below: |
(A)-(IV), (B)-(I), (C)-(III), (D)-(II) (A)-(I), (B)-(IV), (C)-(II), (D)-(III) (A)-(II), (B)-(III), (C)-(IV), (D)-(I) (A)-(III), (B)-(IV), (C)-(II), (D)-(I) |
(A)-(II), (B)-(III), (C)-(IV), (D)-(I) |
The correct answer is Option (3) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
Matching the items with explanation: (A) $\frac{d^2y}{dx^2} = e^{\frac{dy}{dx}}$: The degree is not defined because the derivative $\frac{dy}{dx}$ is inside the exponential function, which makes the equation non-polynomial. → (II) not defined. (B) $\left(\frac{dy}{dx}\right)^2 + \frac{d^3y}{dx^3} = 0$: The order is the highest derivative present, which is $\frac{d^3y}{dx^3}$. The degree is the exponent of this highest-order derivative, which is 1. Thus, the order is $3$. → (III) 3. (C) $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 - 5x^2 = 0$: The highest-order derivative is $\frac{d^2y}{dx^2}$, which appears with power 1. Hence, the degree is 2 due to the squared first derivative. → (I) 2. (D) $\frac{dy}{dx} + 3y = e^x$: The order is 1, and the degree is also 1 since the first derivative appears linearly. $p + q = 1 + 1 = 2$. → (IV) 1. |