Match List-I with List-II
| List-I | List-II | ||
| Differential Equation | Order and degree of differential Equation | ||
| (A) | \( \left(\displaystyle\frac{dy}{dx}\right)^2=10x \) | (I) | Order = 3 and degree = 1 |
| (B) | \( \displaystyle\frac{dy}{dx}+x\displaystyle\frac{d^2y}{dx^2}+4y=0 \) | (II) | Order = 1 and degree = 1 |
| (C) | \( \sqrt{\displaystyle\frac{dy}{dx}}+y=4x \) | (III) | Order = 2 and degree = 1 |
| (D) | \( 3\sqrt{\displaystyle\frac{d^3y}{dx^3}}-3y=x\displaystyle\frac{dy}{dx} \) | (IV) | Order = 1 and degree = 2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
The order is the order of the highest derivative. The degree is the power of that highest derivative after the equation is made free of radicals and fractional powers of the derivatives.
(A) \( \left(\frac{dy}{dx}\right)^2=10x \)
This matches (IV) Order = 1 and degree = 2. The highest derivative is dy/dx, and it is squared.
(B) \( \frac{dy}{dx}+x\frac{d^2y}{dx^2}+4y=0 \)
This matches (III) Order = 2 and degree = 1. The highest derivative is \( \frac{d^2y}{dx^2} \), to the power 1.
(C) \( \sqrt{\frac{dy}{dx}}+y=4x \)
This matches (II) Order = 1 and degree = 1. Removing the root:
\( \displaystyle \sqrt{\frac{dy}{dx}}=4x-y \)
\( \displaystyle \frac{dy}{dx}=(4x-y)^2 \)
Now dy/dx appears to the power 1.
(D) \( 3\sqrt{\frac{d^3y}{dx^3}}-3y=x\frac{dy}{dx} \)
This matches (I) Order = 3 and degree = 1. Removing the root:
\( \displaystyle 3\sqrt{\frac{d^3y}{dx^3}}=x\frac{dy}{dx}+3y \)
\( \displaystyle \frac{d^3y}{dx^3}=\frac{1}{9}\left(x\frac{dy}{dx}+3y\right)^2 \)
The highest derivative \( \frac{d^3y}{dx^3} \) now appears to the power 1.
So the matching is (A)-(IV), (B)-(III), (C)-(II), (D)-(I).