Match List-I with List-II
| List-I | List-II | ||
| (Differential Equation) | (Degree) | ||
| (A) | \( \left(\displaystyle\frac{dy}{dx}\right)^3+\left(\displaystyle\frac{d^2y}{dx^2}\right)^2=0 \) | (I) | Not defined |
| (B) | \( \left(\displaystyle\frac{d^2y}{dx^2}\right)^3+\left(\displaystyle\frac{dy}{dx}\right)^2+1=0 \) | (II) | 2 |
| (C) | \( \sqrt{1+\displaystyle\frac{d^2y}{dx^2}}=\displaystyle\frac{dy}{dx}+x \) | (III) | 1 |
| (D) | \( \displaystyle\frac{d^2y}{dx^2}+e^{\frac{dy}{dx}}=0 \) | (IV) | 3 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (II), (B) - (IV), (C) - (III), (D) - (I)
The degree is the highest power of the highest order derivative, once the equation is a polynomial in its derivatives. If a derivative sits inside a function such as \( e^{y'} \), the degree is not defined.
(A) \( \left(\frac{dy}{dx}\right)^3+\left(\frac{d^2y}{dx^2}\right)^2=0 \)
This matches (II) 2.
The highest order derivative is \( \frac{d^2y}{dx^2} \), and its power is 2.
(B) \( \left(\frac{d^2y}{dx^2}\right)^3+\left(\frac{dy}{dx}\right)^2+1=0 \)
This matches (IV) 3.
The highest order derivative is \( \frac{d^2y}{dx^2} \), and its power is 3.
(C) \( \sqrt{1+\frac{d^2y}{dx^2}}=\frac{dy}{dx}+x \)
This matches (III) 1.
Squaring both sides removes the radical:
\( \displaystyle 1+\frac{d^2y}{dx^2}=\left(\frac{dy}{dx}+x\right)^2 \)
Now \( \frac{d^2y}{dx^2} \) appears with power 1.
(D) \( \frac{d^2y}{dx^2}+e^{\frac{dy}{dx}}=0 \)
This matches (I) Not defined.
\( \frac{dy}{dx} \) appears in an exponent, so the equation is not a polynomial in its derivatives.
So the matching is (A)-(II), (B)-(IV), (C)-(III), (D)-(I).