Match each differential equation in List-I with its degree in List-II.
| List-I (Differential equation) | List-II (Degree) | ||
| (A) | \(\displaystyle\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^5 + y^3 = 0\) | (I) | not defined |
| (B) | \(\displaystyle\frac{d^3y}{dx^3} = \sqrt{1 + \frac{dy}{dx}}\) | (II) | 3 |
| (C) | \(\left(\displaystyle\frac{d^2y}{dx^2}\right)^3 + \left(\frac{dy}{dx}\right)^4 = 0\) | (III) | 1 |
| (D) | \(\displaystyle\frac{d^2y}{dx^2} + x\,e^{\frac{dy}{dx}} = y\) | (IV) | 2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
The degree is the highest power of the highest derivative, once the equation is a polynomial in the derivatives.
(A) \(\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^5 + y^3 = 0\)
This matches (III) 1.
The highest derivative \(\frac{d^2y}{dx^2}\) has power 1; the powers 5 and 3 are on \(\frac{dy}{dx}\) and y.
(B) \(\frac{d^3y}{dx^3} = \sqrt{1 + \frac{dy}{dx}}\)
This matches (IV) 2.
Squaring both sides removes the root:
\( \displaystyle \left(\frac{d^3y}{dx^3}\right)^2 = 1 + \frac{dy}{dx} \)
The highest derivative \(\frac{d^3y}{dx^3}\) now has power 2.
(C) \(\left(\frac{d^2y}{dx^2}\right)^3 + \left(\frac{dy}{dx}\right)^4 = 0\)
This matches (II) 3.
The highest derivative \(\frac{d^2y}{dx^2}\) is cubed; the power 4 is on the lower derivative.
(D) \(\frac{d^2y}{dx^2} + x\,e^{\frac{dy}{dx}} = y\)
This matches (I) not defined.
\(e^{\frac{dy}{dx}}\) is not a polynomial in \(\frac{dy}{dx}\), so the equation is not a polynomial in the derivatives.