Match List-I with List-II
| List-I | List-II | ||
| Differential equation | Order and degree | ||
| (A) | \((y'')^3 + (y')^4 - 6 = (y''')^2\) | (I) | Order = 1, Degree = 2 |
| (B) | \(\sqrt{(y')^2 + 5} = y''\) | (II) | Order = 2, Degree = 3 |
| (C) | \((y')^2 = (2 + y'')^{3/2}\) | (III) | Order = 2, Degree = 2 |
| (D) | \(y = xy' + \sqrt{a^2(y')^2 + b^2}\) | (IV) | Order = 3, Degree = 2 |
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) - (II), (D) - (I)
The order is the highest derivative present. The degree is the power of that derivative once the equation is a polynomial in the derivatives (free of radicals and fractional powers).
(A) \((y'')^3 + (y')^4 - 6 = (y''')^2\)
This matches (IV) Order = 3, Degree = 2. The highest derivative is \(y'''\), and it appears squared; the equation is already a polynomial in the derivatives.
(B) \(\sqrt{(y')^2 + 5} = y''\)
This matches (III) Order = 2, Degree = 2. Squaring removes the radical:
\((y')^2 + 5 = (y'')^2\)
The highest derivative \(y''\) appears with power 2.
(C) \((y')^2 = (2 + y'')^{3/2}\)
This matches (II) Order = 2, Degree = 3. Squaring removes the fractional power:
\((y')^4 = (2 + y'')^3\)
The highest derivative \(y''\) now appears with power 3.
(D) \(y = xy' + \sqrt{a^2(y')^2 + b^2}\)
This matches (I) Order = 1, Degree = 2. Isolating the radical and squaring:
\((y - xy')^2 = a^2(y')^2 + b^2\)
Only \(y'\) appears, and its highest power is 2.
So the matching is (A) - (IV), (B) - (III), (C) - (II), (D) - (I).