Match List - I with List - II
| List - I (Differential Equation) | List - II (Degree) | ||
| A. | \(\left[1 + (y')^2\right] = y''\) | I. | 2 |
| B. | \(\left[1 + (y'')^3\right]^{\frac{1}{2}} = (y')^3\) | II. | 4 |
| C. | \((y''')^2 + y'' + 3y' + 5y = e^x\) | III. | 1 |
| D. | \(\left[1 + (y')^3\right]^{\frac{1}{2}} = (y'')^2\) | IV. | 3 |
Choose the correct answer from the option given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-III, B-IV, C-I, D-II
The degree is the highest power of the highest-order derivative, once the equation is a polynomial in the derivatives. A fractional power must first be removed by squaring.
(A) \(\left[1 + (y')^2\right] = y''\)
This matches (III) 1. The highest-order derivative y'' appears with power 1.
(B) \(\left[1 + (y'')^3\right]^{\frac{1}{2}} = (y')^3\)
This matches (IV) 3. Squaring both sides:
\(1 + (y'')^3 = (y')^6\)
The highest-order derivative y'' appears with power 3.
(C) \((y''')^2 + y'' + 3y' + 5y = e^x\)
This matches (I) 2. The highest-order derivative y''' appears with power 2.
(D) \(\left[1 + (y')^3\right]^{\frac{1}{2}} = (y'')^2\)
This matches (II) 4. Squaring both sides:
\(1 + (y')^3 = (y'')^4\)
The highest-order derivative y'' appears with power 4.
So the matching is A-III, B-IV, C-I, D-II.