Match the LIST-I with LIST-II
| LIST-I Differential equation | LIST-II Order and degree | ||
| A. | \( y''=\left[1+(y')^{2}\right]^{3/2} \) | I. | 3, 4 |
| B. | \( y'''+y'+y=2x \) | II. | 2, 2 |
| C. | \( 2(y'')^{3}-(y')^{4}+y=0 \) | III. | 3, 1 |
| D. | \( (y''')^{4}+(y'')^{3}+y'+y=0 \) | IV. | 2, 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 - III, C - IV, D - I
The order is the order of the highest derivative. The degree is the power of that derivative once the equation is free of radicals and fractional powers.
(A) \( y''=[1+(y')^{2}]^{3/2} \)
The highest derivative is y″, so the order is 2. Squaring both sides removes the fractional power:
\( \displaystyle (y'')^{2}=[1+(y')^{2}]^{3} \)
So the degree is 2. This matches (II) 2, 2.
(B) \( y'''+y'+y=2x \)
The order is 3 and the degree is 1. This matches (III) 3, 1.
(C) \( 2(y'')^{3}-(y')^{4}+y=0 \)
The highest derivative is y″ with power 3, so the order is 2 and the degree is 3. This matches (IV) 2, 3.
(D) \( (y''')^{4}+(y'')^{3}+y'+y=0 \)
The highest derivative is y′′′ with power 4, so the order is 3 and the degree is 4. This matches (I) 3, 4.
So the matching is (A)-(II), (B)-(III), (C)-(IV), (D)-(I).
Tip: The degree is the power of the highest-order derivative only. In (C) it is 3, not 4, because the power 4 belongs to y′.