Match List-I with List-II List-I Differential equation List-II Order…

The correct answer is Option (3) → (A) - (IV), (B) - (III), (C) - (II), (D) - (I)

List-I Differential equation	List-II Order and Degree
(A) $(y'')^3 + (y')^4 - 6 = (y''')^2$	(IV) Order = 3, Degree = 2
(B) $\sqrt{(y')^2 + 5} = y''$	(III) Order = 2, Degree = 2
(C) $(y')^2 =(2 + y'')^{3/2}$	(II) Order = 2, Degree = 3
(D) $y = xy' + \sqrt{a^2(y')^2 + b^2}$	(I) Order = 1, Degree = 2

Given differential equations and definitions:

Order: The highest order derivative present in the equation.

Degree: The power of the highest order derivative after the equation is free from radicals and fractions involving derivatives.

Analyze each:

(A)

Highest order derivative: $y'''$ (third derivative).

Power of highest order derivative: 2 (since $(y''')^2$).

Order = 3

Degree = 2

(B)

Equation: $\sqrt{(y')^2 + 5} = y''$

Highest order derivative: $y''$ (second derivative).

Remove the square root by squaring both sides:

$(y')^2 + 5 = (y'')^2$

Now free from radicals.

Power of highest order derivative $y''$ is 2.

Order = 2

Degree = 2

(C)

Equation: $(y')^2 = (2 + y'')^{3/2}$

Highest order derivative: $y''$ (second derivative).

Remove fractional power by raising both sides to the power 2:

$((y')^2)^2 = (2 + y'')^{3}$

Or

$(y')^4 = (2 + y'')^{3}$

Power of highest order derivative $y''$ is 3.

Order = 2

Degree = 3

(D)

Equation: $y = x y' + \sqrt{a^2 (y')^2 + b^2}$

Highest order derivative: $y'$ (first derivative).

Expression inside the square root involves $(y')^2$.

Remove the square root by isolating and squaring if necessary, but here the derivative is inside a radical only in one term.

Power of highest order derivative in the original form is 1 inside the square root (because $(y')^2$ is under root).

After squaring:

$(y - x y')^2 = a^2 (y')^2 + b^2$

Highest order derivative $y'$ appears as $(y')^2$.

Order = 1

Degree = 2