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:

Before finding the degree, remove all square roots and fractional powers involving derivatives by squaring or raising both sides to appropriate powers.

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$\text{Order}$ $=$ the highest order derivative in the equation.

$\text{Degree}$ $=$ the power of the highest order derivative, after the equation is made free of radicals/fractional powers in derivatives.

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$(A) \quad (y'')^3 + (y')^4 - 6 = (y''')^2$

No radicals here, so we read directly:

Highest order derivative is $y'''$ $\Rightarrow$ $\text{Order} = 3$

Power of $y'''$ is $2$ $\Rightarrow$ $\text{Degree} = 2$

$(A) \rightarrow (IV)$

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$(B) \quad \sqrt{(y')^2 + 5} = y''$

Squaring both sides:

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

Highest order derivative is $y''$ $\Rightarrow$ $\text{Order} = 2$

Power of $y''$ is $2$ $\Rightarrow$ $\text{Degree} = 2$

$(B) \rightarrow (III)$

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$(C) \quad (y')^2 = (2 + y'')^{3/2}$

Raising both sides to the power $2$:

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

Highest order derivative is $y''$ $\Rightarrow$ $\text{Order} = 2$

Power of $y''$ is $3$ $\Rightarrow$ $\text{Degree} = 3$

$(C) \rightarrow (II)$

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$(D) \quad y = xy' + \sqrt{a^2(y')^2 + b^2}$

Isolate the radical:

$y - xy' = \sqrt{a^2(y')^2 + b^2}$

Squaring both sides:

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

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

Highest order derivative is $y'$ $\Rightarrow$ $\text{Order} = 1$

Power of $y'$ is $2$ $\Rightarrow$ $\text{Degree} = 2$

$(D) \rightarrow (I)$