Match List-I with List-II

List-I	List-II
Differential Equations	Order and degree
(A) $\frac{dy}{dx} + e^y = 0$	(I) order 2, degree not defined
(B) $\frac{d^2y}{dx^2} = \left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2}$	(II) order 2, degree 1
(C) $\left(\frac{d^2y}{dx^2}\right)^2 + e^{(\frac{dy}{dx})} = 0$	(III) order 1, degree 1
(D) $\frac{d^2y}{dx^2} + x\frac{dy}{dx} - 2y = logx; x > 0$	(IV) order 2, degree 2

Choose the correct answer from the options given below:

Order is determined by the highest derivative present in the equation.

Degree is the power of the highest order derivative when the equation is polynomial in derivatives. If derivatives appear inside exponentials, logarithms, or fractional powers that cannot be removed, the degree is not defined.

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For $\frac{dy}{dx} + e^y = 0$:

The highest derivative is $\frac{dy}{dx}$.

Order $= 1$

The derivative $\frac{dy}{dx}$ appears to power $1$. The term $e^y$ involves the function, not a derivative.

Degree $= 1$

(A) matches with (III)

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For $\frac{d^2y}{dx^2} = \left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2}$:

The highest derivative is $\frac{d^2y}{dx^2}$.

Order $= 2$

The equation has a fractional power $\frac{3}{2}$. Squaring both sides:

$\left(\frac{d^2y}{dx^2}\right)^2 = \left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3}$

The equation becomes polynomial in derivatives. The highest derivative $\frac{d^2y}{dx^2}$ appears to power $2$.

Degree $= 2$

(B) matches with (IV)

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For $\left(\frac{d^2y}{dx^2}\right)^2 + e^{(\frac{dy}{dx})} = 0$:

The highest derivative is $\frac{d^2y}{dx^2}$.

Order $= 2$

The derivative $\frac{dy}{dx}$ appears inside an exponential function $e^{(\frac{dy}{dx})}$. This cannot be removed by algebraic manipulation, so the equation is not polynomial in derivatives.

Degree is not defined

(C) matches with (I)

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For $\frac{d^2y}{dx^2} + x\frac{dy}{dx} - 2y = \log x$:

The highest derivative is $\frac{d^2y}{dx^2}$.

Order $= 2$

The equation is polynomial in derivatives. The highest derivative $\frac{d^2y}{dx^2}$ appears to power $1$. The term $\log x$ is a function of $x$ only.

Degree $= 1$

(D) matches with (II)

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Final matching:

(A) - (III): Order 1, Degree 1

(B) - (IV): Order 2, Degree 2

(C) - (I): Order 2, Degree not defined

(D) - (II): Order 2, Degree 1