Given differential equation, (1 + y²)dx = (tan⁻¹y - x)dy, then which of the following is/are true?

(A) Integrating factor = tan⁻¹x

(B) Integrating factor = tan⁻¹y

(C) Integrating factor = $e^{tan⁻¹y}$

(D) Degree = 1

Choose the correct answer from the options given below:

Given differential equation: $(1 + y^2)dx = (\tan^{-1}y - x)dy$

Dividing both sides by $dy$:

$(1 + y^2) \cdot \frac{dx}{dy} = \tan^{-1}y - x$

$\frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{\tan^{-1}y}{1 + y^2}$

This is in the standard linear form: $\frac{dx}{dy} + P(y) \cdot x = Q(y)$

Where $P(y) = \frac{1}{1 + y^2}$ and $Q(y) = \frac{\tan^{-1}y}{1 + y^2}$

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The integrating factor is given by:

$I.F. = e^{\int P(y)dy}$

$= e^{\int \frac{1}{1+y^2} dy}$

Since $\int \frac{1}{1+y^2} dy = \tan^{-1}y$:

$I.F. = e^{\tan^{-1}y}$

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From the standard form $\frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{\tan^{-1}y}{1 + y^2}$:

The derivative $\frac{dx}{dy}$ appears to the power $1$.

There are no radicals or fractions involving the derivative.

Therefore, Degree $= 1$

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Option (A): Integrating factor $= \tan^{-1}x$ is incorrect.

Option (B): Integrating factor $= \tan^{-1}y$ is incorrect.

Option (C): Integrating factor $= e^{\tan^{-1}y}$ is correct.

Option (D): Degree $= 1$ is correct.

Therefore, statements (C) and (D) are true.