Given differential equation, $(1+ y^2)dx = (\tan^{-1}y-x)dy$, then which of the following is/are true?
(A) Integrating factor = $\tan^{-1}x$
(B) Integrating factor = $\tan^{-1}y$
(C) Integrating factor = $e^{\tan^{-1}}y$
(D) Degree = 1
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (C) and (D) only
Given differential equation:
$(1+y^2)\,dx = (\tan^{-1}y - x)\,dy$
Rewrite as linear in $x$ with independent variable $y$:
$\displaystyle \frac{dx}{dy} + \frac{1}{1+y^2}\,x = \frac{\tan^{-1}y}{1+y^2}$
Check for an integrating factor depending only on $y$ by computing the coefficient of $x$ (which is $\frac{1}{1+y^2}$). An integrating factor is $\displaystyle \mu(y)=\exp\!\bigg(\int \frac{1}{1+y^2}\,dy\bigg)=\exp(\tan^{-1}y)=e^{\tan^{-1}y}\,$, so a valid integrating factor is $e^{\tan^{-1}y}$.
Therefore:
- (A) Integrating factor = $\tan^{-1}x$ — False.
- (B) Integrating factor = $\tan^{-1}y$ — False (not of the form $\mu(y)$; missing the exponential).
- (C) Integrating factor = $e^{\tan^{-1}y}$ — True.
- (D) Degree = 1 — True (first degree in $\frac{dx}{dy}$ or $\frac{dy}{dx}$).