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
Dividing by \((1 + y^2)\,dy\), the equation is linear in x, with y as the independent variable:
\( \displaystyle \frac{dx}{dy} = \frac{\tan^{-1}y - x}{1 + y^2} \)
\( \displaystyle \frac{dx}{dy} + \frac{1}{1 + y^2}\,x = \frac{\tan^{-1}y}{1 + y^2} \)
This is of the form \(\frac{dx}{dy} + P x = Q\) with \(P = \frac{1}{1 + y^2}\), so the integrating factor is
\( \displaystyle \text{I.F.} = e^{\int P\,dy} = e^{\int \frac{1}{1 + y^2}\,dy} = e^{\tan^{-1}y} \)
So the integrating factor is \(e^{\tan^{-1}y}\), and the only derivative \(\frac{dx}{dy}\) appears with power 1.
Checking the statements
(A) Integrating factor = \(\tan^{-1}x\) — Incorrect. The integrating factor is \(e^{\tan^{-1}y}\), a function of y.
(B) Integrating factor = \(\tan^{-1}y\) — Incorrect. \(\tan^{-1}y\) is only the exponent \(\int P\,dy\); the integrating factor is \(e^{\tan^{-1}y}\).
(C) Integrating factor = \(e^{\tan^{-1}y}\) — Correct. \(e^{\int \frac{1}{1 + y^2}\,dy} = e^{\tan^{-1}y}\).
(D) Degree = 1 — Correct. The equation is a polynomial in \(\frac{dx}{dy}\), with highest power 1.