Question

Solution of the differential equation \(\left(1+y^{2}\right) \mathrm{d}x=\left(\tan^{-1}y-x\right) \mathrm{d}y\) is :
(where \(c\) is an arbitrary constant)

• A. \(x=1-\tan^{-1}y+ce^{-\tan^{-1}y}\)
• B. \(y=\tan^{-1}y-1+ce^{-\tan^{-1}x}\)
• C. \(x=\tan^{-1}y-1+ce^{-\tan^{-1}y}\)
• D. \(x=-\tan^{-1}y-1-ce^{\tan^{-1}x}\)

Hint:

Treat \(x\) as the dependent variable: write it as \(\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{\tan^{-1}y}{1+y^2}\) and use the integrating factor \(e^{\tan^{-1}y}\).

Answer 1

The Correct Option is C

Rewrite: \(\frac{dx}{dy}=\frac{\tan^{-1}y-x}{1+y^2}\Rightarrow\frac{dx}{dy}+\frac{1}{1+y^2}x=\frac{\tan^{-1}y}{1+y^2}\), a linear equation in \(x\).
Integrating factor: \(\text{I.F.}=e^{\int\frac{dy}{1+y^2}}=e^{\tan^{-1}y}\).
Solution: \(x\,e^{\tan^{-1}y}=\int\frac{\tan^{-1}y}{1+y^2}e^{\tan^{-1}y}dy\).
Put \(t=\tan^{-1}y\), \(dt=\frac{dy}{1+y^2}\): \(\int te^t dt=(t-1)e^t+c\).
So \(x\,e^{t}=(t-1)e^{t}+c\Rightarrow x=\tan^{-1}y-1+ce^{-\tan^{-1}y}\).
Answer: (3) \(x=\tan^{-1}y-1+ce^{-\tan^{-1}y}\)