$\tan ^{-1} x+\tan ^{-1} y=c$ is the general solution of the differential equation

$\tan ^{-1} x+\tan ^{-1} y=c$ is the general solution of the differential equation
  1. $\frac{d y}{d x}=-\left(\frac{1+y^{2}}{1+x^{2}}\right)$
  2. $\frac{d y}{d x}=\left(\frac{1+y^{2}}{1+x^{2}}\right)$
  3. $\frac{d y}{d x}=-\left(\frac{1+x^{2}}{1+y^{2}}\right)$
  4. $\frac{d y}{d x}=\left(\frac{1+x^{2}}{1+y^{2}}\right)$

Solution

Given $\tan ^{-1} x+\tan ^{-1} y=c$ $\therefore \frac{1}{1+x^{2}}+\frac{1}{1+y^{2}} \frac{d y}{d x}=0$ $\frac{1}{1+y^{2}} \frac{d y}{d x}=-\frac{1}{1+x^{2}} \quad \Rightarrow \frac{d y}{d x}=-\left(\frac{1+y^{2}}{1+x^{2}}\right)$

Asked in: MHT CET 2020 (13 Oct Shift 2)

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