tan−1x + tan−1y = c is the general solution of the differential equation

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  1. \( \frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}} \)
  2. \(\frac{d y}{d x}=\frac{1+x^{2}}{1+y^{2}}\)
  3. (1 + x2)dy + (1 + y2)dx = 0
  4. (1 + x2)dx + (1 + y2)dy = 0

Answer (Detailed Solution Below)

Option 3 : (1 + x2)dy + (1 + y2)dx = 0
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Detailed Solution

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Calculation

Let us find the differential equation by differentiating y with respect to x

⇒ tan-1x + tan-1y = c

Differentiating with respect to x

\(\Rightarrow \frac{1}{1+x^2} + \frac{1}{1+y^2} (\frac{dy}{dx}) = 0\)

\(\Rightarrow \frac{(1+y^2)dx + (1+x^2)dy}{(1+x^2)(1+y^2)dx} = 0\)

\(\Rightarrow (1+y^2)dx + (1+x^2)dy = 0\)

Hence option 3 is correct

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