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MHT CET · Maths · Differential Equations

The solution of the differential equation \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1+y^2}{1+x^2}\) is

  1. A \(x+y=\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
  2. B \(x-y=\mathrm{c}(x y)\), where \(\mathrm{c}\) is a constant of integration.
  3. C \(x+y=\mathrm{c}(1+x y)\), where \(\mathrm{c}\) is a constant of integration.
  4. D \(y-x=\mathrm{c}(1+x y)\), where \(\mathrm{c}\) is a constant of integration.
Verified Solution

Answer & Solution

Correct Answer

(D) \(y-x=\mathrm{c}(1+x y)\), where \(\mathrm{c}\) is a constant of integration.

Step-by-step Solution

Detailed explanation

\(\int \frac{\mathrm{d} y}{1+y^2} = \int \frac{\mathrm{d} x}{1+x^2}\) \(\arctan(y) = \arctan(x) + C\)