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
- A \(x+y=\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
- B \(x-y=\mathrm{c}(x y)\), where \(\mathrm{c}\) is a constant of integration.
- C \(x+y=\mathrm{c}(1+x y)\), where \(\mathrm{c}\) is a constant of integration.
- D \(y-x=\mathrm{c}(1+x y)\), where \(\mathrm{c}\) is a constant of integration.
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
\(\int \frac{\mathrm{d} y}{1+y^2} = \int \frac{\mathrm{d} x}{1+x^2}\) \(\arctan(y) = \arctan(x) + C\)
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