CUET · MATHS · PYQ PAPER 2025
The solution of the differential equation \(\frac{d y}{d x}=\left(1+x^2\right)\left(1+y^2\right)\) is (Here \(C\) is an arbitrary constant)
- A \(\tan ^{-1} y+x+\frac{x^3}{3}=C\)
- B \(\tan ^{-1} y-x-\frac{x^3}{3}=C\)
- C \(\tan ^{-1} x-y-\frac{y^3}{3}=C\)
- D \(\tan ^{-1} x+y+\frac{y^3}{3}=C\)
Answer & Solution
Correct Answer
(B) \(\tan ^{-1} y-x-\frac{x^3}{3}=C\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{1+y^2} = (1+x^2) d x\) \(\int \frac{d y}{1+y^2} = \int (1+x^2) d x\) \(\tan^{-1} y = x + \frac{x^3}{3} + C\) \(\tan^{-1} y - x - \frac{x^3}{3} = C\)
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