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CUET · MATHS · PYQ PAPER 2025

The general solution of the differential equation \(x\left(1+y^2\right) d x+y\left(1+x^2\right) d y=0\) is

  1. A \(\left(1+y^2\right)\left(1+x^2\right)=C\) : where \(C\) is an arbitrary constant
  2. B \(\left(1-y^2\right)\left(1-x^2\right)=C\) : where \(C\) is an arbitrary constant
  3. C \(\left(1-y^2\right)\left(1+x^2\right)=C:\) where \(C\) is an arbitrary constant
  4. D \(\left(1+y^2\right)\left(1-x^2\right)=C:\) where \(C\) is an arbitrary constant
Verified Solution

Answer & Solution

Correct Answer

(A) \(\left(1+y^2\right)\left(1+x^2\right)=C\) : where \(C\) is an arbitrary constant

Step-by-step Solution

Detailed explanation

\(\frac{x}{1+x^2} d x = -\frac{y}{1+y^2} d y\) \(\int \frac{x}{1+x^2} d x = -\int \frac{y}{1+y^2} d y\) \(\frac{1}{2} \ln(1+x^2) = -\frac{1}{2} \ln(1+y^2) + \frac{1}{2} \ln C\) \(\ln(1+x^2) + \ln(1+y^2) = \ln C\) \((1+x^2)(1+y^2) = C\)
From CUET
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