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Solve the following.
Solve the differential equation: (1 + y²)(1 + log x) dx + x dy = 0
Solve the differential equation: $(1 + y^2)(1 + \log x) \, dx + x \, dy = 0$
Answer
Answer
Separating the variables: $\frac{dy}{1 + y^2} = -\frac{1 + \log x}{x}\, dx$
Integrating: $\tan^{-1} y = -\left(\log x + \frac{(\log x)^2}{2}\right) + C$
Hence $\tan^{-1} y + \log x + \frac{(\log x)^2}{2} = C$.
Final answer: $\tan^{-1} y + \log x + \frac{(\log x)^2}{2} = C$