The solution of the differential equation \[xy\frac{dy}{dx}=\frac{(1+{{y}^{2}})(1+x+{{x}^{2}})}{(1+{{x}^{2}})}\] is [AISSE 1983]

\[\log (1+{{y}^{2}})=\log x-{{\tan }^{-1}}x+c\]

\[\frac{1}{2}\log (1+{{y}^{2}})=\log x-{{\tan }^{-1}}x+c\]

\[\frac{1}{2}\log (1+{{y}^{2}})=\log x+{{\tan }^{-1}}x+c\]

\[\log (1+{{y}^{2}})=\log x+{{\tan }^{-1}}x+c\]

The solution of the differential equation \[xy\frac{dy}{dx}=\frac{(1+{

1.	The solution of the differential equation \[xy\frac{dy}{dx}=\frac{(1+{{y}^{2}})(1+x+{{x}^{2}})}{(1+{{x}^{2}})}\] is [AISSE 1983]
A.	\[\frac{1}{2}\log (1+{{y}^{2}})=\log x-{{\tan }^{-1}}x+c\]
B.	\[\frac{1}{2}\log (1+{{y}^{2}})=\log x+{{\tan }^{-1}}x+c\]
C.	\[\log (1+{{y}^{2}})=\log x-{{\tan }^{-1}}x+c\]
D.	\[\log (1+{{y}^{2}})=\log x+{{\tan }^{-1}}x+c\]
Answer» C. \[\log (1+{{y}^{2}})=\log x-{{\tan }^{-1}}x+c\]

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The solution of the differential equation \[xy\frac{dy}{dx}=\frac{(1+{{y}^{2}})(1+x+{{x}^{2}})}{(1+{{x}^{2}})}\] is [AISSE 1983]

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