The solution of the equation \[\frac{dy}{dx}=\frac{x+y}{x-y}\]is [AI CBSE 1990]

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

\[c{{({{x}^{2}}+{{y}^{2}})}^{1/2}}+{{e}^{{{\tan }^{-1}}(y/x)}}=0\]

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

The solution of the equation \[\frac{dy}{dx}=\frac{x+y}{x-y}\]is �

1.	The solution of the equation \[\frac{dy}{dx}=\frac{x+y}{x-y}\]is [AI CBSE 1990]
A.	\[c{{({{x}^{2}}+{{y}^{2}})}^{1/2}}+{{e}^{{{\tan }^{-1}}(y/x)}}=0\]
B.	\[c{{({{x}^{2}}+{{y}^{2}})}^{1/2}}={{e}^{{{\tan }^{-1}}(y/x)}}\]
C.	\[c({{x}^{2}}-{{y}^{2}})={{e}^{{{\tan }^{-1}}(y/x)}}\]
D.	None of these
Answer» C. \[c({{x}^{2}}-{{y}^{2}})={{e}^{{{\tan }^{-1}}(y/x)}}\]

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