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
Correct Answer: B
Solution :
Given equation, \[\frac{dy}{dx}=\frac{x+y}{x-y}\] It is a homogeneous equation so putting \[y=vx\] and \[\frac{dy}{dx}=v+x\frac{dv}{dx},\] we get \[v+x\frac{dv}{dx}=\frac{x+vx}{x-vx}=\frac{1+v}{1-v}\] Þ \[x\frac{dv}{dx}=\frac{1+{{v}^{2}}}{1-v}\] Þ \[\frac{1}{x}dx=\left( \frac{1}{1+{{v}^{2}}}-\frac{v}{1+{{v}^{2}}} \right)dv\] Þ \[{{\log }_{e}}x={{\tan }^{-1}}v-\frac{1}{2}\log (1+{{v}^{2}})+{{\log }_{e}}c\] Substituting \[v=\frac{y}{x},\]we get \[{{\log }_{e}}x={{\tan }^{-1}}\frac{y}{x}-\frac{1}{2}\log \left[ 1+{{\left( \frac{y}{x} \right)}^{2}} \right]+{{\log }_{e}}c\] Þ \[c{{({{x}^{2}}+{{y}^{2}})}^{1/2}}={{e}^{{{\tan }^{-1}}(y/x)}}\].
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