A) \[(1+{{x}^{2}})(1+{{y}^{2}})=c\]
B) \[(1+{{x}^{2}})(1+{{y}^{2}})=c{{x}^{2}}\]
C) \[(1-{{x}^{2}})(1-{{y}^{2}})=c\]
D) \[(1+{{x}^{2}})(1+{{y}^{2}})=c{{y}^{2}}\]
Correct Answer: B
Solution :
Given differential equation can be rewritten as \[\frac{y}{(1+{{y}^{2}})}dy=\frac{dx}{x(1+{{x}^{2}})}\] \[\Rightarrow \] \[\frac{1}{2}\int{\frac{2y}{(1+{{y}^{2}})}dy=\frac{1}{2}\int{\frac{2x\,\,dx}{{{x}^{2}}(1+{{x}^{2}})}}}\] \[\Rightarrow \] \[\frac{1}{2}\int{\frac{2y}{(1+{{y}^{2}})}dy=\frac{1}{2}\int{\frac{dt}{t(1+t)}}}\] (Put \[{{x}^{2}}=t\] in RHS integral] \[\Rightarrow \] \[\frac{1}{2}\log \,(1+{{y}^{2}})=\frac{1}{2}\left[ \int{\frac{1}{t}dt-\int{\frac{1}{1+t}dt}} \right]\] \[\Rightarrow \] \[\frac{1}{2}\frac{1}{\log }(1+{{y}^{2}})=\frac{1}{2}[logt-log(1+t)]\] \[+\frac{1}{2}\log \,c\] \[\Rightarrow \] \[\log \,\,(1+{{y}^{2}})\,(1+{{x}^{2}})=\log \,c{{x}^{2}}\] \[\Rightarrow \] \[(1+{{y}^{2}})\,(1+{{x}^{2}})=c{{x}^{2}}\]
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