A) \[\sqrt{1+{{x}^{2}}}+\sqrt{1+{{y}^{2}}}=c\]
B) \[\sqrt{1+{{x}^{2}}}-\sqrt{1+{{y}^{2}}}=c\]
C) \[{{(1+{{x}^{2}})}^{3/2}}+{{(1+{{y}^{2}})}^{3/2}}=c\]
D) None of these
Correct Answer: A
Solution :
Given equation is, \[(x\sqrt{1+{{y}^{2}}})dx+(y\sqrt{1+{{x}^{2}}})dy=0\] Þ \[x\sqrt{1+{{y}^{2}}}dx=-y\sqrt{1+{{x}^{2}}}dy\] Þ \[\int_{{}}^{{}}{\frac{x}{\sqrt{1+{{x}^{2}}}}dx+}\int_{{}}^{{}}{\frac{y}{\sqrt{1+{{y}^{2}}}}dy=c}\] Þ \[\sqrt{1+{{x}^{2}}}+\sqrt{1+{{y}^{2}}}=c\].You need to login to perform this action.
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