A) \[x+\sqrt{{{x}^{2}}+{{y}^{2}}}=c{{x}^{2}}\]
B) \[y-\sqrt{{{x}^{2}}+{{y}^{2}}}=cx\]
C) \[x-\sqrt{{{x}^{2}}+{{y}^{2}}}=cx\]
D) \[y+\sqrt{{{x}^{2}}+{{y}^{2}}}=c{{x}^{2}}\]
Correct Answer: D
Solution :
Given that, \[xdy-ydx=\sqrt{{{x}^{2}}+{{y}^{2}}}dx\] \[\therefore \]\[xdy=(\sqrt{{{x}^{2}}+{{y}^{2}}}+y)dx\] \[\Rightarrow \]\[\frac{dy}{dx}=\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}+y}{x}\] Now, put \[y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\] \[\therefore \]\[v+x\frac{dv}{dx}=\frac{\sqrt{{{x}^{2}}+{{v}^{2}}{{x}^{2}}}+vx}{x}\] \[\Rightarrow \]\[x\frac{dv}{dx}=\sqrt{1+{{v}^{2}}}\]\[\Rightarrow \]\[\frac{dv}{\sqrt{1+{{v}^{2}}}}=\frac{dx}{x}\] \[\Rightarrow \]\[\log (v+\sqrt{(1+{{v}^{2}})}=\log x+\log c\] \[\Rightarrow \]\[y+\sqrt{{{x}^{2}}+{{y}^{2}}}=c{{x}^{2}}\]
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