A) \[y-\sqrt{{{x}^{2}}+{{y}^{2}}}=C{{x}^{2}}\]
B) \[y+\sqrt{{{x}^{2}}+{{y}^{2}}}=C{{x}^{2}}\]
C) \[x+\sqrt{{{x}^{2}}+{{y}^{2}}}=C{{y}^{2}}\]
D) \[x-\sqrt{{{x}^{2}}+{{y}^{2}}}=C{{y}^{2}}\]
Correct Answer: B
Solution :
| Writing the given equation as \[\frac{dy}{dx}=-\frac{y+\sqrt{{{x}^{2}}+{{y}^{2}}}}{x}\] and putting y=ux, |
| we have \[v+x\frac{dv}{dx}=v+\sqrt{1+{{v}^{2}}}\] \[\Rightarrow \,\,\,\frac{dv}{\sqrt{1+{{v}^{2}}}}=\frac{dx}{x}\] |
| Integrating, we have |
| \[\log \,\,(v+\sqrt{1+{{v}^{2}}})=\log x+\text{constant}\]\[\Rightarrow y/x+\sqrt{1+({{y}^{2}}/{{x}^{2}})}=Cx\]\[\Rightarrow y+\sqrt{{{x}^{2}}+{{y}^{2}}}=C{{x}^{2}}\] |
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