A) \[y-\sqrt{{{x}^{2}}+{{y}^{2}}}=c\,{{x}^{2}}\]
B) \[y+\sqrt{{{x}^{2}}+{{y}^{2}}}=c\,{{x}^{2}}\]
C) \[y+\sqrt{{{x}^{2}}+{{y}^{2}}}=c\,{{y}^{2}}\]
D) \[x-\sqrt{{{x}^{2}}+{{y}^{2}}}=c\,{{y}^{2}}\]
Correct Answer: B
Solution :
Given equation can be rewritten as \[\frac{dy}{dx}=\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}+y}{x}\] ?(i) Which is a homogeneous differential equation. Put \[y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\] \[\therefore \]From Eq. (i), \[v+x\frac{dv}{dx}=\frac{\sqrt{{{x}^{2}}+{{v}^{2}}{{x}^{2}}}+vx}{x}\] \[\Rightarrow \] \[\frac{dv}{\sqrt{1+{{v}^{2}}}}=\frac{dx}{x}\] On integrating, we get \[\log (v+\sqrt{1+{{v}^{2}}})=\log x+\log c\] \[\Rightarrow \] \[\log \left( \frac{y}{x}+\sqrt{1+\frac{{{y}^{2}}}{{{x}^{2}}}} \right)=\log cx\] \[\Rightarrow \] \[y+\sqrt{{{x}^{2}}+{{y}^{2}}}=c{{x}^{2}}\]
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