A) \[\sqrt{3}\text{/}e\]
B) \[\sqrt{3}\,{{e}^{2}}\]
C) \[\sqrt{3}\,e\]
D) none of these
Correct Answer: C
Solution :
\[({{x}^{2}}+{{y}^{2}})\,dy=xy\,dx\] \[\frac{dy}{dx}=\frac{xy}{{{x}^{2}}+{{y}^{2}}}\] ?(1) Put \[y=vx,\]\[\frac{dy}{dx}=v+x\frac{dv}{dx}\]in (1) \[v+x\frac{dv}{dx}=\frac{v}{1+{{v}^{2}}}\] \[\Rightarrow \] \[x\frac{dv}{dx}=\frac{v}{1+{{v}^{2}}}-v\] \[\Rightarrow \] \[x\frac{dv}{dx}=\frac{v}{1+{{v}^{2}}}-v\] \[\Rightarrow \] \[-\frac{1+{{v}^{2}}}{{{v}^{3}}}dv=\frac{dx}{x}\] \[\Rightarrow \] \[-\frac{1}{2{{v}^{2}}}+\ln \,v=-\ln x+C\] We have \[y(1)=1\] \[\therefore \] \[c=-\frac{1}{2}\] \[\therefore \] \[\ln \,\frac{y}{x}-\frac{1}{2}\frac{{{x}^{2}}}{{{y}^{2}}}=-\,\ln x-\frac{1}{2}\] Putting\[y=e,\]we get \[x=\sqrt{3}\,e\]
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