A) \[\sqrt{2({{e}^{2}}-1)}\]
B) \[\sqrt{2({{e}^{2}}+1)}\]
C) \[\sqrt{3}e\]
D) \[\sqrt{\frac{1}{2}({{e}^{2}}+1)}\]
Correct Answer: C
Solution :
[c] We have \[\frac{dy}{dx}=\frac{xy}{{{x}^{2}}+{{y}^{2}}}\] Put \[y=vx\] \[\Rightarrow \frac{dy}{dx}=v.1+x\frac{dv}{dx}\] \[v+x\frac{dv}{dx}=\frac{v}{1+{{v}^{2}}}\] \[x\frac{dv}{dx}=\frac{v}{1+{{v}^{2}}}-V=\frac{v-v-{{v}^{3}}}{1+{{v}^{2}}}\] \[\frac{1+{{v}^{2}}}{{{v}^{3}}}dv=-\frac{dx}{x}\] \[\int{\left( {{V}^{-3}}+\frac{1}{V} \right)dv=-\int{\frac{dx}{x}}}\] \[\frac{{{V}^{-2}}}{-2}+log\text{ }v=-logx+C\] \[-\frac{1}{2{{\left( \frac{y}{x} \right)}^{2}}}+\log \left( \frac{y}{x} \right)=-\log x+c\] \[-\frac{{{x}^{2}}}{2{{y}^{2}}}+\log y=C\] ??..(1) \[y\left( 1 \right)=1at\,x=1,y=1\] \[C=-1/2\] put in (1) \[-\frac{{{x}^{2}}}{2{{y}^{2}}}+\log y=-\frac{1}{2}\] \[y\left( {{x}_{0}} \right)=e,\] at \[x={{x}_{0}},\text{ }y=e\] \[-\frac{{{x}^{2}}}{2{{e}^{2}}}+1=\frac{-1}{2}\] \[{{X}_{0}}=\sqrt{3}e\]
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