A) \[a{{y}^{2}}={{e}^{{{x}^{2}}/{{y}^{2}}}}\]
B) \[ay={{e}^{x/y}}\]
C) \[y={{e}^{{{x}^{2}}}}+{{e}^{{{y}^{2}}}}+c\]
D) \[y={{e}^{{{x}^{2}}}}+{{y}^{2}}+c\]
Correct Answer: A
Solution :
Given \[\frac{dy}{dx}=\frac{xy}{{{x}^{2}}+{{y}^{2}}}\]. Put \[y=vx\]; \[\frac{dy}{dx}=v+x.\frac{dv}{dx}\] \[\therefore v+x\frac{dv}{dx}=\frac{(x)(vx)}{{{x}^{2}}+{{v}^{2}}{{x}^{2}}}\] Þ \[v+x.\frac{dv}{dx}=\frac{v}{1+{{v}^{2}}}\]Þ \[x\frac{dv}{dx}=\frac{-{{v}^{3}}}{1+{{v}^{2}}}\] Þ \[\frac{(1+{{v}^{2}})}{{{v}^{3}}}dv=-\frac{dx}{x}\]Þ\[\left( \frac{1}{{{v}^{3}}}+\frac{1}{v} \right)dv=-\frac{dx}{x}\] Integrating both sides, \[\int_{{}}^{{}}{\frac{dv}{{{v}^{3}}}}+\int_{{}}^{{}}{\frac{dv}{v}}=-\int_{{}}^{{}}{\frac{dx}{x}}\] Þ \[-\frac{1}{2{{v}^{2}}}+\log v=-\log x-\log c\] Þ \[-\frac{{{x}^{2}}}{2{{y}^{2}}}+\log y=-\log c\]Þ\[\log cy=\frac{{{x}^{2}}}{2{{y}^{2}}}\] Þ \[cy={{e}^{{{x}^{2}}/2{{y}^{2}}}}\] Þ \[{{c}^{2}}{{y}^{2}}={{e}^{{{x}^{2}}/{{y}^{2}}}}\] \[\therefore {{y}^{2}}a={{e}^{{{x}^{2}}/{{y}^{2}}}}\], where \[{{c}^{2}}=a\].
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