A) \[k.{{e}^{\frac{{{x}^{2}}}{2}}}\]
B) \[k.{{e}^{{{y}^{2}}/2}}\]
C) \[k.{{e}^{{{x}^{2}}}}\]
D) \[k.{{e}^{\frac{xy}{2}}}\]
Correct Answer: A
Solution :
Given, \[x.\frac{dy}{dx}+y=x.\frac{f\left( xy \right)}{f'\left( xy \right)}\] i.e \[\frac{d}{dx}\left( xy \right)=x\frac{f\left( x,y \right)}{f'\left( x,y \right)}\] \[\Rightarrow \] \[\frac{f'\left( xy \right)}{f\left( xy \right)}d\left( xy \right)=xdx\] \[\Rightarrow \] \[\int{\frac{f'\left( xy \right)}{f\left( xy \right)}d\left( xy \right)=\int{xdx}}\] \[\Rightarrow \] \[\log \left[ f\left( xy \right) \right]=\frac{{{x}^{2}}}{2}+C\] \[\Rightarrow \] \[f\left( xy \right)={{e}^{\left( {{x}^{2}}/2+C \right)}}\] \[={{e}^{\frac{{{x}^{2}}}{2}}}.{{e}^{C}}=k.{{e}^{\frac{{{x}^{2}}}{2}}}\]
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