A) \[k{{e}^{{{x}^{2}}/2}}\]
B) \[k{{e}^{{{y}^{2}}/2}}\]
C) \[k{{e}^{xy/2}}\]
D) \[k{{e}^{xy}}\]
Correct Answer: A
Solution :
[a] Put \[xy=v\] \[\therefore \,\,\,\,y+x\frac{dy}{dx}=\frac{dv}{dx}\Rightarrow \frac{dv}{dx}=x\frac{\phi (v)}{\phi '(v)}\] \[\therefore \frac{\phi '(v)}{\phi (v)}dv=xdx.\] Integrating, we get \[\log \phi (v)=\frac{{{x}^{2}}}{2}+\log k\Rightarrow \log \frac{\phi (v)}{k}=\frac{{{x}^{2}}}{2}\] or \[\phi (v)=k{{e}^{{{x}^{2}}/2}}\Rightarrow \phi (xy)=k{{e}^{{{x}^{2}}/2}}\]
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