A) \[y=\ln \,x+x\]
B) \[y=x\,\ln \,x+{{x}^{2}}\]
C) \[y=x{{e}^{(x-1)}}\]
D) \[y=x\,\,\ln \,x+x\]
Correct Answer: D
Solution :
[d] Given equation is \[\frac{dy}{dx}=1+\frac{y}{x}.\] Let \[y=vx\] \[\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\] \[\Rightarrow v+x\frac{dv}{dx}=1+v\] \[\Rightarrow dv=\frac{dx}{x}\] \[\therefore v=\ln x+c\] \[\Rightarrow \frac{y}{x}=\ln x+c\] Since, y (1) =1, c=1, so we have \[y=x\,\ln x+x\]
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