A) \[y=x{{e}^{\left( x-1 \right)}}\]
B) \[y=x\ln x+x\]
C) \[y=\ln x+x\]
D) \[y=x\ln x+{{x}^{2}}\]
Correct Answer: B
Solution :
Given \[\frac{dy}{dx}=1+\frac{y}{x}\Rightarrow \frac{dy}{dx}-\frac{y}{x}=1\] \[IF={{e}^{-\int{\frac{1}{x}dx}}}=\frac{1}{x}\] \[y.\frac{1}{x}=\int{1.\frac{1}{x}dx+c}\Rightarrow \frac{y}{x}=\ln x+c\] \[\because \,\,y\left( 1 \right)=1\], so \[c=2\Rightarrow y=x\ln x+x\]
You need to login to perform this action.
You will be redirected in
3 sec
