A) \[y=x{{e}^{x}}+c\]
B) \[y={{e}^{x}}+c\]
C) \[y=Ax{{e}^{x-y}}\]
D) \[y=x+A\]
Correct Answer: C
Solution :
\[\frac{dy}{dx}=\frac{xy+y}{xy+x}\] Þ \[\left( \frac{1+y}{y} \right)dy=\left( \frac{1+x}{x} \right)dx\] On integrating both sides, we get \[\log y+y=\log x+x+\log A\] Þ \[\log \left( \frac{y}{Ax} \right)=x-y\]Þ\[y=Ax{{e}^{x-y}}\].
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