Question

Let $y=f\left(x\right)$ satisfies $\frac{dy}{dx}=\frac{x+y}{x}$ and $f\left(e\right)=e$ , then the value of $f\left(1\right)$ is

Answer 1

Answer: 0

$\frac{dy}{dx}=1+\frac{y}{x}$
i.e. $\frac{dy}{dx}-\frac{y}{x}=1$ is a linear differential equation.
I.F. $=e^{-\int \frac{1}{x}dx}=\frac{1}{x}$
Solution of the differential equation is
$\frac{y}{x}=\int \frac{1}{x}dx\Rightarrow y=xlnx+cx=f\left(x\right)$
So, $f\left(e\right)=e+ce$
$\Rightarrow e+ce=e\Rightarrow c=0$
Hence, $f\left(1\right)=0$