Question

The solution of the differential equation $\frac{dy}{dx}=\frac{x+y}{x}$ satisfying the condition $y\left(1\right)=1$ is

• A. $y=\log x+x$
• B. $y=x\log x+x^2$
• C. $y=x{e}^{x-1}$
• D. $y=x\log x+x$

Answer 1

The correct option is A $y=\log x+x$

$dy=(1+\frac{1}{x})dx$
$\int dy=\int (1+\frac{1}{x})dx$
$y=x+\log x+C$
Substituting x=1 and y=1:
$1=1+0+C⟹C=0$
$\therefore y=x+\log x$