Question

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

• A. $y=x{e}^{(x-1)}$
• B. $y=xlnx+x$
• C. $y=lnx+x$
• D. $y=x\ln x+x^2$

Answer 1

The correct option is B $y=xlnx+x$

$\frac{dy}{dx}=\frac{x+y}{x}$
Put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\Rightarrow dv=\frac{dx}{x}\Rightarrow v=\log x+c\Rightarrow \frac{y}{x}=\log x+c$
Since $y\left(1\right)=1$
$y=x\log x+x$