Question

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

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

Answer 1

The correct option is D $y=x\ln x+x$

Putting $y=vx$
$\frac{dy}{dx}=x\frac{dv}{dx}+v$
Now, $x\frac{dv}{dx}+v=1+v$
$\Rightarrow dv=\frac{dx}{x}$
Integrating both sides, we get
$v=\ln x+C$
$\Rightarrow \frac{y}{x}=\ln x+C$
$\Rightarrow y=x\ln x+Cx$
Also, $y=1$ when $x=1$
$\Rightarrow C=1$
$\therefore y=x\ln x+x$ is the required solution.