Question

The solution of the differential equation $x\frac{dy}{dx}=y(logy-logx+1)$ is

• A. $y=x{e}^{cx}$
• B. $y+x{e}^{cx}=0$
• C. $y+e^x=0$
• D. $Noneofthese$

Answer 1

The correct option is A $y=x{e}^{cx}$

Here $\frac{dy}{dx}=\frac{y}{x}(log\frac{y}{x}+1)...\left(i\right)$
It is homogeneous equation
So now put y =vx ~and $\frac{dy}{dx}=v+x\frac{dv}{dx}$, then the equation (i) reduces to $\frac{dv}{vlogv}=\frac{dx}{x}$
On integrating, we get log (log v)=log x +log c
$\Rightarrow log\left(\frac{y}{x}\right)=cx\Rightarrow y=x{e}^{cx}$