Question

The solution of the differential equation $y\ell n\left(y\right)+x{y}^{\prime }=0,$ where $y\left(1\right)=e,$ is [Note: $y^{\prime }$ denotes $\frac{dy}{dx}$ and $e$ is Napier's constant]

• A. $x(\ell ny{)}^2=1$
• B. $x\left(\ell ny\right)=1$
• C. $(\ell ny{)}^2=x$
• D. $(x+\ell ny)=2$

Answer 1

The correct option is B $x\left(\ell ny\right)=1$

$ylny+x\frac{dy}{dx}=0$

$\Rightarrow ylny+x\frac{dy}{dx}=0$

$\Rightarrow x\frac{dy}{dx}=-ylny$

$\Rightarrow \frac{dy}{ylny}=\frac{-dx}{dx}$

integrating both sides :

$\Rightarrow \int \frac{dy}{ylny}=-\int \frac{dx}{x}$

$\Rightarrow ln\left(lny\right)=-lnx+lnA$

Where ln $A$ an integrating constant.

$\Rightarrow ln\left(lny\right)=ln\frac{A}{x}$

$\Rightarrow lny=\frac{A}{x}$

$\Rightarrow y=e^{A/x}$

given

$y(x=1)=e$

$\Rightarrow e^{A/1}=e$

$\Rightarrow A=1$

$\therefore$ $y=e^{1/x}$

$\Rightarrow lny=\frac{1}{x}$

$\Rightarrow xlny=1$

$\therefore$ Answer : Option B.