Question

The solution of the differential equation $\frac{dy}{dx}+\frac{y}{x{\log }_{e}x}=\frac{1}{x}$ under the condition $y=1$ when $x=e$ is

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

Answer 1

Answer: (A)

Given, differential equation is
$\frac{dy}{dx}+\frac{y}{x{\log }_{e}x}=\frac{1}{x}$
It is a linear equation of the form
$\frac{dy}{dx}+Py=Q$
where $P=\frac{1}{x{\log }_{\theta }x}$ and $Q=\frac{1}{x}$
$\therefore$ Integrating factor, IF $=e$
$=e^{\int \frac{1}{x{\log }_{e}x}dx}$
$=e^{\log \left({\log }_{e}x\right)}$
$={\log }_{e}x$
$\therefore$ Solution of differential equation is
$y\times {\log }_{e}x=\int \frac{1}{x}{\log }_{e}xdx$
$\Rightarrow y\times {\log }_{e}x=\frac{{\left({\log }_{\theta }x\right)}^2}{2}+C...(i)$
When $y=1$ and $x=e$, then
$1\times {\log }_{e}e=\frac{{\left({\log }_{\theta }e\right)}^2}{2}+C$
$\Rightarrow 1=\frac{1}{2}+C$
$\Rightarrow C=\frac{1}{2}$
On putting $C=\frac{1}{2}$ in Eq. (i), we get
$Y\times {\log }_{\theta }x=\frac{{\left({\log }_{\theta }x\right)}^2}{2}+\frac{1}{2}$
$\Rightarrow 2y={\log }_{e}x+\frac{1}{{\log }_{e}x}$