Q121 / 200 · WBJEE 2014 · Mathematics
The solution of the differential equation $\frac{d y}{d x}+\frac{y}{x \log _{e} x}=\frac{1}{x}$ under the condition $y=1$ when $x=e$ is
- Correct answerA$2 y=\log _{e} x+\frac{1}{\log _{e} x}$
- B$y=\log _{e} x+\frac{2}{\log _{e} x}$
- C$y \log _{e} x=\log _{2} x+1$
- D$y=\log _{e} x+e$
Solution
Given, differential equation is $\frac{d y}{d x}+\frac{y}{x \log _{e} x}=\frac{1}{x}$
It is a linear equation of the form $\frac{d y}{d x}+P y=Q$
where $\quad P=\frac{1}{x \log _{e} x}$ and $Q=\frac{1}{x}$
$\therefore$ Integrating factor, $\mathrm{IF}=e$
$$
\begin{array}{l}
=e^{\int \frac{1}{x \log _{e} x} d x} \\
=e^{\left.\log \log _{e} x\right)} \\
=\log _{e} x
\end{array}
$$
$\therefore$ Solution of differential equation is $y \times \log _{e} x=\int \frac{1}{x} \log _{e} x d x$
$\Rightarrow y \times \log _{e} x=\frac{\left(\log _{e} x\right)^{2}}{2}+C$
When $y=1$ and $x=e,$ then
$$
\begin{aligned}
& 1 \times \log _{e} e=\frac{\left(\log _{e} e\right)^{2}}{2}+C \\
\Rightarrow \quad 1 &=\frac{1}{2}+C \\
\Rightarrow \quad C=\frac{1}{2}
\end{aligned}
$$
On putting $C=\frac{1}{2}$ in Eq. (i), we get
$$
y \times \log _{e} x=\frac{\left(\log _{e} x\right)^{2}}{2}+\frac{1}{2}
$$
$\Rightarrow \quad 2 y=\log _{e} x+\frac{1}{\log _{e} x}$
Hence, the correct answer is option: 'A'.