Question

The particular solution of the differential equation $y\left(1+logx\right)\frac{dx}{dy}-xlogx=0$ when $x=e,y=e^{2}is$

• A. $y=exlogx$
• B. $ey=xlogx$
• C. $xy=elogx$
• D. $ylogx=ex$

Answer 1

Answer: (A)

Given, differential equation is
$y(1+\log x)\frac{dx}{dy}-x\log x=0$
$\Rightarrow \frac{(1+\log x)dx}{x\log x}=\frac{dy}{y}$
$\Rightarrow \left(\frac{1}{x\log x}+\frac{1}{x}\right)dx=\frac{1}{y}dy$
On integrating both sides, we get
$\int \left(\frac{1}{x\log x}+\frac{1}{x}\right)dx=\int \frac{1}{y}dy$
Put $\log x=t$
$\Rightarrow \frac{1}{x}dx=dt$
$\therefore \int \frac{1}{t}dt+\int \frac{1}{x}dx=\int \frac{1}{y}dy$
$\Rightarrow \log t+\log x=\log y+\log c$
$\Rightarrow \log tx=\log yc$
$\Rightarrow tx=yc$
$\Rightarrow x\log x=yc$
When $x=e$ and $y=e^2$
$\therefore e\log e=e^{2}c$
$\Rightarrow e\times 1=e^{2}c$
$\Rightarrow c\Rightarrow \frac{1}{e}$
$\therefore x\log x=\frac{y}{e}$
$\Rightarrow y=ex\log x$