Question

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

• A. $y=ex\log x$
• B. $ey=x\log x$
• C. $xy=e\log x$
• D. $y\log x=ex$

Answer 1

The correct option is A $y=ex\log x$

Correct equation is,

$y(1+logx)\frac{dx}{dy}-xlogx=0$

$\to \frac{dy}{y}=\frac{1+logx}{logx}dx$$-\left(1\right)$

Let $xlogx=t$

$(1+logx)dx=dt$

$\left(1\right)\to \frac{dy}{y}=\frac{dt}{t}$

integrating Both the sides

$\to \int \frac{dy}{y}=\int \frac{dt}{t}$

$logy=logt+c$

$atx=eandy=e^2$

$log{e}^2=eloge+c$

$2-1=c$

c=1

$logy=logt+1$

$logy=logt+loge$

$logy=log\left(xlogx\right)+loge$

$y=xelogx$