Question

If $y\left(x\right):$ Solution of a $D.E.$
$\left(x\log x\right)\frac{dy}{dx}+y=2x\log x,$ $(x,1)$
$y\left(e\right)=?x=e$

• A. $e$
• B. $0$
• C. $2$
• D. $2e$

Answer 1

The correct option is A $e$

$\begin{array}{c}Fromthegiven,data\\ e^{\int \frac{dx}{x\log x}}\alpha \left(\frac{dy}{dx}+\frac{y}{x\log x}=2\right)---\left(1\right)\\ Now,solving\\ I=\int \frac{dx}{x\log x}let\log x=t,dt=\frac{dx}{x}\\ I=\int \frac{dt}{t}=\log t\\ hence,\left(1\right)becomes\\ \frac{d}{dx}\left(e^{\log t}y\right)=2e^{\log t}\\ \frac{d}{dx}\left(y\log x\right)=2\log x\\ \int d\left(y\log x\right)=2\int \log xdx\\ y\log x=2\left(x\log x-x\right)\\ Hence,\\ y\left(x\right)=f\left(x\right)=\frac{2\left(x\log x-x\right)}{\log x}\\ Now,\\ y\left(e\right)=\frac{2e\left(\log e\right)-e}{\log e}=e\\ Hence,theoptionAisthecorrectanswer.\end{array}$