Question

Let $y=y\left(x\right)$ is the solution of the differential equation
$y\ln y\frac{dx}{dy}+x-\ln y=0$ where $y\left(2\right)=e^2$ and $y>1,$ then the value of $6k$ such that $y\left(k\right)=e^3,$ is

Answer 1

$y\ln y\frac{dx}{dy}+x-\ln y=0$
$\frac{dx}{dy}+\frac{x}{y\ln y}=\frac{1}{y}$
It becomes linear differential equation
$\therefore P=\frac{1}{y\ln y}$and$Q=\frac{1}{y}$
I.F.$=e^{\int \frac{1}{y\ln y}dy}=e^{\ln |\ln y|}=\ln y$
Solution is,
$\Rightarrow x(I.F.)=\int (I.F)\frac{1}{y}dy$
$\Rightarrow x\ln y=\int \left(\ln y\right)\frac{1}{y}dy$
Let $\ln y=t,\frac{1}{y}\cdot dy=dt$
$\Rightarrow x\ln y=\int tdt$
$\Rightarrow x\ln y=\frac{t^2}{2}+c$
$\Rightarrow x\ln y=\frac{(\ln y{)}^2}{2}+c$
$\Rightarrow x=\frac{\ln y}{2}+\frac{c}{\ln y}$
Given, $y\left(2\right)=e^2\Rightarrow c=2$
$\therefore x=\frac{\ln y}{2}+\frac{2}{\ln y}$
Now, $y\left(k\right)=e^3$
$\Rightarrow k=\frac{3}{2}+\frac{2}{3}=\frac{13}{6}$
$\therefore 6k=13$