Question

Find the particular solution of the following differential equation $x{\log }_{e}x\frac{dy}{dx}+y=2{\log }_{e}x$,
given that $y=1$ when $x=e$

Answer 1

Lets take the given differential equation
$x{\log }_{e}x\frac{dy}{dx}+y=2{\log }_{e}x$
Divide by $x\log x$ on both sides, we get,
$\frac{dy}{dx}+\frac{y}{x{\log }_{e}x}=\frac{2}{x}$

This is the linear differential equation. Compare the above equation with $\frac{dy}{dx}+Py=Q$, we get
$P=\frac{1}{x{\log }_{e}x},Q=\frac{2}{x}$

$\therefore$ Integrating factor is given as
$I.F.=e^{\int \frac{1}{x{\log }_{e}x}dx}$

As, $\int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx$ = ${\log }_e\left(f\right(x\left)\right)$

$\therefore I.F.={\log }_{e}x$

The general solution of the given differential equation can be given as

$y\cdot I.F.=\int Q\cdot (I.F.)dx+C$
$\therefore y{\log }_{e}x=\int \frac{2}{x}{\log }_{e}xdx+C$

We know, $\int f^n\left(x\right)f^{\prime }\left(x\right)dx=\frac{\left(f\right(x){)}^{n+1}}{n+1}$

$\Rightarrow y{\log }_{e}x=2\frac{({\log }_{e}x{)}^2}{2}+C$
$\Rightarrow y{\log }_{e}x=({\log }_{e}x{)}^2+C$

At $x=e,y=1$
$\Rightarrow C=0$

The particular solution is
$y{\log }_{e}x=({\log }_{e}x{)}^2$
$\therefore y={\log }_{e}x$