Question

Solution of the differential equation
$\frac{dx}{dy}-\frac{xlogx}{1+logx}=\frac{e^y}{1+logx}$, if $y\left(1\right)=0$, is

• A. $x^x=e^{y{e}^y}$
• B. $e^y=x^{e^y}$
• C. $x^x=y{e}^y$
• D. none of these

Answer 1

Answer: (A)

The given equation can be written as,
$\left(1+logx\right)\frac{dx}{dy}-xlogx=e^y$
Putting $xlogx=t$
$\Rightarrow \left(1+logx\right)dx=dt$
$\therefore \frac{dt}{dy}-t=e^y$
Now, $I.F=e^{\int -1dy}=e^{-y}$
$\Rightarrow t{e}^{-y}=\int e^{-y}{e}^{y}dy+C$
$\Rightarrow t=C{e}^y+y{e}^y$
$\Rightarrow xlogx=\left(C+y\right)e^y$,
Since, $y\left(1\right)=0$, then
$0=\left(C+0\right)1$
$\Rightarrow C=0$
$\therefore y{e}^y=xlogx$
$\Rightarrow x^x=e^{y{e}^y}$