Question

The solution of the differential equation $\frac{dx}{dy}-\frac{x\log x}{1+\log x}=\frac{e^y}{1+\log x}$ 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

The correct option is A $x^x=e^{y{e}^y}$

$\frac{dx}{dy}=\frac{x\log x}{1+\log x}=\frac{e^y}{1+\log x}$
$\Rightarrow (1+\log x)\frac{dx}{dy}-x\log x=e^y$
Put $x\log x=t$
$(1+\log x)\frac{dx}{dt}=\frac{dt}{dy}$
$\Rightarrow \frac{dt}{dt}-t=e^y\Rightarrow P=-1,Q=e^y$
I.F. $=e^{\int -dy}=e^{-y}$
$\Rightarrow t.e^{-y}=\int dy$
$\Rightarrow t.e^{-y}=y+c$
$\Rightarrow x\log x{e}^{-y}=y+c$
$y\left(1\right)=0\Rightarrow c=0$
$\Rightarrow x\log x{e}^{-y}=y$
$\Rightarrow \log x^x=y{e}^y$
$\Rightarrow x^x=e^{y{e}^y}$