Question

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

Answer 1

Answer: (A)

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