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. $x^x=e^{y{e}^y}$
• B. $e^y=x^{ye}$
• C. $x^x{=}^{y{e}^y}$
• D. $y=e^{e^{xy}}$

Answer 1

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

Given

$\frac{dx}{dy}-\frac{xlogx}{1+logx}=\frac{e^y}{1+logx}$

$(1+logx)\frac{dx}{dy}-xlogx=e^y$

let xlogx=t

$\to (1+logx)dx=dt$

Above equation became

$\frac{dt}{dy}-t=e^y$

P=-1 and q=$e^y$

I.f.=$e^{\int p.dy}$

=$e^{-1dy}$

=$e^{-y}$

Complete solution

$t\times e^{-y}=\int e^{-y}.e^{y}dy+c$

$\to xlogx\times e^{-y}=y+c$

at y(1)=0

$1log1=0+c$

so c=0

$\to xlogx=y{e}^y$

$\to log{x}^x=y{e}^y$

$\to x^x=e^{y{e}^y}$