Question

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

Answer 1

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

The given equation can be re-written as $(1+\log x)\frac{dx}{dy}-x\log x=e^y$
Put $x\log x=t$

$\therefore (1+\log x)\frac{dx}{dy}=\frac{dt}{dy}$
$\therefore \frac{dt}{dy}-t.1=e^y$
Above is a linear differential equation.
$\therefore =e^{\int pdy}=e^{\int -dy}=e^{-y}$
Multiplying both sides by I.F. and integrating
$t.e^{-y}=\int e^{-y}.e^{y}dy+c$
or $t.e^{-y}=y+c$ or $x\log x=(y+c)e^y$
Now $y\left(1\right)=0,$ i.e, when $x=1,y=0$
$\therefore 1\log 1=0+c{e}^0$ or 0=c
$\therefore x\log x=y{e}^y$ or $x^x=y{e}^y$
$\therefore x^x=e^{y{e}^y}\Rightarrow \left(a\right)$