Question

If $\frac{dy}{dx}=(e^y-x{)}^{-1},$ where $y\left(0\right)=0,$ then $y$ is expressed explicitly as

• A. $\frac{1}{2}\ln (1+x^2)$
• B. $\ln (1+x^2)$
• C. $\ln (x+\sqrt{1+x^2})$
• D. $\ln (x+\sqrt{1-x^2})$

Answer 1

The correct option is C $\ln (x+\sqrt{1+x^2})$

we have $\frac{dy}{dx}=(e^y-x{)}^{-1}\Rightarrow \frac{dx}{dy}=e^y-x$
$\Rightarrow \frac{dx}{dy}+x=e^y$
On comparing with $\frac{dx}{dy}+Px=Q,$
We get, $P=1$ and $Q=e^y$
So $I.F.=e^{\int 1dy}=e^y$

$\therefore$ General solution is given by
$x{e}^y=\int e^{2y}dy+C$
$x{e}^y=\frac{1}{2}{e}^{2y}+C$
$\Rightarrow x=\frac{e^y}{2}+C{e}^{-y}$
As $y\left(0\right)=0,$So $C=-\frac{1}{2}$
$\therefore x=\frac{e^y}{2}-\frac{1}{2}{e}^{-y}$
$\Rightarrow e^y-e^{-y}=2x$
$\Rightarrow e^{2y}-2x{e}^y-1=0$
$\Rightarrow 2e^y=2x\pm \sqrt{4x^2+4}$
But $e^y=x-\sqrt{x^2+1}\left(\text{Rejected}\right)$
Hence, $y=\ln (x+\sqrt{x^2+1})$