Question

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

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

Answer 1

Answer: (C)

We have $\frac{dy}{dx}={\left(e^y-x\right)}^{-1}\Rightarrow \frac{dx}{dy}=e^y-x\Rightarrow \frac{dx}{dy}+x=e^y;$
So I.F. $=e^{\int dy}=e^y$
$\therefore$ General solution is given by $x{e}^y=\frac{1}{2}{e}^{2y}+C\Rightarrow x=\frac{e^y}{2}+C{e}^{-y}$
As $y(0)=0$, 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 $e^y=x-\sqrt{x^2+1}$
(Rejected)
Hence $y=\ln \left(x+\sqrt{x^2+1}\right)$