Question

The solution of differential equation $(1+e^{2y})e^{{\tan }^{-1}x}dx-(1+x^2)(e^y+{(e^y-1)}^2)dy=0$ is
$($Here, $C$ is a constant of integration$)$

• A. $y=\tan (y-e^{{\tan }^{-1}x}+C)$
• B. $e^y=\tan (y-e^{{\tan }^{-1}x}+C)$
• C. $y=\tan (e^{{\tan }^{-1}x}-y+C)$
• D. $e^y=\tan (e^{{\tan }^{-1}x}-y+C)$

Answer 1

The correct option is B $e^y=\tan (y-e^{{\tan }^{-1}x}+C)$

$(1+e^{2y})e^{{\tan }^{-1}x}dx-(1+x^2)(e^y+{(e^y-1)}^2)dy=0$
$\Rightarrow \frac{e^{{\tan }^{-1}x}}{1+x^2}dx-\frac{e^{2y}-e^y+1}{e^{2y}+1}dy=0$
$\Rightarrow \int \frac{e^{{\tan }^{-1}x}}{1+x^2}dx=\int \frac{e^{2y}+1-e^y}{e^{2y}+1}dy\dots \left(1\right)$

Let $I_1=\int \frac{e^{{\tan }^{-1}x}}{1+x^2}dx$
Put ${\tan }^{-1}x=t$
$\Rightarrow \frac{1}{1+x^2}dx=dt$
$I_1=\int e^{t}dt=e^t+C_1$
$\therefore I_1=e^{{\tan }^{-1}x}+C_1$

Let $I_2=\int (1-\frac{e^y}{e^{2y}+1})dy$
$I_2=y-\int \frac{e^y}{e^{2y}+1}dy$
Put $e^y=u\Rightarrow e^{y}dy=du$
$I_2=y-\int \left(\frac{1}{1+u^2}\right)du=y-{\tan }^{-1}u+C_2$
$\therefore I_2=y-{\tan }^{-1}\left(e^y\right)+C_2$

From equation $\left(1\right),$
$e^{{\tan }^{-1}x}+C_1=y-{\tan }^{-1}{e}^y+C_2$
$\Rightarrow e^{{\tan }^{-1}x}-y+{\tan }^{-1}\left(e^y\right)=C,$ where $C_2-C_1=C$
$\Rightarrow e^y=\tan (y-e^{{\tan }^{-1}x}+C)$