Question

If $y = y ( x )$ is the solution of the differential equation $\left(1+e^{2 x}\right) \frac{d y}{d x}+2\left(1+y^{2}\right) e^{x}=0$ and $y(0)=0$, then $6\left(y^{\prime}(0)+\left(y\left(\log _{e} \sqrt{3}\right)\right)^{2}\right)$ is equal to:

• A. 2
• B. -2
• C. -4
• D. -1

Answer 1

Answer: (C)

$\frac{d y}{1+y^{2}}+\frac{2 e^{x}}{1+e^{2 x}} d x=0....$(i)
on integration
$\tan ^{-1} y+2 \tan ^{-1} e^{x}=c$
$y(0)=0$
so, $C=\frac{\pi}{2} \Rightarrow \tan ^{-1} y+2 \tan ^{-1} e^{x}=\frac{\pi}{4}$
from eq.(i), $\left(\frac{d y}{d x}\right)_{x=0}=-1$
$\arg y (\ln \sqrt{3})=-\frac{1}{\sqrt{3}}$
$6\left[ y ^{\prime}(0)+\left( y (\ln \sqrt{3})^{2}\right]=6\left[-1+\frac{1}{3}\right]=-4\right.$