Question

If $y=y(x)$ is the solution of the differential equation $\left(1+e^{2x}\right)\frac{dy}{dx}+2\left(1+y^2\right)e^x=0$ 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)

$\frac{dy}{1+y^2}+\frac{2e^x}{1+e^{2x}}dx=\mathrm{0....}$(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), ${\left(\frac{dy}{dx}\right)}_{x=0}=-1$
$\arg y(\ln \sqrt{3})=-\frac{1}{\sqrt{3}}$
$6[y^{\prime }(0)+\left(y(\ln \sqrt{3}{)}^2\right]=6\left[-1+\frac{1}{3}\right]=-4$