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

Option: 2
$-2$

Option: 3
$-4$

Option: 4
$-1$

Answers (1)

best_answer

$\mathrm{\left(1+e^{2 x}\right) \frac{d y}{d x}+2\left(1+y^{2}\right) e^{x}=0} \\$

$\mathrm{\left(1+e^{2 x}\right) d y=-2\left(1+y^{2}\right) e^{x} d x} \\$

$\mathrm{\frac{d y}{1+y^{2}}=-\frac{2 e^{x} d x}{1+e^{2 x}}}$

$\mathrm{Let \,\,e^{x}=t \Rightarrow e^{x} d x=d t}$

$\mathrm{\frac{d y}{1+y^{2}}=-2 \frac{d t}{1+t^{2}}}$

$\mathrm{Integrating}$

$\mathrm{\tan ^{-1}(y)=-2 \tan ^{-1}(t)+c }\\$

$\mathrm{\tan ^{-1}(y)=-2 \tan ^{-1}\left(e^{x}\right)+c}$

$\mathrm{\text { As } y(0) =0} \\$

$\mathrm{0 =-2 \cdot \frac{\pi}{4}+c \Rightarrow c=\frac{\pi}{2}} \\$

$\mathrm{\therefore \tan ^{-1} y =-2 \tan ^{-1}\left(e^{x}\right)+\frac{\pi}{2}}$ ...........(1)

From given equation putting $\mathrm{x=0,y=0}$

$\mathrm{\left(1+e^{0}\right) \cdot y^{\prime}(0)+2(1+0) \cdot e^{0}=0} \\$

$\mathrm{2 \cdot y^{\prime}(0)+2=0} \\$

$\mathrm{y^{\prime}(0)=-1}$

$\mathrm{Putting\: x=\log (\sqrt{3}) \,\,in \,\,(i)}$

$\mathrm{\tan ^{-1} y =-2 \tan ^{-1}(\sqrt{3})+\frac{\pi}{2}} \\$

$\mathrm{\tan ^{-1} y =-\frac{2 \pi}{3}+\frac{\pi}{2}} \\$

$\mathrm{\tan ^{-1} y =-\frac{\pi}{6} }\\$

$\mathrm{y =-\frac{1}{\sqrt{3}}}$

$\mathrm{\therefore 6\left(y^{\prime}(0)+(y(\log \sqrt{3}))^{2}\right)} \\$

$\mathrm{= 6\left(-1+\frac{1}{3}\right)} \\$

$\mathrm{= 6\left(-\frac{2}{3}\right)} \\$

$\mathrm{=-4}$

Hence answer is option 3

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