The solution of $\frac{dy}{dx}=e^{-2x},y\left(\log 2\right)=\frac{1}{16}$ is $y=$

Given, $\frac{dy}{dx}=e^{-2x}...............\left(i\right)$

$y\left(\log 2\right)=\frac{1}{16}$

From equation $\left(i\right)$

$dy=e^{-2x}dx$

Integrating on both sides

$\int dy=\int e^{-2x}dx$

$y=-\frac{1}{2}{e}^{-2x}+C................\left(ii\right)$

Now, $y\left(\log 2\right)=\frac{1}{16}$

Then, $\frac{1}{16}=-\frac{1}{2}{e}^{-2\log 2}+C$

$C=\frac{3}{16}$

Putting value of $C$ in equation $\left(ii\right)$

$y=-\frac{1}{2}{e}^{-2\log 2}+\frac{3}{16}$

Hence, the solution of given equation, $y=\frac{3-8e^{-2x}}{16}$