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Q.

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

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a

$\frac{4 e^{- 2 x}}{16}$

b

$\frac{4 - 12 e^{- 2 x}}{16}$

c

$\frac{\log x}{16}$

d

$\frac{3 - 8 e^{- 2 x}}{16}$

answer is D.

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Detailed Solution

$d y = e^{- 2 x} d x$
Integration
$y = \frac{e^{- 2 x}}{- 2} + C \Rightarrow y = \frac{- e^{- 2 x}}{2} + \frac{3}{16} = \frac{3 - 8 e^{- 2 x}}{16}$

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The solution of dydx=e−2x, y(log2)=116 is y =