The solution of frac{dy}{dx} = e^{-2x} with t | vedclass

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Question

(D) Given the differential equation $\frac{dy}{dx} = e^{-2x}$.Integrating both sides with respect to $x$:$\int dy = \int e^{-2x} dx$$y = -\frac{1}{2}e

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given the differential equation $\frac{dy}{dx} = e^{-2x}$.Integrating both sides with respect to $x$:$\int dy = \int e^{-2x} dx$$y = -\frac{1}{2}e^{-2x} + C$  $(i)$Given the initial condition $y(\log 2) = \frac{1}{16}$:$\frac{1}{16} = -\frac{1}{2}e^{-2(\log 2)} + C$Since $e^{-2\log 2} = e^{\log(2^{-2})} = 2^{-2} = \frac{1}{4}$:$\frac{1}{16} = -\frac{1}{2}(\frac{1}{4}) + C$$\frac{1}{16} = -\frac{1}{8} + C$$C = \frac{1}{16} + \frac{2}{16} = \frac{3}{16}$Substituting $C$ back into equation $(i)$:$y = -\frac{1}{2}e^{-2x} + \frac{3}{16}$$y = \frac{-8e^{-2x} + 3}{16} = \frac{3 - 8e^{-2x}}{16}$

The solution of $\frac{dy}{dx} = e^{-2x}$ with the condition $y(\log 2) = \frac{1}{16}$ is $y =$

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