If frac{dy}{dx} = e^{-2y} and y = 0 when x = | vedclass

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(C) Given the differential equation $\frac{dy}{dx} = e^{-2y}$.Separating the variables,we get $e^{2y} dy = dx$.Integrating both sides,we have $\int e^

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$\frac{e^6 + 9}{2}$

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(C) Given the differential equation $\frac{dy}{dx} = e^{-2y}$.Separating the variables,we get $e^{2y} dy = dx$.Integrating both sides,we have $\int e^{2y} dy = \int dx$,which gives $\frac{e^{2y}}{2} = x + C$.Using the initial condition $y = 0$ when $x = 5$,we substitute these values into the equation: $\frac{e^{2(0)}}{2} = 5 + C$.$\frac{1}{2} = 5 + C$,which implies $C = \frac{1}{2} - 5 = -\frac{9}{2}$.So,the equation becomes $\frac{e^{2y}}{2} = x - \frac{9}{2}$.Now,to find $x$ when $y = 3$,we substitute $y = 3$ into the equation: $\frac{e^{2(3)}}{2} = x - \frac{9}{2}$.$\frac{e^6}{2} = x - \frac{9}{2}$,which gives $x = \frac{e^6}{2} + \frac{9}{2} = \frac{e^6 + 9}{2}$.

If $\frac{dy}{dx} = e^{-2y}$ and $y = 0$ when $x = 5$,the value of $x$ for $y = 3$ is

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