If 2 f(x) = f^{prime}(x) and f(0) = 3,then th | vedclass

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(D) Given the differential equation $f^{\prime}(x) = 2 f(x)$.Rearranging the terms,we get $\frac{f^{\prime}(x)}{f(x)} = 2$.Integrating both sides with

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$3 e^{4}$

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(D) Given the differential equation $f^{\prime}(x) = 2 f(x)$.Rearranging the terms,we get $\frac{f^{\prime}(x)}{f(x)} = 2$.Integrating both sides with respect to $x$:$\int \frac{f^{\prime}(x)}{f(x)} dx = \int 2 dx$$\ln |f(x)| = 2x + C$.Using the initial condition $f(0) = 3$:$\ln |f(0)| = 2(0) + C \Rightarrow \ln 3 = C$.Substituting $C$ back into the equation:$\ln |f(x)| = 2x + \ln 3$.To find $f(2)$,substitute $x = 2$:$\ln |f(2)| = 2(2) + \ln 3 = 4 + \ln 3$.Taking the exponential on both sides:$f(2) = e^{4 + \ln 3} = e^{4} \cdot e^{\ln 3} = 3 e^{4}$.

If $2 f(x) = f^{\prime}(x)$ and $f(0) = 3$,then the value of $f(2)$ is

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