Discussion
If \(y(x) = 2y'(x), y(x) \geq 0\) and \(y(0) = e^{2}\) then \(y(x) =\)
Answer: D
\(y(x) = \) \(e^{2}e^{x/2}\).
The equation is \(\frac{dy}{dx} = \frac{y}{2}\). Separating variables:
\[ \begin{aligned} \frac{dy}{y} &= \frac{dx}{2} \\ \ln y &= \frac{x}{2} + c \\ y &= Ce^{x/2} \end{aligned} \]
\(y(0) = C = e^{2}\), so \(y = e^{2}e^{x/2}\). \(e^{x/2}\) has the right shape but gives \(y(0) = 1\); \(e^{2x}\) solves \(y' = 2y\), which reverses the given relation; the options with \(+ 2\) do not satisfy the equation at all.
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