Discussion
The solution of the differential equation \(\dfrac{d^{2}y}{dx^{2}} + 3\dfrac{dy}{dx} = 0\), given that \(y = 0\) and \(\dfrac{dy}{dx} = 1\) at \(x = 0\) is
Answer: B
The solution is \(\frac{1}{3}(1 - e^{-3x})\).
The characteristic equation of \(y'' + 3y' = 0\) is \(m^{2} + 3m = 0\), with roots 0 and -3, so
\[ \begin{aligned} y &= A + Be^{-3x} \\ y(0) = 0 &\Rightarrow A + B = 0 \\ y'(0) = 1 &\Rightarrow -3B = 1,\ B = -\tfrac{1}{3} \end{aligned} \]
Then \(A = \frac{1}{3}\), giving \(y = \frac{1}{3}(1 - e^{-3x})\). Check: \(y' = e^{-3x}\) and \(y'' = -3e^{-3x}\), so \(y'' + 3y' = 0\), with \(y(0) = 0\) and \(y'(0) = 1\).
\(\frac{1}{3}(1 + e^{-3x})\) gives \(y(0) = 2/3\), not 0. \(x(1 - e^{-3x})\) has \(y'(0) = 0\), and \(\frac{1}{3}xe^{-3x/2}\) contains \(x\,e^{-3x/2}\), a form that would need a repeated root at \(-3/2\), which this equation does not have.
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