Discussion
The general solution to the second order linear homogeneous differential equation \(y'' - 6y' + 25y = 0\) is
Answer: A
The general solution is \(e^{3x}(a\cos 4x + b\sin 4x)\).
The characteristic equation of \(y'' - 6y' + 25y = 0\) is
\[ \begin{aligned} m^{2} - 6m + 25 &= 0 \\ m &= \frac{6 \pm \sqrt{36 - 100}}{2} = 3 \pm 4i \end{aligned} \]
Complex roots \(\alpha \pm i\beta\) give \(y = e^{\alpha x}(a\cos\beta x + b\sin\beta x)\), here \(\alpha = 3\), \(\beta = 4\).
The option with \(e^{4x}\) and \(\cos 3x\) swaps the real and imaginary parts, and the options with \(e^{3ix}\) or \(e^{4ix}\) put an imaginary exponent where the real part of the root belongs, which is not a real solution form.
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