Discussion
Given \(y = e^{px} \sin qx\), where \(p\) and \(q\) are non-zero real numbers, the value of the differential expression
\[ \frac{d^{2}y}{dx^{2}} - 2 p\frac{dy}{dx} + (p^{2} + q^{2})y \]
is
Answer: A
The expression equals 0.
Differentiate \(y = e^{px}\sin qx\):
\[ \begin{aligned} y' &= e^{px}(p\sin qx + q\cos qx) \\ y'' &= e^{px}\left[(p^{2} - q^{2})\sin qx + 2pq\cos qx\right] \end{aligned} \]
Then
\[ \begin{aligned} y'' - 2py' &= e^{px}\left[(-p^{2} - q^{2})\sin qx\right] \\ &= -(p^{2} + q^{2})\,y \end{aligned} \]
so adding \((p^{2} + q^{2})y\) leaves 0. This is expected: the characteristic equation \(m^{2} - 2pm + p^{2} + q^{2} = 0\) has roots \(p \pm iq\), and \(e^{px}\sin qx\) is one of its solutions. The other options would require a forcing term that the equation does not have.
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