Discussion
Consider the differential equation
\[ \frac{d^{2}y}{dx^{2}} + 3\frac{dy}{dx} + 2y = \sin(e^{x}). \]
Which ONE of the following is a particular solution of this differential equation?
Answer: B
A particular solution is \(-e^{-2x}\sin(e^{x})\).
The operator factors as \((D + 1)(D + 2)\). Put \(u = y' + 2y\); then \(u' + u = y'' + 3y' + 2y = \sin(e^{x})\). Solve the two first-order equations in turn:
\[ \begin{aligned} (e^{x}u)' &= e^{x}\sin(e^{x}) \\ e^{x}u &= -\cos(e^{x}) \\ (e^{2x}y)' &= e^{2x}u = -e^{x}\cos(e^{x}) \\ e^{2x}y &= -\sin(e^{x}) \end{aligned} \]
Taking both constants of integration as zero gives \(y = -e^{-2x}\sin(e^{x})\). The substitution \(w = e^{x}\), with \(dw = e^{x}dx\), is what makes each integral elementary.
The other options do not satisfy the equation. Substituting \(-e^{-x}\sin(e^{x})\) gives \(e^{x}\sin(e^{x}) - 2\cos(e^{x})\) on the left, not \(\sin(e^{x})\). Substituting \(-e^{-2x}(\sin(e^{x}) + \cos(e^{x}))\) gives \(\sin(e^{x}) + \cos(e^{x})\): the added cosine term survives as an extra \(\cos(e^{x})\).
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