Discussion
The expression \(\frac{1}{D_{x}^{2} - D_{y}^{2}} \sin(x - y)\) is equal to
Answer: A
The particular integral is \(-\frac{x}{2}\cos(x - y)\).
For \(\sin(x - y)\), replacing \(D_{x}^{2}\) by \(-1\) and \(D_{y}^{2}\) by \(-1\) makes the denominator \(-1 - (-1) = 0\), the case of failure. The rule is then to multiply by \(x\) and differentiate the denominator with respect to \(D_{x}\):
\[ \begin{aligned} \frac{1}{D_{x}^{2} - D_{y}^{2}}\sin(x - y) &= x\cdot\frac{1}{2D_{x}}\sin(x - y) \\ &= \frac{x}{2}\int \sin(x - y)\,dx \\ &= -\frac{x}{2}\cos(x - y). \end{aligned} \]
Check: with \(u = -\frac{x}{2}\cos(x - y)\), \(u_{xx} - u_{yy} = \sin(x - y)\). The option that adds \(\sin(x - y)\) differs only by a term the operator annihilates, so it is not what the inverse operator produces; the particular integral carries no complementary-function part. \(-\frac{x}{2}\sin(x - y) + \cos(x - y)\) gives \(-\cos(x - y)\) under the operator, and \(\frac{3x}{2}\sin(x - y)\) also produces a cosine term, so neither satisfies the equation.
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