Answer to Question #280794 in Differential Equations for Bikerkid

Question

The particular integral of (D³ - 2D²D')z = sin (x + 2y) is

Answer 1

$Given\mathrm{:}\ \ \left(D^{\mathrm{3}}-\mathrm{2}D^{\mathrm{2}}D'\right)z\ \ \ =\ \mathrm{sin}\left(x+\mathrm{2}y\right) \\ Let\ \ \ P.I.\ \ =\frac{\mathrm{1}}{\left(D^{\mathrm{3}}-\mathrm{2}D^{\mathrm{2}}D'\right)}\ \mathrm{sin}\left(x+\mathrm{2}y\right) \\ \\ P.I.\ \ =\frac{\mathrm{1}}{\left(D^{\mathrm{2}}D-\mathrm{2}D^{\mathrm{2}}D'\right)}\ \mathrm{sin}\left(x+\mathrm{2}y\right) \\ \\ Put\ \ D^{\mathrm{2}}=-\mathrm{1} \\ \\ P.I.\ \ =\frac{\mathrm{1}}{D\left(\left(-\mathrm{1}\right)-\mathrm{2}DD'\right)}\ \mathrm{sin}\left(x+\mathrm{2}y\right) \\ \\ Put\ \ DD\mathrm{'}\ \ =\ -\mathrm{2} \\ \\ P.I.\ \ =\frac{\mathrm{1}}{D\left(-\mathrm{1}+\mathrm{4}\right)}\ \mathrm{sin}\left(x+\mathrm{2}y\right) \\ \\ \\ P.I.\ \ =\frac{\mathrm{1}}{\mathrm{3}D}\ \mathrm{sin}\left(x+\mathrm{2}y\right) \\ \\ P.I.\ \ =-\frac{\ \mathrm{cos}\left(x+\mathrm{2}y\right)}{\mathrm{3}}$