It's auxiliary equation is m² + m + 1 = 0

⇒ \(m = \frac{-1\pm \sqrt{1-4}}{2}= \frac{-1\pm\sqrt3 C}{2}\)

\(C.F. = e^{-\frac x2} \left(C_1 cos\left(\frac{\sqrt3}2x\right)+ C_2sin\left(\frac{\sqrt3}2x\right)\right)\)

\(P.I. = \frac1{D^2 + D + 1}xe^{2x}cos 3x \)

\(= e^{2x} \frac1{(D + 2)^2 + (D + 2) + 1}x \,cos 3x\)

\(= e^{2x} \frac1{D^2 + 5D + 7}x\, cos3x\)

\(= e^{2x} \left(x\frac1{D^2 + 5D + 7}cos 3x + \frac{-(2D + 5)}{(D^2 + 5D + 7)^2}cos3x\right)\)

\(= e^{2x} \left(x\frac{1}{-9 + 5D + 7}cos3x - \frac{2D + 5}{(-9 + 5D + 7)^2}cos 3x\right)\)

\(= e^{2x}\left(x\frac{5D+2}{25D^2 -4}cos3x- \frac{(2D+5)}{25D^2 - 20D + 4}cos3x\right)\)

\(= e^{2x}\left(x\frac{5D \, cos3x + 2cos3x}{-25 \times 9 - 4}+ \frac{(2D + 5)(20D - 221)}{400D^2- 48841}cos 3x\right)\)

\(= e^{2x}\left(\frac{-x}{229}(-15 sin3x + 2cos 3x) - \frac{1}{52441}(40D^2 - 342D - 1105)cos3x\right)\)

\(= e^2x\left(\frac{-x}{229} (-15 sin3x + 2 cos 3x) - \frac1{52441} (-360cos3x + 1026 sin3x - 1105 cos 3x)\right)\)

\(= e^{2x} \left(\frac x{229} (15sin3x - 2 cos 3x) + \frac1{52441} (1465 cos3x - 1026 sin 3x)\right)\)

y = C.F. + P.I is solution of given differential equation.