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Question Number 141849 by iloveisrael last updated on 24/May/21

$$y^{″}-6y^{\prime }+9y=9x^2{e}^{3x}\cos \left(3x\right)$$

Answered by qaz last updated on 24/May/21

$$y^{″}-6y^{\prime }+9y=9x^2{e}^{3x}\cos \left(3x\right)$$$$y_p=\frac{1}{D^2-6D+9}9x^2{e}^{3x}\cos \left(3x\right)$$$$=9\frac{1}{{(D-3)}^2}{x}^2{e}^{3x}\cos \left(3x\right)$$$$=9e^{3x}\mathrm{\Re }\frac{1}{D^2}{x}^2{e}^{3ix}$$$$=9e^{3x}\mathrm{\Re }{e}^{3ix}\frac{1}{{(D+3i)}^2}{x}^2$$$$=9e^{3x}\mathrm{\Re }{e}^{3ix}\frac{1}{D^2+6iD-9}{x}^2$$$$=-9e^{3x}\mathrm{\Re }{e}^{3ix}\frac{1}{9[1-(\frac{D^2}{9}+\frac{2iD}{3})]}{x}^2$$$$=-e^{3x}\mathrm{\Re }{e}^{3ix}[1+(\frac{D^2}{9}+\frac{2iD}{3})+{(\frac{D^2}{9}+\frac{2iD}{3})}^2+...]x^2$$$$=-e^{3x}\mathrm{\Re }{e}^{3ix}(x^2+\frac{4i}{3}x-\frac{2}{3})$$$$=-e^{3x}[x^2\cos \left(3x\right)-\frac{4}{3}x\sin \left(3x\right)-\frac{2}{3}\cos \left(3x\right)]$$$$\Rightarrow y=e^{3x}(C_1+C_{2}x)-e^{3x}[x^2\cos \left(3x\right)-\frac{4}{3}x\sin \left(3x\right)-\frac{2}{3}\cos \left(3x\right)]$$$$=e^{3x}[C_1+C_{2}x-x^2\cos \left(3x\right)+\frac{4}{3}x\sin \left(3x\right)+\frac{2}{3}\cos \left(3x\right)]$$

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