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Question Number 170998 by 2407 last updated on 06/Jun/22
Answered by ali009 last updated on 06/Jun/22
$${r}^{\mathrm{2}} +\mathrm{3}{r}+\mathrm{2}=\mathrm{0} \\ $$$${r}=−\mathrm{1}\:\:\:\:{r}=−\mathrm{2} \\ $$$${y}_{{h}} ={C}_{\mathrm{1}} {e}^{−{x}} ×{C}_{\mathrm{2}} {e}^{−\mathrm{2}{x}} \\ $$$${W}=\begin{vmatrix}{{e}^{−{x}} }&{{e}^{−\mathrm{3}{x}} }\\{−{e}^{−{x}} }&{−\mathrm{3}{e}^{−\mathrm{3}{x}} }\end{vmatrix}=−\mathrm{3}{e}^{−\mathrm{4}{x}} +{e}^{−\mathrm{4}{x}} =−\mathrm{2}{e}^{−\mathrm{4}{x}} \\ $$$${v}_{\mathrm{1}} =\int\frac{{sin}\left({x}\right){e}^{−\mathrm{3}{x}} }{\mathrm{2}{e}^{−\mathrm{4}{x}} }{dx}=\frac{{e}^{{x}} }{\mathrm{4}}\left({sin}\left({x}\right)−{cos}\left({x}\right)\right) \\ $$$${v}_{\mathrm{2}} =−\int\frac{{sin}\left({x}\right){e}^{−{x}} }{\mathrm{2}{e}^{−\mathrm{4}{x}} }{dx}=\frac{{e}^{\mathrm{3}{x}} }{\mathrm{20}}\left({cos}\left({x}\right)−\mathrm{3}{sin}\left({x}\right)\right) \\ $$$${y}_{{p}} =\frac{\mathrm{1}}{\mathrm{4}}\left({sin}\left({x}\right)−{cos}\left({x}\right)\right)+\frac{\mathrm{1}}{\mathrm{20}}\left(−\mathrm{3}{sin}\left({x}\right)+{cos}\left({x}\right)\right) \\ $$$${y}={y}_{{p}} +{y}_{{h}} \\ $$$$ \\ $$$$ \\ $$