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Question Number 158309 by cortano last updated on 02/Nov/21

$$y^{″}+y^{\prime }-2y=-18{te}^{-2t}$$

Answered by TheSupreme last updated on 02/Nov/21

$${\lambda }^2+\lambda -2=0$$$${\lambda }_{1,2}=-\frac{1}{2}\pm \frac{\sqrt{9}}{2}=-2,1$$$$y_0\left(t\right)=c_1{e}^t+c_2{e}^{-2t}$$$$\stackrel{\sim }{y}=\left(At\right)e^{-2t}$$$$y^{\prime }=(-2At+A)e^{-2t}$$$$y^{″}=(-2A+4At-2A)$$$$(4At-4A-2At+A-2At)=-18$$$$-3A=-18$$$$A=6$$$$y=(6t+c_1)e^{-2t}+c_2{e}^t$$

Answered by tounghoungko last updated on 03/Nov/21

$$particularintegral$$$$y_p=e^{-2t}({At}^2+Bt)$$$$y_p^{\prime }=-2e^{-2t}({At}^2+Bt)+e^{-2t}(2At+B)$$$$y_p^{″}=4e^{-2t}({At}^2+Bt)-4e^{-2t}(2At+B)+2{Ae}^{-2t}$$$$comparingcoefficient$$$$-3e^{-2t}(2At+B)+2{Ae}^{-2t}=-18{te}^{-2t}$$$$\Rightarrow -6{Ate}^{-2t}+(2A-3B)e^{-2t}=-18{te}^{-2t}$$$$\{\begin{array}{l}A=3\\ B=2\end{array}\Rightarrow y_p=e^{-2t}(3t^2+2t).$$

Answered by qaz last updated on 03/Nov/21

$${\mathrm{y}}_{\mathrm{p}}=\frac{1}{(\mathrm{D}+2)(\mathrm{D}-1)}(-{18\mathrm{te}}^{-2\mathrm{t}})$$$$=-{18\mathrm{e}}^{-2\mathrm{t}}\frac{1}{\mathrm{D}(\mathrm{D}-3)}\mathrm{t}$$$$=-{9\mathrm{e}}^{-2\mathrm{t}}\frac{1}{\mathrm{D}-3}{\mathrm{t}}^2$$$$={3\mathrm{e}}^{-2\mathrm{t}}(1+\frac{\mathrm{D}}{3}+\frac{{\mathrm{D}}^2}{3^2}+...){\mathrm{t}}^2$$$$={3\mathrm{e}}^{-2\mathrm{t}}({\mathrm{t}}^2+\frac{2}{3}\mathrm{t}+\frac{2}{9})$$$$\Rightarrow \mathrm{y}={\mathrm{C}}_1{\mathrm{e}}^{-2\mathrm{t}}+{\mathrm{C}}_2{\mathrm{e}}^{\mathrm{t}}+{3\mathrm{e}}^{-2\mathrm{t}}({\mathrm{t}}^2+\frac{2}{3}\mathrm{t})$$

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