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Question Number 91848 by jagoll last updated on 03/May/20

$$y^{{}^{″}}-4y^{{}^{\prime }}+4y=(x+1)e^{2x}$$$$$$

Commented by john santu last updated on 03/May/20

$$complementarysolution$$$$m^2-4m+4=0$$$$m_{1,2}=2\Rightarrow y_c=A_1{e}^{2x}+A_2{e}^{2x}$$$$particularintegral$$$$y_p=(B_1{x}^3+B_2{x}^2)e^{2x}$$$$y_p^{{}^{\prime }}=(3B_1{x}^2+2B_{2}x)e^{2x}+2(B_1{x}^3+B_2{x}^2)e^{2x}$$$$y_p^{″}=(6B_{1}x+2B_2)e^{2x}+2(3B_1{x}^2+2B_{2}x)e^{2x}$$$$2(3B_1{x}^2+2B_{2}x)e^{2x}+4(B_1{x}^3+B_2{x}^2)e^{2x}$$$$comparingcoeff$$$$B_2=\frac{1}{2};B_1=\frac{1}{6}$$$$\therefore y=(A_1+A_2)e^{2x}+(\frac{1}{6}{x}^3+\frac{1}{2}{x}^2)e^{2x}$$

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