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I have to solve $$y''' - 3y'' +3y' - y = x-4e^x$$ I use $y_p$ in the form $Ax+B+Ce^x$. I almost get the right answer, but just one of my terms is incorrect, and one of my signs is incorrect. I am getting $$y = c_1e^x + c_2xe^x + c_3x^2e^x - 2e^x - x + 3$$ but the answer is $$y = c_1e^x + c_2xe^x + c_3x^2e^x - 2/3x^3e^x - x - 3$$ I'm not sure what I'm doing wrong...

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The associated homogeneous auxiliary equation is $$m^3-3m^2+3m-1=0$$ $$(m-1)^3=0$$ $$m=1$$ Hence the complementary function is $$y_c=(A+Bx+Cx^2)e^x$$ So the particular solution is of the form $$y_p=Dx+E+Fx^3e^x$$

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  • $\begingroup$ How did you get Fx^3e^x? @peter foreman ? $\endgroup$ – Melanie Apr 17 at 20:48
  • $\begingroup$ In general, if $f(x)$ is in the original equation and the complementary function then we can try $xf(x), x^2f(x)$ etc. until we have a unique function that is not in the complementary function or the original equation. $\endgroup$ – Peter Foreman Apr 17 at 20:49
  • $\begingroup$ oh I understand now! Thank you! $\endgroup$ – Melanie Apr 17 at 21:08

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