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I'm trying an odd problem involving finding the proper form of the particular solution for several ODEs, using the method of undetermined coefficients.

The four equations are:

$$y'''-y''+4y'-4y=(x^2+11)e^x$$

$$y''-2y'+5y=(4x^2+2x)e^x$$

$$y''-11y'+18y=x^2e^{2x}$$

$$y''-2y'=1+xe^{-6x}$$

The aim is to find $y_p$ with the undetermined coefficients as P, Q, R, S etc. I tried out a couple, finding $Pe^x+Qxe^x +Rx^2e^x$ for equation 2 and $e^{2x}(Px^2+Qx+R)$ for 3 to no avail. Any help is appreciated, thank you.

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  • $\begingroup$ I do not believe the first has a solution (is it written correctly)? For the second, let $y_p = a e^x + b x e^x + c x^2 e^x$. You will get $a = -\dfrac{1}{2}, b = \dfrac{1}{2}, c = 1$. You should work the other two, which both have solutions too. $\endgroup$ – Amzoti Mar 2 '15 at 23:07
  • $\begingroup$ @Amzoti You're correct, I wrote the question wrong, will edit immediately $\endgroup$ – jofl Mar 2 '15 at 23:16
  • $\begingroup$ That is better. For the first, the homogeneous solution has an $e^x$ term, so choose $y_p = x(a e^x + b x e^x + c x^2 e^x)$. Hopefully now, you can work the other two. $\endgroup$ – Amzoti Mar 2 '15 at 23:21
  • $\begingroup$ @Amzoti Should these be the others then: (1) $x(Pe^x+Qe^xx+Re^xx^2$. (2) $Pe^x+Qe^xx+Re^xx^2$. (3) $Pe^{2x}x+Qe^{2x}x^2+Re^{2x}x^3$. (4) Pe^{-6x}x+Qx+Re^{-6x}$? $\endgroup$ – jofl Mar 2 '15 at 23:28
  • $\begingroup$ For the last, you could choose $y_p = a x + b e^{-6x} + c x e^{-6x}$. Yes, correct. $\endgroup$ – Amzoti Mar 2 '15 at 23:33
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i will make a change of dependent variable $$y = ue^x, y' = (u' + u)e^x, y'' = (u'' + 2u' + u)e^x, y'''=(u'''+3u'' + 3u' + u)e^x. $$ with this, the differential equation $ y'''-y''+4y'-4y=(x^2+11)e^x $ turns into $$ (u'''+3u'' + 3u' + u) - (u'' + 2u' + u) +4(u' + u) -4u = (x^2+11) $$ this can be simplified to give $$u'''+2u''+5u'=x^2 + 11 $$ i hope you can take it from here. if you need help completing, let me know.

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