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I need to use the method of undetermined coefficients to find the general solution of the following equation:

$$y''+2y'+10y=5$$

I know I have to find $y_c$ and $y_p$. For $y_c$, which is the characteristic equation, I got:

$$y_c=e^{-x}(c_1\cos(3x) + c_2\sin(3x))$$

I am not sure what to choose for $y_p$ to derive twice and then plug back into the equation to determine the coefficients of $y_p$

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  • $\begingroup$ Hi and welcome to math.stackexchange. In the long term you are encouraged to learn using MathJax and LaTeX typesetting for your questions and answers, but since you seem new I helped you do it this once. There are tutorials both on this site and elsewhere how to use them. $\endgroup$ – mathreadler Jul 8 '16 at 22:55
  • $\begingroup$ would choose yp = A be a good yp, so then y'p would be 0 and y"p = 0 $\endgroup$ – mp12345 Jul 8 '16 at 22:56
  • $\begingroup$ @mathreadler thanks $\endgroup$ – mp12345 Jul 8 '16 at 22:57
  • $\begingroup$ That $y_p$ is perfect and you end up with $10 A = 5 \implies A = \dfrac{1}{2}$. $\endgroup$ – Moo Jul 8 '16 at 23:23
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    $\begingroup$ yes thanks I solved that to get y+e^(-x)(c1cos3x + c2sin3x) +1/2 $\endgroup$ – mp12345 Jul 8 '16 at 23:24
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When you have a differential equation such as $$ay'''+by''+cy'+dy=e$$ in which $d$ and $e$ are constants, define $$y=z+\frac ed\implies y'=z'\implies y''=z''\implies y'''=z'''$$ and the equation becomes $$az'''+bz''+cz'+dz=0$$ and you are back to a simpler form.

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