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I need to find the solution of $$x''-2x'+x=\sum_{n=1}^Ne^{-nt}$$ I was thinking undetermined coefficients. Is there another way?

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$$x''-2x'+x=\sum_{n=1}^Ne^{-nt}$$ $$e^{-t}(x''-2x'+x)=\sum_{n=1}^Ne^{-(n+1)t}$$ $$(e^{-t}x)''=\sum_{n=1}^Ne^{-(n+1)t}$$ Now define $y(t):=e^{-t}x(t)$ $$y''=\sum_{n=1}^Ne^{-(n+1)t}$$ Now you can integrate twice.

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  • $\begingroup$ Nice trick and the solution seems to be much simpler than the brute force answer from Wolfram Alpha. $\endgroup$ – gammatester Jul 1 '14 at 11:55
  • $\begingroup$ Nice answer!!! Thanks $\endgroup$ – Valerin Jul 1 '14 at 11:56
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The characteristic equation of the homogeneous equation is $$r^2-2r+1=0$$ and has the double root $r=1$ so the homogeneous solution is $$x_h(t)=(at+b)e^t$$ Now we look for a particular solution to the ODE $$x''-2x'+x=e^{-kt}$$ on the form $x_p(t)=\alpha_k e^{-kt}$ and we find $$(k^2+2k+1)\alpha_k=1\iff\alpha_k=(k^2+2k+1)^{-1}$$ and finally by the superposition method the general solution of the given equation is $$x(t)=(at+b)e^t+\sum_{n=1}^N\frac{e^{-nt}}{n^2+2n+1}$$

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Solution using the method of operators

The equation can be written as
$(D^2 - 2D + 1)x = \sum\limits_{n = 1}^N e^{-nt} \Leftrightarrow\\ (D - 1)^2 x = \sum\limits_{n = 1}^N e^{-nt}$

The auxiliary equation is $(m - 1)^2 = 0 \Rightarrow m = 1, 1$, so the complementary function (solution of the homogeneous equation $(D - 1)^2 x = 0$) is $x_c = (c_1 + c_2 t)e^t$.

To find the particular integral, $x_p$, we use $\dfrac{1}{f(D)}e^{at} = \dfrac{1}{f(a)}e^{at}$ when $f(a) \ne 0$.

Here, $\dfrac{1}{(D - 1)^2}e^{-nt} = \dfrac{1}{(-n - 1)^2}e^{-nt} = \dfrac{1}{(n + 1)^2}e^{-nt}$ (as $n \ne -1)$.

Thus, $x_p = \sum\limits_{n = 1}^N \dfrac{1}{(n + 1)^2}e^{-nt}$.

Therefore, the complete solution is obtained by adding $x_c$ and $x_p$:

$\boxed{x = \sum\limits_{n = 1}^N \dfrac{1}{(n + 1)^2}e^{-nt} + c_1 + c_2 t}$

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First you can solve $x''-2x'+x=0$. We are looking solution in form $x(t)=e^{\alpha t}$, if you substitute to equation you have $\alpha^2-2\alpha+1=(\alpha-1)^2=0$, $\alpha=1$ is double root, so you have two linear independent solution:$e^{t}$ and $te^{t}$, so general solution is:

$x(t)=Ce^t+Dte^t$

Now find special solutions of $x''-2x'+x=e^{-nt}$. It is in form $x(t)=C_ne^{-nt}$, substitute to equation and find $C_n$:

$C_nn^2+2nC_n+1+C_n=1$, so:

$C_n=\frac{1}{n^2+2n+1}$

Now you know special solution for $x''-2x'+x=\sum_{n=1}^{N}e^{-nt}$, it is: $x(t)=\sum_{n=1}^{N}C_ne^{-nt}$, so finally:

$x(t)=Ce^t+te^{t}+\sum_{n=1}^{N}C_ne^{-nt}$

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