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I am asked to find the solution of the following system of nonhomogeneus differential equations using elimination method:

$$ \begin{cases} x'(t)=4x(t)-2y(t)+t^{-3} \\ y'(t)=8x(t)-4y(t)-t^{-2} \end{cases} $$

By differentiating the first equation and substituing in values from second one, I find the characteristic value which is $ \lambda_{1,2} = 0 $, but from there on I am not sure how to proceed.

I believe the solution of the homogeneus equation for $x(t)$ is $ x_{0}(t) = C_1e^{0t} = C_1 $, but I do not know in what form should I seek a particular solution. Some help or hints on what to do would be greatly appreciated. Thank you.

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  • $\begingroup$ There is a mistake in the system an x instead of an y $\endgroup$
    – user577215664
    May 15, 2020 at 18:59
  • $\begingroup$ Oh yes, thank you, I corrected it now. $\endgroup$
    – grinch
    May 15, 2020 at 19:04

1 Answer 1

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$$\begin{cases} x'(t)=4x(t)-2y(t)+t^{-3} \\ y'(t)=8x(t)-4y(t)-t^{-2} \end{cases}$$ Substract both DEs $$2x'-y'=2t^{-3}+t^{-2}$$ Integrate: $$2x-y=-t^{-2}-t^{-1}+C_1$$ $$4x-2y=-2t^{-2}-2t^{-1}+2C_1$$ Now eliminate $4x-2y$ from first equation and solve for $x(t)$. It's a first DE.

If you have 2 times an eignanvalue equals zero then the solution should be $c_1+c_2t$ But you didnt post your attempt so we can check it.

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    $\begingroup$ Oh, I think I will manage it now, thank you very much. $\endgroup$
    – grinch
    May 15, 2020 at 19:23
  • $\begingroup$ You're welcome @grinch $\endgroup$
    – user577215664
    May 15, 2020 at 19:25

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