I have a system of differential equations like this:

$$ \left\{ \begin{array}{c} \begin{aligned} x'&= x-4y +1\\ y'&=-x+y+e^{-t} \end{aligned} \end{array} \right. $$

I have found the complimentary solution (eigenvalues are -1 and 3), but I'm having a hard time finding the particular solution. I have tried guessing $$e^{at}(v_1+v_2t) $$ where $v_1$ and $v_2$ are vectors, but when I use the method of undetermined coefficients (yes we have to use this method) and construct the equations to find the coefficients, everything is fine but I am left with only one term with $e^{-t}$ as a factor, so I get $1/e^{-t}=0$ or $e^{-t}=0$. Am I guessing the wrong particular solution? Or I am supposed to just ignore this equation?

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    $\begingroup$ How about trying (constants are vectors of course) $$y_p = a + b~ e^{-t} + c~ t ~e^{-t}$$ instead? $\endgroup$ – Moo Apr 6 '18 at 13:13

$$\left\{ \begin{array}{c} \begin{aligned} x'&= x-4y +1\\ y'&=-x+y+e^{-t} \end{aligned} \end{array} \right. \implies X'=AX+e^{-t}\binom 01 +\binom 1 0 $$ $$X_h=C_1e^{-t}\binom 21 +C_2e^{3t}\binom {-2}1$$ Since you already have $e^{-t}$ in the general solution of the system you should as Moo has suggested try this:

$$ X_p=K_1te^{-t}+K_2e^{-t}+K_3 $$

Another approach..

Summing $L_1+2L_2$

$$x'+2y'=-(x+2y)+1+2e^{-t}$$ Substitute $z=x+2y$ $$z'=-z+1+2e^{-t}$$ $$ze^t=\int e^t+2dt$$ $$x+2y=e^{-t}(e^t+2t+K_1)$$ $$\boxed{ \begin{cases} x(t)=1-2y+e^{-t}(2t+K_1)\\ y'=-x+y+e^{-t} \end{cases}} $$ Easy to solve now

  • $\begingroup$ Yes true @moo I just added it as another approach since you have already given a hint $\endgroup$ – Isham Apr 6 '18 at 13:38

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