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[tex]y'' + 2y' + y = e^{-x}[/tex]

I started by solving it like it was y'' + 2y' +y = 0 (Instert y = e^ax and so on) and got the following equation and solution:

[tex]e^{ax}(a^2 + 2a + 1) = 0 => a = -1[/tex]

[tex]y'' + 2y' + y = 0, y = e^{-x}(k_1x + k_2)[/tex]

Since the right side of the original equation is of the form Me^hx, and h is a solution to the equation above (a = -1), I tried inserting f(x) = Axe^(-x).

From that I got that f'(x) = Ae^(-x) - Axe^(-x) and f''(x) = Axe^(-x) -2Ae^(-x).

But when I insert that into the original equation everything cancels out and I'm left with e^(-x) = 0. Since that never applies I'm inclined to think that there are no solutions to the equation. What do you think?