Solve the differential equation y'' + 2y' + y = e^(-x).

Step-by-step solution

1. 1

   Find the characteristic roots. For $y'' + 2y' + y = 0$ the characteristic equation is

   $r^2 + 2r + 1 = (r+1)^2 = 0$

   with the repeated root $r = -1$. A double root contributes both $e^{-x}$ and $xe^{-x}$, so

   $y_h = (C_1 + C_2x)e^{-x}$

2. 2

   Count how deep the resonance goes. The forcing term $e^{-x}$ matches the root $-1$, and that root has multiplicity $2$. The rule is: multiply the trial by $x^{m}$ where $m$ is the multiplicity. Here $m = 2$, so both $Ae^{-x}$ and $Axe^{-x}$ would substitute to zero — only $y_p = Ax^2e^{-x}$ can survive.

3. 3

   Differentiate the trial twice.

   $y_p' = A e^{-x}\left(2x - x^2\right), \qquad y_p'' = A e^{-x}\left(x^2 - 4x + 2\right)$

4. 4

   Substitute and solve for A.

   $y_p'' + 2y_p' + y_p = Ae^{-x}\left[(x^2 - 4x + 2) + (4x - 2x^2) + x^2\right] = 2Ae^{-x}$

   The $x^2$ terms cancel ($1 - 2 + 1 = 0$) and so do the $x$ terms ($-4 + 4 = 0$), leaving only the constant $2$. Setting $2Ae^{-x} = e^{-x}$ gives $A = \tfrac12$.

5. 5

   Write and check the general solution.

   $y = e^{-x}\left(C_1 + C_2x + \tfrac{1}{2}x^2\right)$

   Verification with $y_p = \tfrac12x^2e^{-x}$: $y_p' = \left(x - \tfrac{x^2}{2}\right)e^{-x}$ and $y_p'' = \left(1 - 2x + \tfrac{x^2}{2}\right)e^{-x}$, so the left side is $e^{-x}\left[\left(1 - 2x + \tfrac{x^2}{2}\right) + \left(2x - x^2\right) + \tfrac{x^2}{2}\right] = e^{-x}$. Correct.