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**danny arrigo** Yes, your complimentary solution is right (via your characteristic equation) except you want x's and not t's. Let me explain a little bit more about the construction of this particular solution. There are primary two methods to do this (1) the method of undetermined coefficients and (2) the variation of parameters. The first method is usually easier to use but involves some educated guessing. The method basic seeks the form of the particular solution by looking at the rhs of the ODE. If it has power's guess power's, exponential's guess exponential's, sin's and cos's guess sin's and cos's. In your example

$\displaystyle y'' + y' -2y = 2 - 2e^{-2x}$

it has powers $\displaystyle x^0 $ and exponential's $\displaystyle e^{-2x}. $ Therefore one would try a guess of

$\displaystyle y = A + B e^{-2x}$

Taking derivatives

$\displaystyle y' = -2 B e^{-2x},\;\;\;y'' = 4 B e^{-2x},$

and substituting gives

$\displaystyle y'' + y' - 2y = 4 B e^{-2x} + \left( -2 B e^{-2x}\right) - 2 \left( A + B e^{-2x}\right) = 2 - 2e^{-2x} $

Expanding gives $\displaystyle -2A = 2 - 2e^{-2x} $ (yes, the B's cancel but why?) The reason is that $\displaystyle y = Be^{-2x}$ is part of the complimentary solution so

$\displaystyle y'' + y' - 2y = 0 $ identically! So in this case, what do we do for a guess. Let us onsider the following first order example

$\displaystyle (1)\;\;\; y' - y = e^x $

The complimentary equation is

$\displaystyle y' - y = 0 $

the characteristic equations is

$\displaystyle m - 1 = 0 $

the complimentary solution is

$\displaystyle y = e^x $

However, since eqn. (1) is linear we solve giving

$\displaystyle y = c e^x + x\,e^x $

In this solution we have two parts, the first the complimentary and the second the particular. We notice the extra x so in order to guess a particular solution when rhs contains the complimentary solution we "bump" up the exponential in our first guess and use

$\displaystyle y = A + Bx e^{-2x} $

$\displaystyle y' = B e^{-2x} -2x Bx e^{-2x} $

$\displaystyle y'' = -4B e^{-2x} + 4x Bx e^{-2x} $

substituting gives

$\displaystyle y'' + y' - 2y = -4B e^{-2x} + 4 Bx e^{-2x} + \left( B e^{-2x} -2 Bx e^{-2x} \right) - 2 \left( A + Bx e^{-2x} \right)=$

$\displaystyle = 2 - 2e^{-2x} $

and when we expand we get

$\displaystyle -3B e^{-2x} - 2 A = 2 - 2e^{-2x} $

from which we obtain $\displaystyle A = -1,\;\;B = \frac{2}{3}$ and the particular solution

$\displaystyle y_p = - 1 +\frac{2}{3} \,e^{-2x} $