I am stuck trying to solve the following second order differential equation:
$$ 2y'' + 3y' - 2y = te^{-2t} $$
The problem is from Paul's Online Notes in the Laplace Transform section, however in order to gain some practice I am also calculating the Method of Underdetermined Coefficients as well as the Variation of Parameters since for the IVP the solution should be unique and hence all three different methods should lead to the same solution.
My solution attempt:
(1) Normalize \begin{align} 2y'' + 3y' - 2y &= te^{-2t} \\ \frac{1}{2}(2y'' + 3y' - 2y) &= \frac{1}{2}te^{-2t} \\ y'' + \frac{3}{2}y' - y &= \frac{1}{2}te^{-2t} \end{align}
As the particular solution I try:
$$ y_p = (At + B)e^{-2t} $$
and calculate the derivatives: \begin{align} y_p' &= -2(At + B)e^{-2t} + Ae^{-2t} \\ y_p'' &= 4(At + B)e^{-2t} - 2Ae^{-2t} - 2Ae^{-2t} = 4(At + B)e^{-2t} - 4Ae^{-2t} \\ \end{align}
thus:
\begin{align} y_p'' + \frac{3}{2}y_p' - y_p &= 4(At + B)e^{-2t} - 4Ae^{-2t} + \frac{3}{2}(-2(At + B)e^{-2t} + Ae^{-2t}) - (At + B)e^{-2t} \\ &= e^{-2t}((At + B)(4 - 3 - 1) + A(-4 + \frac{3}{2})) \\ &= e^{-2t}((At + B)0 - A\frac{5}{2})) = t e^{-2t}\\ \end{align}
Now form here we can see that:
$$ 0A = \frac{1}{2} $$
which is not valid. I know that there is a solution, so where is my mistake? My guess is that I chose the wrong function for the particular solution, but I just followed the same pattern as Paul's Online Notes Example 5.
The method of undetermined coefficients due to Lagrange, follows:
The homogeneous ODE has as solutions
$$ y_h = c_1 e^{\frac t2}+c_2 e^{-2t} $$ now considering
$$ y_p = c_1(t) e^{\frac t2}+c_2(t) e^{-2t} $$
after submitting to the complete ODE we have
$$ (2c_1''+5c_1')e^{\frac t2}+(-2c_2''+5c_2'+t)e^{-2t}=0 $$
now as $c_1,c_2$ are independent we can choose
$$ \cases{ 2c_1''+5c_1'=0\\ -2c_2''+5c_2'+t=0 } $$
thus we obtain as solutions
$$ \cases{ c_1 = -\frac 25e^{-5\frac t2}\mu_1+\mu_2\\ c_2 = -t\left(\frac{2}{25}+\frac{t}{10}\right)+e^{5\frac t2}\mu_3+\mu_4 } $$
here $\mu_1,\cdots,\mu_4$ are integration constants. As we are choosing particular solutions, we choose $\mu_1=\cdots=\mu_4=0$ and finally we have
$$ \cases{ c_1(t) = 0\\ c_2(t) = -t\left(\frac{2}{25}+\frac{t}{10}\right) } $$
and a particular solution is
$$ y_p = -t\left(\frac{2}{25}+\frac{t}{10}\right)e^{-2t} $$