First, find the characteristic solution. The characteristic equation for this ODE is
\(r^2+2r+1=(r+1)^2=0\)
which has one root at \(r=-1\) with multiplicity 2. This means the characteristic solution takes the form
\(x_c=C_1e^{-t}+C_2te^{-t}\)
There's no conflict with the nonhomogeneous part, which means you can guess a particular solution with undetermined coefficients of the form
\(y_p=ae^{-2t}+bt+c\)
which has derivatives
\({y_p}'=-2ae^{-2t}+b\)
\({y_p}''=4ae^{-2t}\)
Substituting the particular solution into the ODE yields
\(4ae^{-2t}+2(-2ae^{-2t}+b)+ae^{-2t}+bt+c=5e^{-2t}+t\)
\(ae^{-2t}+bt+2b+c=5e^{-2t}+t\)
Matching up coefficients gives a system of equations with solution
\(\begin{cases}a=5\\b=1\\2b+c=0\end{cases}\implies a=5,b=1,c=-2\)
so that the particular solution is
\(x_p=5e^{-2t}+t-2\)
which in turn means the general solution is
\(x=x_c+x_p\)
\(x=C_1e^{-t}+C_2te^{-t}+5e^{-2t}+t-2\)
Use the initial conditions to solve for the remaining constants.
\(\begin{cases}2=C_1+3&x(0)=2\\1=-C_1+C_2-9&x'(0)=1\end{cases}\implies C_1=-1,C_2=9\)
Therefore the solution to this IVP is
\(x=-e^{-t}+9te^{-t}+5e^{-2t}+t-2\)
