Question

X"+2x'+x=5e^-2t+t, x(0)=2, x'(0)=1

Answer 1

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

$x=-e^{-t}+9te^{-t}+5e^{-2t}+t-2$