Respuesta :

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First, find the characteristic solution. The characteristic equation for this ODE is

[tex]r^2+2r+1=(r+1)^2=0[/tex]

which has one root at [tex]r=-1[/tex] with multiplicity 2. This means the characteristic solution takes the form

[tex]x_c=C_1e^{-t}+C_2te^{-t}[/tex]

There's no conflict with the nonhomogeneous part, which means you can guess a particular solution with undetermined coefficients of the form

[tex]y_p=ae^{-2t}+bt+c[/tex]

which has derivatives

[tex]{y_p}'=-2ae^{-2t}+b[/tex]
[tex]{y_p}''=4ae^{-2t}[/tex]

Substituting the particular solution into the ODE yields

[tex]4ae^{-2t}+2(-2ae^{-2t}+b)+ae^{-2t}+bt+c=5e^{-2t}+t[/tex]
[tex]ae^{-2t}+bt+2b+c=5e^{-2t}+t[/tex]

Matching up coefficients gives a system of equations with solution

[tex]\begin{cases}a=5\\b=1\\2b+c=0\end{cases}\implies a=5,b=1,c=-2[/tex]

so that the particular solution is

[tex]x_p=5e^{-2t}+t-2[/tex]

which in turn means the general solution is

[tex]x=x_c+x_p[/tex]
[tex]x=C_1e^{-t}+C_2te^{-t}+5e^{-2t}+t-2[/tex]

Use the initial conditions to solve for the remaining constants.

[tex]\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[/tex]

Therefore the solution to this IVP is

[tex]x=-e^{-t}+9te^{-t}+5e^{-2t}+t-2[/tex]

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