Answer:

The complementary equation is

\(\frac{d^{2}x}{dt^{2}}+4\frac{dx}{dt}+4x=0.\)

Its auxiliary equation is

\(m^{2}+4m+4=0,\qquad (m+2)^{2}=0,\)

so the complementary function is

\(x_c=Ae^{-2t}+Bte^{-2t}.\)

For a particular integral, try \(x_p=at^{2}+bt+c\). Then

\(x_p'=2at+b,\qquad x_p''=2a.\)

Substitute into the differential equation:

\(2a+4(2at+b)+4(at^{2}+bt+c)=7-2t^{2}.\)

Compare coefficients:

\(4a=-2,\qquad 8a+4b=0,\qquad 2a+4b+4c=7.\)

Thus \(a=-\frac12\), \(b=1\), and \(c=1\). Therefore

\(x=Ae^{-2t}+Bte^{-2t}-\frac12t^{2}+t+1.\)