Answer:

The complementary function comes from

\(m^2+0.16m+0.0064=0.\)

Since \(m^2+0.16m+0.0064=(m+0.08)^2\), the repeated root is \(m=-0.08\). Hence

\(x_c=(A+Bt)e^{-0.08t}.\)

For a particular integral, try \(x_p=p+qt\). Then \(\dot{x}_p=q\) and \(\ddot{x}_p=0\), so substituting gives

\(0.16q+0.0064(p+qt)=8.64+0.32t.\)

Comparing coefficients,

\(0.0064q=0.32\quad\Rightarrow\quad q=50,\)

and

\(0.16(50)+0.0064p=8.64\quad\Rightarrow\quad p=100.\)

Thus

\(x=(A+Bt)e^{-0.08t}+100+50t.\)

Using \(x=0\) when \(t=0\) gives \(A=-100\), so

\(x=(Bt-100)e^{-0.08t}+100+50t.\)

Differentiate:

\(\dot{x}=B e^{-0.08t}-0.08(Bt-100)e^{-0.08t}+50.\)

Using \(\dot{x}=0\) when \(t=0\),

\(0=B+8+50,\)

so \(B=-58\). Therefore

\(x=100+50t-(100+58t)e^{-0.08t}.\)

Now

\(\dot{x}=50+(-50+4.64t)e^{-0.08t}.\)

As \(t\to\infty\), the exponential term tends to \(0\), so for large positive \(t\),

\(\dot{x}\approx50.\)