Answer: \(x=\mathrm{e}^{-t/2}+3t\mathrm{e}^{-t/2}+\frac{2}{3}\mathrm{e}^{-2t}\)

\(\lim_{t\to\infty}x=0\)

We solve

\(4\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}+4\dfrac{\mathrm{d}x}{\mathrm{d}t}+x=6\mathrm{e}^{-2t}.\)

For the complementary function, let \(x=\mathrm{e}^{mt}\). Then

\(4m^2+4m+1=0\)

so

\((2m+1)^2=0,\)

giving a repeated root \(m=-\dfrac12\).

Hence

\(x_c=A\mathrm{e}^{-t/2}+Bt\mathrm{e}^{-t/2}.\)

For a particular integral, try

\(x_p=k\mathrm{e}^{-2t}.\)

Then

\(\dfrac{\mathrm{d}x_p}{\mathrm{d}t}=-2k\mathrm{e}^{-2t},\qquad \dfrac{\mathrm{d}^2x_p}{\mathrm{d}t^2}=4k\mathrm{e}^{-2t}.\)

Substituting into the differential equation gives

\(4(4k\mathrm{e}^{-2t})+4(-2k\mathrm{e}^{-2t})+k\mathrm{e}^{-2t}=6\mathrm{e}^{-2t},\)

so

\((16k-8k+k)\mathrm{e}^{-2t}=6\mathrm{e}^{-2t}.\)

Thus \(9k=6\), so \(k=\dfrac23\).

Therefore the general solution is

\(x=A\mathrm{e}^{-t/2}+Bt\mathrm{e}^{-t/2}+\dfrac23\mathrm{e}^{-2t}.\)

Using \(x(0)=\dfrac53\):

\(\dfrac53=A+\dfrac23,\)

so \(A=1\).

Differentiate:

\(\dfrac{\mathrm{d}x}{\mathrm{d}t}=-\dfrac12A\mathrm{e}^{-t/2}+B\mathrm{e}^{-t/2}-\dfrac12Bt\mathrm{e}^{-t/2}-\dfrac43\mathrm{e}^{-2t}.\)

Now use \(\dfrac{\mathrm{d}x}{\mathrm{d}t}(0)=\dfrac76\):

\(\dfrac76=-\dfrac12(1)+B-\dfrac43.\)

So \(B=3\).

Hence

\(x=\mathrm{e}^{-t/2}+3t\mathrm{e}^{-t/2}+\dfrac23\mathrm{e}^{-2t}.\)

As \(t\to\infty\), all exponential terms tend to \(0\), so

\(\lim_{t\to\infty}x=0.\)