Answer: \(x=\mathrm{e}^{-\frac14 t}\left(-\frac19\cos\frac{\sqrt{15}}4 t+\frac{1}{3\sqrt{15}}\sin\frac{\sqrt{15}}4 t\right)+\frac19\mathrm{e}^{-t}.\)

We solve

\(6\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}+3\dfrac{\mathrm{d}x}{\mathrm{d}t}+6x=\mathrm{e}^{-t}\), with \(x(0)=0\) and \(x'(0)=0\).

First solve the homogeneous equation:

\(6x''+3x'+6x=0.\)

Its auxiliary equation is

\(6m^2+3m+6=0.\)

So

\(m=\dfrac{-3\pm\sqrt{9-144}}{12}=-\frac14\pm\frac{\sqrt{15}}4 i.\)

Hence the complementary function is

\(x_c=\mathrm{e}^{-\frac14 t}\left(A\cos\frac{\sqrt{15}}4 t+B\sin\frac{\sqrt{15}}4 t\right).\)

For a particular integral, try

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

Then

\(x_p'=-k\mathrm{e}^{-t},\qquad x_p''=k\mathrm{e}^{-t}.\)

Substituting into the differential equation gives

\(6k\mathrm{e}^{-t}-3k\mathrm{e}^{-t}+6k\mathrm{e}^{-t}=\mathrm{e}^{-t},\)

so \(9k=1\), hence \(k=\frac19\).

Therefore

\(x=\mathrm{e}^{-\frac14 t}\left(A\cos\frac{\sqrt{15}}4 t+B\sin\frac{\sqrt{15}}4 t\right)+\frac19\mathrm{e}^{-t}.\)

Differentiate:

\(x'=\frac{\sqrt{15}}4\mathrm{e}^{-\frac14 t}\left(-A\sin\frac{\sqrt{15}}4 t+B\cos\frac{\sqrt{15}}4 t\right)-\frac14\mathrm{e}^{-\frac14 t}\left(A\cos\frac{\sqrt{15}}4 t+B\sin\frac{\sqrt{15}}4 t\right)-\frac19\mathrm{e}^{-t}.\)

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

\(0=A+\frac19\), so \(A=-\frac19\).

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

\(0=\frac{\sqrt{15}}4B-\frac14A-\frac19.\)

Substitute \(A=-\frac19\):

\(0=\frac{\sqrt{15}}4B+\frac1{36}-\frac19=\frac{\sqrt{15}}4B-\frac1{12}.\)

Hence

\(\frac{\sqrt{15}}4B=\frac1{12}\Rightarrow B=\frac{1}{3\sqrt{15}}.\)

So the required particular solution is

\(x=\mathrm{e}^{-\frac14 t}\left(-\frac19\cos\frac{\sqrt{15}}4 t+\frac{1}{3\sqrt{15}}\sin\frac{\sqrt{15}}4 t\right)+\frac19\mathrm{e}^{-t}.\)