Answer: \(x=\frac{4}{3}e^{t}-\frac{1}{12}e^{-2t}-t^{2}-\frac{3}{2}t-\frac{5}{4}\).

We solve

\(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}+\dfrac{\mathrm{d}x}{\mathrm{d}t}-2x=2t^2+t-1\),

with \(x=0\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t}=0\) when \(t=0\).

First solve the homogeneous equation

\(x''+x'-2x=0\).

The auxiliary equation is

\(m^2+m-2=0\).

Factorising,

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

so the roots are \(m=1\) and \(m=-2\).

Hence the complementary function is

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

For a particular integral, since the right-hand side is a quadratic, try

\(x_p=pt^2+qt+r\).

Then

\(x_p'=2pt+q\),

\(x_p''=2p\).

Substitute into the differential equation:

\(2p+(2pt+q)-2(pt^2+qt+r)=2t^2+t-1\).

Expand and collect terms:

\(-2pt^2+(2p-2q)t+(2p+q-2r)=2t^2+t-1\).

Equating coefficients gives

\(-2p=2\), so \(p=-1\),

\(2p-2q=1\),

\(2p+q-2r=-1\).

Using \(p=-1\):

\(-2-2q=1\), so \(q=-\frac{3}{2}\).

Then

\(-2-\frac{3}{2}-2r=-1\),

so \(-\frac{7}{2}-2r=-1\),

\(-2r=\frac{5}{2}\),

\(r=-\frac{5}{4}\).

Therefore

\(x_p=-t^2-\frac{3}{2}t-\frac{5}{4}\).

So the general solution is

\(x=Ae^t+Be^{-2t}-t^2-\frac{3}{2}t-\frac{5}{4}\).

Differentiate:

\(x'=Ae^t-2Be^{-2t}-2t-\frac{3}{2}\).

Now apply the conditions at \(t=0\).

From \(x(0)=0\):

\(A+B-\frac{5}{4}=0\).

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

\(A-2B-\frac{3}{2}=0\).

Solve these simultaneously.

From the first, \(A=\frac{5}{4}-B\).

Substitute into the second:

\(\frac{5}{4}-B-2B-\frac{3}{2}=0\).

\(-3B-\frac{1}{4}=0\), so \(B=-\frac{1}{12}\).

Then

\(A=\frac{5}{4}+\frac{1}{12}=\frac{16}{12}=\frac{4}{3}\).

Hence the particular solution satisfying the given conditions is

\(x=\frac{4}{3}e^t-\frac{1}{12}e^{-2t}-t^2-\frac{3}{2}t-\frac{5}{4}\).