Answer: \(x=\mathrm{e}^{6t}\left(t-\frac{12}{37}\right)+\frac{35}{37}\sin t+\frac{12}{37}\cos t\).

We solve

\(\displaystyle \frac{\mathrm{d}^2x}{\mathrm{d}t^2}-12\frac{\mathrm{d}x}{\mathrm{d}t}+36x=37\sin t\), with \(x=0\) and \(\frac{\mathrm{d}x}{\mathrm{d}t}=0\) when \(t=0\).

First solve the homogeneous equation:

\(m^2-12m+36=0\)

\((m-6)^2=0\)

So the complementary function is

\(x_c=\mathrm{e}^{6t}(At+B)\).

For a particular integral, try

\(x_p=p\sin t+q\cos t\).

Then

\(x_p'=p\cos t-q\sin t\),

\(x_p''=-p\sin t-q\cos t\).

Substitute into the differential equation:

\((-p\sin t-q\cos t)-12(p\cos t-q\sin t)+36(p\sin t+q\cos t)=37\sin t\).

Collect coefficients:

\((35p+12q)\sin t+(-12p+35q)\cos t=37\sin t\).

Hence

\(35p+12q=37\),

\(-12p+35q=0\).

From \(-12p+35q=0\), we get \(q=\frac{12}{35}p\).

Substitute into the first equation:

\(35p+12\left(\frac{12}{35}p\right)=37\)

\(35p+\frac{144}{35}p=37\)

\(\frac{1369}{35}p=37\)

\(p=\frac{35}{37}\).

Then

\(q=\frac{12}{35}\cdot\frac{35}{37}=\frac{12}{37}\).

So the general solution is

\(x=\mathrm{e}^{6t}(At+B)+\frac{35}{37}\sin t+\frac{12}{37}\cos t\).

Differentiate:

\(x'=A\mathrm{e}^{6t}+6\mathrm{e}^{6t}(At+B)+\frac{35}{37}\cos t-\frac{12}{37}\sin t\).

Use the initial conditions.

When \(t=0\), \(x=0\):

\(B+\frac{12}{37}=0\)

so \(B=-\frac{12}{37}\).

When \(t=0\), \(x'=0\):

\(A+6B+\frac{35}{37}=0\).

Substitute \(B=-\frac{12}{37}\):

\(A-\frac{72}{37}+\frac{35}{37}=0\)

\(A-1=0\)

so \(A=1\).

Therefore, the required solution is

\(x=\mathrm{e}^{6t}\left(t-\frac{12}{37}\right)+\frac{35}{37}\sin t+\frac{12}{37}\cos t\).