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