Answer: The particular solution is

\(x = 25\sin t - 5\cos t + 5e^{-t/3}\).

We solve the differential equation

\(9\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}+6\dfrac{\mathrm{d}x}{\mathrm{d}t}+x=50\sin t\)

with the conditions \(x(0)=0\) and \(x'(0)=0\).

First divide through by 9:

\(x''+\frac{2}{3}x'+\frac{1}{9}x=\frac{50}{9}\sin t.\)

We look for a particular solution of the form

\(x_p=A\sin t+B\cos t.\)

Then

\(x_p'=A\cos t-B\sin t,\)

\(x_p''=-A\sin t-B\cos t.\)

Substituting into the differential equation gives

\(9(-A\sin t-B\cos t)+6(A\cos t-B\sin t)+(A\sin t+B\cos t)=50\sin t.\)

Collecting coefficients of \(\sin t\) and \(\cos t\):

\(( -8A-6B )\sin t + ( 6A-8B )\cos t = 50\sin t.\)

So

\(-8A-6B=50,\qquad 6A-8B=0.\)

From \(6A-8B=0\), \(3A=4B\), so \(A=\frac{4}{3}B\).

Substitute into the first equation:

\(-8\left(\frac{4}{3}B\right)-6B=50\)

\(\Rightarrow -\frac{32}{3}B-\frac{18}{3}B=50\)

\(\Rightarrow -\frac{50}{3}B=50\)

\(\Rightarrow B=-3\), hence \(A=-4\).

So a particular solution is \(x_p=-4\sin t-3\cos t\).

The complementary function satisfies

\(9r^2+6r+1=0\),

which gives \((3r+1)^2=0\), so \(r=-\frac13\) is a repeated root. Therefore

\(x_c=(C_1+C_2t)e^{-t/3}.\)

Hence the general solution is

\(x=(C_1+C_2t)e^{-t/3}-4\sin t-3\cos t.\)

Now use the conditions.

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

\(C_1-3=0\), so \(C_1=3\).

Differentiate:

\(x' = \left(C_2-\frac13(C_1+C_2t)\right)e^{-t/3}-4\cos t+3\sin t.\)

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

\(C_2-\frac13C_1-4=0.\)

Since \(C_1=3\), this gives \(C_2-1-4=0\), so \(C_2=5\).

Therefore the required solution is

\(x=(3+5t)e^{-t/3}-4\sin t-3\cos t.\)