While I agree with Chu that you can save yourself some pain by using phasors/Fourier transform, it is possible to solve these equations.
$$\begin{align}
&v_x)\quad &\frac{v_s-v_x}{1} + \frac{v_1-v_x}{2} + \frac{1}{3}\frac{d(v-v_x)}{dt} &= 0 \\
&v_1)\quad &\frac{v_x - v_1}{2} - i_1 &= 0 \\
&v)\quad &\frac{1}{3}\frac{d(v_x-v)}{dt} + i_2 - \frac{v}{5} &= 0\\
&tf)\quad &v &= 4v_1\\
& & i_1 &= 4i_2
\end{align}$$
You can eliminate \$\frac{1}{3}\frac{d(v-v_x)}{dt}\$ using the third equation by substituting it in the first equation.
$$\frac{1}{3}\frac{d(v-v_x)}{dt} = i_2 - \frac{v}{5}$$
You can then solve all equations but the third to all unknowns except \$v\$ to find that (I used a CAS like Maxima to solve it)
$$\begin{align}
i_1 &= \frac{20v_s-9v}{55} \\
i_2 &= \frac{20v_s-9v}{220} \\
v_1 &= \frac{v}{4} \\
v_x &= \frac{160v_s - 17v}{220}
\end{align}$$
which allows us to plug \$i_2\$ and \$v_x\$ back in:
$$\begin{align}
\frac{1}{3}\frac{d(v-v_x)}{dt} &= i_2 - \frac{v}{5} \\
&\Downarrow \\
79\frac{dv}{dt} + 53v &= 80\cos(3t) - 640\sin(3t)
\end{align}$$
The homogeneous solution is not important, as we're only interested in the steady-state solution. The particular solution is given by
$$\begin{align}
v(t) &= \frac{77960}{29489}\cos(3t) -\frac{7480}{29489}\sin(3t) \\
&= 2.656\cos(3t + 5.48^\circ)
\end{align}$$
Now, I'm not sure how you would get the reference answer of your book. When I verified using LTSpice, it seemed to support this answer rather than the one from your book.