这是一份UCL伦敦大学 MATH003509作业代写的成功案例
show that
$$
\dot{\mathbf{x}}=\mathbf{A}(t) \mathbf{x}+\mathbf{B}(t)
$$
has solution
$$
x(t)=\varphi\left(t, t_{0}\right)\left(x_{0}+\int_{C_{0}}^{t} \varphi\left(t_{0}, t\right) B(\tau) d \tau\right)
$$
when $\mathbf{x}=\mathrm{x}{0}$ at $t=t{0}$.
Find the solution of
$$
\dot{x}=\left(\begin{array}{rr}
0 & -1 \
1 & 0
\end{array}\right) x+\left(\begin{array}{l}
\sin t \
\cos t
\end{array}\right)
$$
when $\mathbf{x}(0)=\mathbf{x}_{0}$.
MATH003509 COURSE NOTES ：
$$
\mathbf{f}(\pi(\mathbf{x}))=\pi(\mathrm{f}(\mathbf{x})) \text {, }
$$
for each $x \in \mathbb{R}^{n}$, where $\pi: \mathbb{R}^{n} \rightarrow T^{n}$ is given by
$$
\begin{aligned}
\pi(\mathbf{x})=\pi\left(\left(x_{1}, \ldots, x_{n}\right)^{\mathrm{T}}\right) &=\left(x_{1} \bmod 1, \ldots, x_{n} \bmod 1\right)^{\mathrm{T}} \
&=\left(\theta_{1}, \ldots, \theta_{n}\right)^{\mathrm{T}}=\theta
\end{aligned}
$$
(see Figure 3.10). If $k \in \mathbb{Z}^{\mathrm{n}}$, then (3.4.1) implies
$$
\pi(\bar{f}(\mathbf{x}+\mathbf{k}))=\mathbf{f}(\pi(\mathbf{x}+\mathbf{k}))=\mathbf{f}(\pi(\mathbf{x}))=\pi(\bar{f}(\mathbf{x})),
$$
for all $\mathbf{x} \in \mathbb{R}^{n}$. Continuity of $\overline{\mathbf{f}}$ then gives,
$$
\overline{\mathbf{f}}(\mathbf{x}+\mathbf{k})=\overline{\mathbf{f}}(\mathbf{x})+\mathbf{1}(\mathbf{k})
$$