= Solution
Termwise differentiation of the <matrix exponential> gives
$$
\begin{aligned}
\frac d{dt}e^{tA}
&=\sum_{m=1}^{\infty}\frac{m t^{m-1}A^m}{m!}\\
&=A\sum_{j=0}^{\infty}\frac{t^jA^j}{j!}
=Ae^{tA}.
\end{aligned}
$$
Thus $y(t)=e^{tA}y_0$ satisfies $y'=Ay$ and $y(0)=y_0$. If two solutions existed, multiplying their difference by $e^{-tA}$ would give a vector with zero derivative and zero initial value, proving uniqueness.
For the inhomogeneous equation, multiply by the integrating factor $e^{-tA}$:
$$
\frac d{dt}(e^{-tA}x)=e^{-tA}f(t).
$$
Integration and multiplication by $e^{tA}$ give the variation-of-constants formula
$$
\boxed{
x(t)=e^{tA}x_0
+\int_0^te^{(t-s)A}f(s)\,ds}.
$$
For the stated matrix,
$$
A^2=
\begin{pmatrix}
12&-4&-4\\
12&-4&-4\\
24&-8&-8
\end{pmatrix},
\qquad
A^3=0,
$$
so
$$
e^{tA}=I+tA+\frac{t^2}{2}A^2.
$$
Writing $f(t)=v\sin t$ with $v=(1,3,0)^T$, one finds
$$
Ax_0=0,\qquad Av=(8,8,16)^T,\qquad A^2v=0.
$$
Therefore
$$
x(t)=x_0+v(1-\cos t)+Av(t-\sin t),
$$
or explicitly
$$
\boxed{
x(t)=
\begin{pmatrix}
8t-8\sin t-\cos t+2\\
8t-8\sin t-3\cos t+4\\
16t-16\sin t+2
\end{pmatrix}}.
$$
Back to article page
