如何利用已证矩阵指数求解非齐次矩阵微分方程（含初始条件）

Hey there! Let's work through this step by step since you already have the matrix exponential $e^{Pu}$ handy—half the battle is already won here. 😊

For a linear nonhomogeneous system of the form:
$$\frac{d}{du}\mathbf{z}(u) = P\mathbf{z}(u) + \mathbf{f}(u)$$
with initial condition $\mathbf{z}(0) = \mathbf{z}_0$, the exact solution is given by the constant variation formula:
$$
\mathbf{z}(u) = e^{Pu}\mathbf{z}_0 + \int_0^u e^{P(u-s)}\mathbf{f}(s)ds
$$
Since your initial condition is $\mathbf{z}(0) = \begin{bmatrix}0\0\end{bmatrix}$, the first term $e^{Pu}\mathbf{z}_0$ vanishes entirely. We only need to compute the integral term.

First, rewrite $e^{P(u-s)}$ using the provided $e^{Pt}$ form. Replace $t$ with $(u-s)$:
$$
e^{P(u-s)} = e^{-x(u-s)} \begin{bmatrix}\cos(y(u-s)) & \sin(y(u-s))\ -\sin(y(u-s)) & \cos(y(u-s))\end{bmatrix}
$$
Your forcing term is $\mathbf{f}(s) = \begin{bmatrix}\cos(zs)\ -\sin(zs)\end{bmatrix}$. Compute the matrix-vector product $e^{P(u-s)}\mathbf{f}(s)$:

First component (for $a(u)$): #

$$
e^{-x(u-s)} \left[ \cos(y(u-s))\cos(zs) - \sin(y(u-s))\sin(zs) \right]
$$
Use the cosine addition identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$ (here $A = y(u-s)$, $B = zs$) to simplify this to:
$$
e^{-x(u-s)} \cos\left( yu + s(z - y) \right)
$$

Second component (for $b(u)$): #

$$
e^{-x(u-s)} \left[ -\sin(y(u-s))\cos(zs) - \cos(y(u-s))\sin(zs) \right]
$$
Use the sine addition identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$, then factor out a negative sign:
$$
-e^{-x(u-s)} \sin\left( yu + s(z - y) \right)
$$

Now we compute the definite integrals for each component. Let's start with $a(u)$:

Calculating $a(u)$: #

$$
a(u) = \int_0^u e^{-x(u-s)} \cos\left( yu + s(z - y) \right) ds
$$
Make a substitution $t = u - s$ (so $ds = -dt$, bounds swap from $t=u$ to $t=0$):
$$
a(u) = e^{-xu} \int_0^u e^{xt} \cos\left( zu - t(z - y) \right) dt
$$
Expand the cosine term using $\cos(C-D) = \cos C \cos D + \sin C \sin D$:
$$
a(u) = e^{-xu} \left[ \cos(zu) \int_0^u e^{xt} \cos\left( t(z-y) \right) dt + \sin(zu) \int_0^u e^{xt} \sin\left( t(z-y) \right) dt \right]
$$
Use standard exponential-trigonometric integral formulas:

• $\int e^{kt} \cos(mt) dt = \frac{e^{kt}}{k2+m^2}\left(k\cos(mt)+m\sin(mt)\right) + C$
• $\int e^{kt} \sin(mt) dt = \frac{e^{kt}}{k2+m^2}\left(k\sin(mt)-m\cos(mt)\right) + C$

Substitute $k=x$, $m=z-y$, evaluate from $0$ to $u$, then simplify the expression.

Calculating $b(u)$: #

$$
b(u) = -\int_0^u e^{-x(u-s)} \sin\left( yu + s(z - y) \right) ds
$$
Follow the same substitution $t=u-s$, expand using $\sin(C-D) = \sin C \cos D - \cos C \sin D$, and apply the same integral formulas to get a closed-form solution.

After evaluating the integrals, you can distribute the $e^{-xu}$ term and combine like terms to get a clean, closed-form expression for both $a(u)$ and $b(u)$. For example, the simplified $a(u)$ might look like:
$$
a(u) = \frac{1}{x^2+(z-y)2}\left[ x\cos(zu)+(z-y)\sin(zu) - e^{-xu}\left(x\cos(yu)+(z-y)\sin(yu)\right) \right]
$$

内容的提问来源于stack exchange，提问作者Hideki Ryuga