Finding the solution to a non-homogeneous matrix exponential.

Consider

$$\frac{d}{du} \begin{bmatrix}a\\b \end{bmatrix} = \begin{bmatrix}-x& y\\ -y&-x\end{bmatrix} \begin{bmatrix}a\\ b\end{bmatrix} + \begin{bmatrix}\cos(zu)\\ -\sin(zu)\end{bmatrix}$$ where $P = \begin{bmatrix}-x& y\\ -y&-x\end{bmatrix}$. And I have it for $u=0:$ $$\begin{bmatrix}a\\b \end{bmatrix} = \begin{bmatrix}0\\0 \end{bmatrix}$$

Having proved that $\exp\left[Pu\right] = \exp\left[-xu\right]$

$$\begin{bmatrix}\cos(yu)& \sin(yu)\\ -\sin(yu) & \cos(yu)\end{bmatrix}.$$

My question is:

How do I use this proof to solve this problem using the initial conditions?

Having thought about it - I consider that using the double angle formulae may help although I haven't been able to formulate this.

Any help would be welcome.

So you solved the homogeneous system. To now get the solution of the inhomogeneous system you can use the variation of constant technique. Let

$$ \begin{bmatrix}a\\b \end{bmatrix}= e^{Pu}c(u) \longrightarrow \frac{d}{du}\begin{bmatrix}a\\b \end{bmatrix} = Pe^{Pu}c(u) + e^{Pu}c'(u)$$

Thus we have $e^{Pu}c'(u) = \begin{bmatrix}\cos(zu)\\ -\sin(zu)\end{bmatrix}$ and $c(0) = 0$. All that is left to do is to solve this linear system and integrate w.r.t. $u$:

$$c'(u) = e^{-Pu}\begin{bmatrix}\cos(zu)\\ -\sin(zu)\end{bmatrix} = e^{xu}\begin{bmatrix}\cos(yu)& \sin(yu)\\ -\sin(yu) & \cos(yu)\end{bmatrix}^{-1} \cdot\begin{bmatrix}\cos(zu)\\ -\sin(zu)\end{bmatrix} $$

Since the matrix here is orthonormal its inverse is simply its transpose:

$$c'(u) = e^{xu}\begin{bmatrix}\cos(yu)& -\sin(yu)\\ \sin(yu) & \cos(yu)\end{bmatrix} \cdot\begin{bmatrix}\cos(zu)\\ -\sin(zu)\end{bmatrix} = e^{xu}\begin{bmatrix}\cos(yu-zu) \\\sin(yu-zu)\end{bmatrix} $$

Were I used the addition theorems to simplify the expression. Now integrate, using the formulas

\begin{align} \int \sin(\alpha u) e^{\beta u} du &= \frac{\beta\sin(\alpha u) - \alpha\cos(\alpha u)}{\alpha^2+\beta^2}e^{\beta u} +\text{const.}\\ \int \cos(\alpha u) e^{\beta u} du &= \frac{\beta\cos(\alpha u) + \alpha\sin(\alpha u)}{\alpha^2+\beta^2}e^{\beta u} +\text{const.}\\ \end{align}

to get

$$ c(u) = \frac{1}{x^2 + (y-z)^2}e^{xu} \begin{bmatrix} x\cos((y-z)u) + (y-z)\sin((y-z) u)\\ x\sin((y-z) u) - (y-z)\cos((y-z) u) \end{bmatrix} + \text{const.} $$

And to satify the initial condition $c(0)=0$ we need

$$ \text{const.} = -\frac{1}{x^2 + (y-z)^2}\begin{bmatrix} x \\- (y-z) \end{bmatrix}$$

I will use standard notation for differential equations. Your notation is quite unusual and it might lead to unwanted mixing up of formulas.

$$\frac{d}{dt} \begin{bmatrix}x_1\\x_2 \end{bmatrix} = \begin{bmatrix}-a& b\\ -b&-a\end{bmatrix} \begin{bmatrix}x_1\\ x_2\end{bmatrix} + \begin{bmatrix}\cos(\omega t)\\ -\sin(\omega t)\end{bmatrix}$$

In order to determine the solution to the problem, we will compare this to the standard linear control problem. Given a system

$$\dot{\boldsymbol{x}}(t)=\boldsymbol{Ax}(t)+\boldsymbol{Bu}(t)$$

with the initial condition $\boldsymbol{x}(t=t_0)=\boldsymbol{x}_0$.

The solution to this equation is given by

$$\boldsymbol{x}(t)=\exp\left[\left(t-t_0 \right)\boldsymbol{A} \right]\boldsymbol{x}_0+\int_{t=t_0}^{t}\exp\left[(t-\tau)\boldsymbol{A}\right]\boldsymbol{B}\boldsymbol{u}(\tau)d\tau.$$

This formula can be verified by differentiating it and invoking the fundamental theorem of calculus for the integral.

In your case

$$\boldsymbol{A}=\begin{bmatrix}-a& b\\ -b&-a\end{bmatrix},$$ $$\boldsymbol{B}=\begin{bmatrix}1& 0\\ 0 & 1\end{bmatrix},$$ $$\boldsymbol{u}=\begin{bmatrix}\cos\left(\omega t\right)\\-\sin\left(\omega t\right)\end{bmatrix}, \text{ and }$$ $$\boldsymbol{x}_0=\begin{bmatrix}0\\ 0 \end{bmatrix} .$$