Applying a linear transformation to a system of differential equations I am reading the book:  Nonlinear Oscillations, Dynamical System and Bifurcations of Vector Fields (Guckenheimer and Holmes), chapter 2: An introduction to chaos.
About Van der Pol's equation, it can be expressed as the system
$(*) \ \ \begin{align} \dot{x} &= y-\alpha \Phi(x)\\ \dot{y} &= -x + \beta p(\theta) \\ \dot{\theta} &= 1  \end{align}$
Taking $\alpha\ll 1$, $\beta = 0$(unforced) and $\Phi(x)=\frac{x^3}{3}-x$, the system $(*)$ can be considered as a perturbation of: 
$$\dot{x}=y, \ \ \ \dot{y}=-x$$
Now, using the transformation: 
$$\begin{pmatrix}u\\v\end{pmatrix}
=\begin{pmatrix}\cos(t)& -\sin(t)\\ -\sin(t)& -\cos(t)\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} $$
The system $(*)$ becomes: 
$$\dot{u} = -\alpha \cos(t)\left[(u\cos(t)-v\sin(t))^3/3-(u\cos(t)-v\sin(t))\right]$$
$$\dot{v} = -\alpha \sin(t)\left[(u\cos(t)-v\sin(t))^3/3-(u\cos(t)-v\sin(t))\right]$$
The goal of applying this transformation is, in the next step to apply average method to the system. 
My problem is:
I already applied the transformation to the system: 
$$\begin{pmatrix}\dot{u}\\ \dot{v}\end{pmatrix}
=\begin{pmatrix}\cos(t)& -\sin(t)\\ -\sin(t)& -\cos(t)\end{pmatrix}\begin{pmatrix}\dot{x}\\ \dot{y}\end{pmatrix}, $$
and using the expressions for $\dot{x}, \dot{y}$,  I obtained:  
$$\dot{u} = -v -\alpha \cos(t)\left((u\cos(t)-v\sin(t))^3/3-(u\cos(t)-v\sin(t))\right)$$
$$\dot{v} = u -\alpha \sin(t)\left((u\cos(t)-v\sin(t))^3/3-(u\cos(t)-v\sin(t))\right)$$
What I am doing wrong? Are the authors just considering the non-linear part of the "new system"? Thanks.
 A: When you differentiate
$$\begin{pmatrix}u\\v\end{pmatrix}=\begin{pmatrix}\cos(t)& -\sin(t)\\ -\sin(t)& -\cos(t)\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$$
You must also differentiate the $\sin$ and $\cos$, for example
$$u=\cos(t) x - \sin(t) y$$ leads to
$$\dot{u}=\cos(t) \dot{x} -\sin(t) \dot{y} -\sin(t) x -\cos(t) y.$$

On a side note, are you sure of the transformation ?
The goal of such transform is to obtain $\dot{u}=\dot{v}=0$ in the case $\alpha=0$.
As in the case $\alpha=0$ the solution is of the form
\begin{equation*}
\begin{pmatrix}x \\ y \end{pmatrix} = \begin{pmatrix}\cos(t) &  \sin(t) \\- \sin(t)& \cos(t) \end{pmatrix} \begin{pmatrix}x_0 \\ y_0 \end{pmatrix}
\end{equation*}
so we obtain
\begin{equation*}
\begin{pmatrix}u \\ v \end{pmatrix} = \begin{pmatrix}\cos(t)²-\sin(t)^2 &  -2 \cos(t)\sin(t) \\-2 \cos(t)\sin(t)& \sin(t)^2-\cos(t)^2 \end{pmatrix} \begin{pmatrix}x_0 \\ y_0 \end{pmatrix}
\end{equation*}
which is not a constant.
As in the method of the variation of parameters the change of variables should be
\begin{equation*}
\begin{pmatrix}x \\ y \end{pmatrix} = \begin{pmatrix}\cos(t) &  \sin(t) \\- \sin(t)& \cos(t) \end{pmatrix} \begin{pmatrix}u \\ v \end{pmatrix}
\end{equation*}
i.e
\begin{equation*}
\begin{pmatrix}u \\ v \end{pmatrix} = \begin{pmatrix}\cos(t) &  \sin(t) \\- \sin(t)& \cos(t) \end{pmatrix}^{-1} \begin{pmatrix}x \\ y \end{pmatrix}=\begin{pmatrix}\cos(t) &  -\sin(t) \\ \sin(t)& \cos(t) \end{pmatrix} \begin{pmatrix}x \\ y \end{pmatrix}.
\end{equation*}
We obtain in this case
$$\dot{u}=-\sin(t) x +\cos(t) \dot{x} -\cos(t) y -\sin(t) \dot{y}$$
and using the equation
$$\dot{u}=-\sin(t) x +\cos(t) (y-\alpha \Phi(x)) -\cos(t) y -\sin(t) (-x)=0-\alpha \cos(t) \Phi(x)$$
