Converting an ODE in polar form

Convert the ODE system
\dot{x}=\begin{pmatrix}a(t) & b(t)\\c(t) & d(t)\end{pmatrix}x
into polar form. You should get two equations
\frac{d}{dt}\Phi(t)=…\\ \frac{d}{dt}\ln r(t)=….

I set
x_1:=r(t)\cos\Phi(t)\\ x_2:=r(t)\sin\Phi(t)
and got
\frac{d}{dt}\Phi(t)=b(t)+\frac{\frac{d}{dt}r(t)\cos\Phi(t)}{r(t)\sin\Phi(t)}-\frac{a(t)\cos\Phi(t)}{\sin\Phi(t)}\\ \frac{d}{dt}\ln r(t)=d(t)+\frac{c(t)\cos\Phi(t)}{\sin\Phi(t)}-\frac{\cos\Phi(t)\frac{d}{dt}\Phi(t)}{\sin\Phi(t)}

Would like to know if this is right.

With greetings

Solutions Collecting From Web of "Converting an ODE in polar form"

I have the following:

$$x_1:=r(t) \cos \Phi(t),$$
$$x_2:=r(t) \sin \Phi(t)$$

and the derivatives are:

$$\dot{x_1}=\dot r \cos \Phi – r \sin(\Phi)\, \dot \Phi,$$
$$\dot{x_2}=\dot r \sin \Phi + r \cos(\Phi)\, \dot \Phi.$$

with $\dot x \equiv \dfrac{d x}{dt}.$

So you get:

$$\dot r \cos \Phi – r \sin(\Phi)\, \dot \Phi =a r\cos \Phi+br\sin\Phi,$$
$$\dot r \sin \Phi + r \cos(\Phi)\, \dot \Phi =c r\cos \Phi+r d\sin\Phi.$$

The equation for $\dot\Phi$ is:

$$\dot\Phi =\dfrac{(a r\cos \Phi+br\sin\Phi)\sin\Phi-(c r\cos \Phi+rd\sin\Phi)\cos\Phi}{-r},$$

and for $\dot r$ is :

$$\dot r = (a r\cos \Phi+br\sin\Phi)\cos\Phi – (c r\cos \Phi+rd\sin\Phi)\sin\Phi, $$


$$\dfrac{\dot r}{r(t)} = \dfrac{d}{dt}\ln[r(t)],$$


$$\dfrac{d}{dt}\ln[r(t)]=(a \cos \Phi+b\sin\Phi)\cos\Phi – (c\cos \Phi+d\sin\Phi)\sin\Phi. $$