When solving a physics problem, I had the following system of differential equations ($\rho$ and $\varphi$ are functions of $t$, $c$ and $v$ are constants)
$$ \begin{cases} \rho\,\dot{\varphi} = c \cdot \cos \varphi \\ \dot{\rho} = c \cdot \sin\varphi - v \end{cases}$$
I integrated the second equation wrt $t$
$$\rho = \int(c \cdot \sin\varphi - v) dt = - \frac{c \cos \varphi} {\dot\varphi} - vt + \gamma$$
I wanted to find out the constant $\gamma$, so I substituted $c \cos \varphi = \rho \dot\varphi$ and got
$$\rho = -\rho - vt + \gamma$$
But this formula contradicts with the physical interpretation of the problem, since $\rho$ must depend on $c$ outside the constant.
What have I done wrong?
I do not know the physics of this DE, but if we will try to search for the solution in the form $$ \rho(t)=\eta(\phi(t)). $$ then we get $\dot\rho=\eta'(\phi)\dot\phi$ and dividing the second equation by the first one gives $$ \frac{\eta'}{\eta}=(\ln\eta)'=\frac{c\sin\phi-v}{c\cos\phi}=\tan\phi-\frac{v}{c}\frac{1}{\cos\phi} $$ which is possible to integrate $$ \ln\eta(\phi)=-\ln|\cos\phi|-\frac{v}{c}\ln\Big|\frac{\sin\phi+1}{\cos\phi}\Big|+\gamma. $$ Does it make sense from the physical interpretation?
P.S. Quantity $\frac{v}{c}$ makes me think that it is kind of relativity equation.