How would I solve these differential equations? $R : I \rightarrow (0,\infty), v: I \rightarrow (0,\infty)$
$q: I \rightarrow \mathbb{R}
$ where $I$ is a real interval.
$$R' = (\alpha + \frac{q}{\alpha})R\cos v \sin v $$
$$v' = \frac{q}{a}\cos^2 v - \alpha \sin^2 v$$
I have no idea, what so ever where to even begin. Even a push will do.
I need a solution $R$ and $v$, the idea is to get a solution $u = Rv$
 A: Hint
Considering the second equation $$v' = \frac{q}{a}\cos^2( v) - \alpha \sin^2( v)$$ define $v=\tan^{-1}(z)$ and the equation reduces to $$ z'+ \alpha  z^2=\frac q a$$ which seems to be easy to integrate.
A: The variable for derivation isn't defined. For example, say $t$ , so that $R'=\frac{dR}{dt}$ and $v'=\frac{dv}{dt}$
Second ODE:
$$\frac{dv}{dt}=\frac{q}{a}\cos^2(v)-\alpha\sin^2(v)$$
This is a separable ODE. The usual way to solve it is :
$$dt=\frac{dv}{\frac{q}{a}\cos^2(v)-\alpha\sin^2(v)}$$
$$t=\int \frac{dv}{\frac{q}{a}\cos^2(v)-\alpha\sin^2(v)}=
\sqrt{\frac{\alpha}{aq}}\tanh^{-1} \left(\sqrt{\frac{a\alpha}{q}}\tan(v) \right) +\text{constant}$$
The solution of the second ODE is :
$\quad v(t)=\tan\left(\sqrt{\frac{q}{a\alpha}}\tanh\left(\sqrt{\frac{aq}{\alpha}}(t+c_1) \right) \right)$
First ODE :
$$\frac{dR}{dt}=\left(\alpha+\frac{q}{\alpha}\right)R\cos(v)\sin(v) $$
$\cos(v)\sin(v)=\frac{\tan^{-1}(v) }{1+\left(\tan^{-1}(v) \right)^2}
=\frac{\sqrt{\frac{q}{a\alpha}}\tanh\left(\sqrt{\frac{aq}{\alpha}}(t+c_1) \right) }{1+\frac{q}{a\alpha}\tanh^2\left(\sqrt{\frac{aq}{\alpha}}(t+c_1) \right) }$
$$\frac{dR}{dt}=\left(\alpha+\frac{q}{\alpha}\right)R
\frac{\sqrt{\frac{q}{a\alpha}}\tanh\left(\sqrt{\frac{aq}{\alpha}}(t+c_1) \right) }{1+\frac{q}{a\alpha}\tanh^2\left(\sqrt{\frac{aq}{\alpha}}(t+c_1) \right) } $$
This is a separable ODE :
$$\int\frac{dR}{R}=\left(\alpha+\frac{q}{\alpha}\right)\sqrt{\frac{q}{a\alpha}}
\int\frac{  \tanh\left(\sqrt{\frac{aq}{\alpha}}(t+c_1) \right) }{1+\frac{q}{a\alpha}\tanh^2\left(\sqrt{\frac{aq}{\alpha}}(t+c_1) \right) }dt$$
$$\ln(R)=\frac{\alpha}{2}\frac{\alpha+\frac{q}{\alpha} }{\alpha+\frac{q}{a} }\sqrt{\frac{q}{a\alpha}}\ln\left( (a\alpha+q)\cosh\left(2\sqrt{\frac{aq}{\alpha}}(t+c_1) \right)+a\alpha-q \right)+\text{constant}$$
The solution of the first ODE is :
$$R=c_2 \left((a\alpha+q)\cosh\left(2\sqrt{\frac{aq}{\alpha}}(t+c_1) \right)+a\alpha-q \right)^{\frac{\alpha}{2}\frac{\alpha+\frac{q}{\alpha} }{\alpha+\frac{q}{a} }\sqrt{\frac{q}{a\alpha}}}$$
I take my hat off with the greatest respect to the one who dare check all this stuff.
