Question

dr/dΩ+rTanΩ=cos2Ω

Answer 1

Let us solve the differential equation $\frac{dr}{dΩ}+r\tanΩ=\cos2Ω.\\$

Firstly, let us divide both parts by $\cos\Omega.$ Then we have the differential equation

$\frac{1}{\cos\Omega}\frac{dr}{dΩ}+r\frac{1}{\cos\Omega}\tanΩ=\frac{1}{\cos\Omega}\cos2Ω,$

which is equivalent to the equation

$\frac{1}{\cos\Omega}\frac{dr}{dΩ}+r\frac{\sin\Omega}{\cos^2\Omega}=\frac{1}{\cos\Omega}(\cos^2Ω-\sin^2Ω),$

and hence $(r\frac{1}{\cos\Omega})'=\cosΩ-\frac{\sin^2Ω}{\cos\Omega}.$

It follows that

$r\frac{1}{\cos\Omega}=\int(\cosΩ-\frac{\sin^2Ω}{\cos\Omega})d\Omega$

$=\sin\Omega-\int\frac{\sin^2Ω}{\cos^2\Omega}d(\sin\Omega)$

$= \sin\Omega+\int\frac{\sin^2Ω}{\sin^2\Omega-1}d(\sin\Omega)$

$= \sin\Omega+\int\frac{\sin^2Ω-1+1}{\sin^2\Omega-1}d(\sin\Omega)$

$= \sin\Omega+\sin\Omega+\int\frac{1}{\sin^2\Omega-1}d(\sin\Omega)$

$= 2\sin\Omega+\frac{1}{2}\int(\frac{1}{\sin\Omega-1}-\frac{1}{\sin\Omega+1})d(\sin\Omega)$

$= 2\sin\Omega+\frac{1}{2}(\ln|\sin\Omega-1|-\ln|\sin\Omega+1|)+C$

$= 2\sin\Omega+\frac{1}{2}\ln|\frac{\sin\Omega-1}{\sin\Omega+1}|+C.$

Therefore, the general solution is

$r= 2\sin\Omega\cos\Omega+\frac{1}{2}\cos\Omega\ln|\frac{\sin\Omega-1}{\sin\Omega+1}|+C\cos\Omega$

or

$r= \sin2\Omega+\frac{1}{2}\cos\Omega\ln|\frac{\sin\Omega-1}{\sin\Omega+1}|+C\cos\Omega.$