Answer to Question #220727 in Differential Equations for Jowes

Question

dr/dΩ+tanΩ=cos2Ω

Answer 1

Solution;

$\frac{dr}{d\Omega}$ =cos2$\Omega$ -tan$\Omega$

dr=(cos2$\Omega$ -tan$\Omega$ )d$\Omega$

Integrate both sides;

$\int{dr}=\int{(cos2\Omega-tan\Omega)}d\Omega$

$\int{dr}=\int{cos2\Omega}d\Omega-\int{tan\Omega}d\Omega$

$\int dr=\int{cos2\Omega}d\Omega-\int{\frac{sin\Omega}{cos\Omega}}d\Omega$

But for the integration of;

$\int{\frac{sin\Omega}{cos\Omega}}$

take u=cos$\Omega$ ;$\frac{du}{d\Omega}=-sin\Omega$

$d\Omega=\frac{-1}{sin\Omega}du$

Replace back

$\int{\frac{-1}{u}}du=-ln(u)=-ln(cos\Omega)$ =ln$(sec\Omega)$

Hence,

$r=\frac{sin2\Omega}{2}-ln(|sec\Omega|)+C$