Answer to Question #232409 in Differential Equations for john

Question

dr = b(cos ϕdr + rsin ϕdϕ) note: b is a constant 

Separable DE

Answer 1

$dr\ \ =\ \ b\left(\mathrm{cos}\left(\phi \right)dr\ \ \ +\ \ r\mathrm{sin}\left(\phi \right)d\phi \right) \\ \\ \mathrm{1}\ \ =\ \ b\left(\mathrm{cos}\left(\phi \right)\ \ \ +\ \ r\mathrm{sin}\left(\phi \right)\frac{d\phi }{dr}\right) \\ \\ \frac{\left(\mathrm{1}-b\mathrm{cos}\left(\phi \right)\right)}{b\ \mathrm{sin}\left(\phi \right)}\ \ =\ \ \left(\ r\frac{d\phi }{dr}\right) \\ \\ b\frac{\mathrm{sin}\left(\phi \right)\ \ d\phi }{\left(\mathrm{1}-b\mathrm{cos}\left(\phi \right)\right)}\ \ =\ \ \left(\ \frac{dr}{r}\right) \\ \\ \int{\frac{d\left(\mathrm{1}-b\mathrm{cos}\left(\phi \right)\ \right)\ }{\left(\mathrm{1}-b\mathrm{cos}\left(\phi \right)\right)}}\ \ =\ \ \int{d\left(\mathrm{ln}\left(r\right)\right)} \\ \\ \mathrm{ln}\left(\mathrm{1}-b\mathrm{cos}\left(\phi \right)\ \right)\ \ \ =\ \ \mathrm{ln}\left(r\right)+\mathrm{ln}\left(C\right) \\ \\ \mathrm{ln}\left(\mathrm{1}-b\mathrm{cos}\left(\phi \right)\ \right)\ \ \ =\ \ \mathrm{ln}\left(Cr\right) \\ \\ \mathrm{1}-b\mathrm{cos}\left(\phi \right)\ =Cr\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(where\ \ C\ \ is\ \ an\ \ abitrarily\ \ \ cons\mathrm{tan}t\right)$