Answer to Question #144878 in Differential Equations for Divine Grace Jaravata

Question

y’ = Csc x -yCot x

Answer 1

$y' = csc x - ycotx\\ y' = \dfrac{1}{sinx} - y\dfrac{cosx}{sinx}\\ y = u(x)*v(x)\\ y' = u'v+uv'\\ u'v+uv'+uv\dfrac{cosx}{sinx} = \dfrac{1}{sinx}\\ \space\\ u'v +u(v'+v\dfrac{cosx}{sinx}) =\dfrac{1}{sinx}\\ \space\\ v' +v\dfrac{cosx}{sinx}=0\\ \space\\ \dfrac{dv}{dx} =-v\dfrac{cosx}{sinx}\\ \space\\ \dfrac{dv}{v} = -cotxdx\\ \space\\ \int \dfrac{dv}{v} =- \int cotxdx\\ ln v = -ln(sinx)\\ v = \dfrac{1}{sinx} = cscx\\ u' * \dfrac{1}{sinx} = \dfrac{1}{sinx}\\ \dfrac{du}{dx} = 1\\ \space\\ du = dx\\ \space\\ \int du =\int dx\\ \space\\ u = x +C, \space C-constant\\ y = u*v =csc*(x+C) = Ccscx+xcscx\\ answer: y = Ccscx+xcscx$