Question

write the ordinary differential equation (1+sin y)dx = (2ycosy-x(secy+tan y))dy

Answer 1

$(1+\sin y)x'+x\frac{1+\sin y}{\cos y}=2y\cos y\\ 1) (1+\sin y)x'+x\frac{1+\sin y}{\cos y}=0\\ \frac{dx}{dy}=-\frac{x}{\cos y}\\ \frac{dx}{x}=-\frac{dy}{\cos y}\\ \ln|x|=-\ln|\tan(\frac{y}{2}+\frac{\pi}{4})|+\ln|c|\\ x_1=\frac{c}{\tan(\frac{y}{2}+\frac{\pi}{4})}\\ 2) x=\frac{c(y)}{\tan(\frac{y}{2}+\frac{\pi}{4})}\\$

substitute x in equation

$(1+\sin y)\frac{c'(y)\tan(\frac{y}{2}+\frac{\pi}{4})-c(y)\frac{1}{2}\frac{1}{\cos^2(\frac{y}{2}+\frac{\pi}{4})}}{\tan^2(\frac{y}{2}+\frac{\pi}{4})}+\\ +\frac{c(y)}{\tan(\frac{y}{2}+\frac{\pi}{4})}\frac{1+\sin y}{\cos y}=2y\cos y\\ \frac{(1+\sin y)c'(y)}{\tan(\frac{y}{2}+\frac{\pi}{4})} - \frac{(1+\sin y)c(y)}{2\tan^2(\frac{y}{2}+\frac{\pi}{4})\cos^2(\frac{y}{2}+\frac{\pi}{4})} +\\ +\frac{(1+\sin y)c(y)}{\tan(\frac{y}{2}+\frac{\pi}{4})\cos y}=2y\cos y\\ \frac{(1+\sin y)c'(y)}{\tan(\frac{y}{2}+\frac{\pi}{4})} -\frac{(1+\sin y)c(y)}{2\sin^2(\frac{y}{2}+\frac{\pi}{4})}+\\ +\frac{(1+\sin y)c(y)}{\tan(\frac{y}{2}+\frac{\pi}{4})\cos y}=2y\cos y$

show that

$2\sin^2(\frac{y}{2}+\frac{\pi}{4})=2\tan(\frac{y}{2}+\frac{\pi}{4}) \cos y\\ 2\sin(\frac{y}{2}+\frac{\pi}{4})=\frac{\cos y}{\cos(\frac{y}{2}+\frac{\pi}{4})}\\ 2\sin(\frac{y}{2}+\frac{\pi}{4})\cos(\frac{y}{2}+\frac{\pi}{4})=\cos y\\ \sin2(\frac{y}{2}+\frac{\pi}{4})=\cos y\\ \sin(y+\frac{\pi}{2})=\cos y\\ \cos y=\cos y$

so

$\frac{(1+\sin y)c'(y)}{\tan(\frac{y}{2}+\frac{\pi}{4})} =2y\cos y\\ c'(y)=\frac{2y\cos y\tan(\frac{y}{2}+\frac{\pi}{4})}{1+\sin y} =\\ =\frac{2y\cos y\tan(\frac{y}{2}+\frac{\pi}{4})}{\cos^2\frac{y}{2}+2\cos\frac{y}{2}\sin\frac{y}{2}+ \sin^2\frac{y}{2}} =\\ =\frac{2y(\cos^2\frac{y}{2}-\sin^2\frac{y}{2})\tan(\frac{y}{2}+\frac{\pi}{4})}{(\cos\frac{y}{2}+\sin\frac{y}{2})^2} =\\ =\frac{2y(\cos\frac{y}{2}-\sin\frac{y}{2})\sin(\frac{y}{2}+\frac{\pi}{4})}{\cos(\frac{y}{2}+\frac{\pi}{4})(\cos\frac{y}{2}+\sin\frac{y}{2})} =\\ =\frac{2y(\cos\frac{y}{2}-\sin\frac{y}{2})(\sin\frac{y}{2}\cdot\frac{\sqrt{2}}{2}+\cos\frac{y}{2}\cdot\frac{\sqrt{2}}{2})}{(\cos\frac{y}{2}\cdot\frac{\sqrt{2}}{2}-\sin\frac{y}{2}\cdot\frac{\sqrt{2}}{2})(\cos\frac{y}{2}+\sin\frac{y}{2})} =2y\\ c'(y)=2y\\ c(y)=y^2+c_1$

So solution of equation is

$x=\frac{y^2+c_1}{\tan(\frac{y}{2}+\frac{\pi}{4})}$