Questions about this specific non linear problem

Question

I have some questions concerning the following first order ODE $$y=y' \sin(y')+\cos(y')$$ 1- what is the degree of this equation ?

2- I solved it for y by differentiating wrt x . I got $$p(1-\cos(p)\frac{dp}{dx})=0$$ the general solution is determined by 2 equations since we can not eliminate p : $$x=\sin(p)+c$$ $$y=p\ \sin(p)+\cos(p)$$ where the last equation forming the solution is the same as the DE . Is this right ? and how do we say that the solution to the DE is the DE itself ?

3- Concerning the singular solution , I plugged p=0 in the DE , I got: $$y=1$$ But when trying to get the singular solution from the envelope by differentiating the general solution wrt c and equating by zero :

differentiate wrt c the equation $$x=\sin(p)+c$$ we got $$1=0$$ So does the singular solution exist or not ?

Answer 1

1. There is no degree as the equation is not of a polynomial type.

2. The solution of a DE is any means that allows you to compute the independent vaiable in terms of the dependent one. In this case, this is indirect as you have an additional independent variable, hence the solution is described by parametric equations. But whatever you call/write the terms of the equations, the identity holds.

Note that you can absolutely eliminate $p$ by $p=\arcsin(x-c)$. This yields

$$y=\arcsin(x-c)\,(x-c)\pm\sqrt{1-(x-c)^2}.$$

3. $y=1$ is indeed a solution, it does verify the DE.