Solve the given differential equation

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Homework Statement
##(2xy+\cos y+\dfrac{2y}{x}-\sin x)dx +(x^2-y\sin y+1)dy=0##
Relevant Equations
hybrid ode
Ok, i just came up with this ode. How does one go about to solve it?

My attempt

Let ##M(x,y) =(2xy+\cos y+\dfrac{2y}{x}-\sin x)## and ##N(x,y) = (x^2-y\sin y+1)##

##\dfrac{\partial M}{\partial y}= 2x - \sin y + \dfrac{2}{x}## and ##\dfrac{\partial N}{\partial x}= 2x##

Not exact.
 
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Say we have a point (x,y)=(x_0,y_0) on the solution, (x_0+\Delta x, y_0+\Delta y)
is also on the solution, where
\frac{\Delta y}{\Delta x}=\frac{dy}{dx}|_{x=x_0,y=y_0}=-\frac{2xy+\cos y+\frac{2y}{x}-\sin x}{x^2-y\sin y+1}|_{x=x_0,y=y_0}
In this way we can find successive solution points numerically. I have no idea how to get analytic solution if there is.
 
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chwala said:
Ok, i just came up with this ode.
There's no guarantee that an ODE devised more-or-less at random will have an analytical solution.
chwala said:
How does one go about to solve it?
Your approach of checking for exactness is a good start, but as you found, the equation isn't exact. Many textbooks on differential equations take this approach farther by making additional assumptions about the possible solutions.
@anuttarasammyak's suggestion of going for a numerical solution might be the only way forward here.
 
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Thanks for the thoughtful input — actually, I’ve taken this as a personal challenge! I’ll be applying the Runge-Kutta method to this problem and working through it manually as far as possible. I’m also committed to generating more of these hybrid-style problems that push the boundaries of how we see mathematics, rather than sticking to traditional textbook paths.

Your suggestion of checking for exactness was solid, and even though it didn’t yield a direct solution here, it shows the right instinct. I completely agree with @anuttarasammyak — numerical methods may indeed be our best route for a meaningful solution. Let’s stretch the limits a little!
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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