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$$\left(x+\sqrt{y^2+1}\right)dx-\left(y-\frac{xy}{\sqrt{y^2+1}}\right)dy=0$$

This first-order equation is obviously not separable nor linear and not homogenous, and I'm not sure if this equation is exact too, is it exact differential equation? and if it's what's the solution of this with its steps? and if it's not exact differential equation then what is it? explain please, Thanks!

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let $$P(x,y)=x+\sqrt{y^2+1},Q(x,y)=-y+\frac{xy}{\sqrt{y^2+1}}$$ then we get $$\frac{\partial P(x,y)}{\partial y}=\frac{y}{y^2+1}=\frac{\partial Q(x,y)}{\partial x}$$ So we get $$f(x,y)=\int x+\sqrt{y^2+1}dx=\frac{x^2}{2}+x\sqrt{y^2+1}+g(y)$$ differentiating with respect to $y$ $$\frac{\partial f(x,y)}{\partial y}=\frac{xy}{\sqrt{y^2+1}}+\frac{d g(y)}{dy}$$ and we get $$\frac{xy}{\sqrt{y^2+1}}+g'(y)=-y+\frac {xy}{\sqrt{y^2+1}}$$ so $$g(y)=-\frac{y^2}{2}$$ Can you finish?

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