Determine affine linear f $\in C^1(R)$ such that the differential form is exact

Question

Determine affine linear function $f$ such that the differential form $\omega$ is exact, where $$ \omega (x,y) = (6x^2y+2y^3+2yf(x^2+y^2))dx + (3y^2-2xf(x^2+y^2))dy. $$ I started the exercise but at some point I can't continue. I calculated the partial derivatives and imposed them to be equal. $$ \begin{split} \frac{\partial}{\partial y}F_1 & = 6x^2+6y^2+2f(x^2+y^2)+4y^2f'(x^2+y^2) \\ & = -2f(x^2+y^2)-4x^2f'(x^2+y^2) = \frac{\partial}{\partial x}F_2 \end{split} $$ Doing the math I found $$ 6(x^2+y^2)+4f(x^2+y^2)+4(x^2+y^2)f'(x^2+y^2)= 0 $$ The problem is in solving this differential equation. I have placed $x^2+y^2 = s$ so $$ 6(s)+4f(s)+4(s)f'(s)= 0 $$ can someone help me?

Answer 1

Starting from your last line: $$6s+4f(s)+4sf'(s)= 0$$ $$f(s)+sf'(s)= -\dfrac 32s$$ $$(sf(s))'=-\dfrac 32 s$$ Integrate. $$sf(s)=-\dfrac 34 s^2+C$$ $$f(s)=-\dfrac 34 s+\dfrac Cs$$

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$f$ is required to be linear affine, I bet $C=0$ is the right choice. – @Daniele Tampieri