Q:

Solve a Exact quation, simplify

Show that $ (4bxy+3x+5)y'+3x^2 + 8ax+2by^2+3y=0$ is exact. Find F and find the solution.
Hello,
it is clear that $ \frac{\partial 4bxy+3x+5}{\partial y}= 4by+3= \frac{\partial 3x^2 + 8ax+2by^2+3y}{\partial x}$. So the equation is exact.
I also find F as $F(x,y)=x^3+4ax^2+2by^2 x+3yx+5y$ which satisfy $ D_y F= 4bxy+3x+5$ and $D_x F=3x^2 + 8ax+2b^2+3y$
y(x) is a solution of the equation if and only if y(x) satisfy $F(x,y(x))=c$
So i become $y_{1,2}= \frac{-3x+5 \pm \sqrt{-8bx^4+32bax^3+9x^2-(30+8bc)x+25}}{4bx}$
Can i simplifly the solution or say something about the validity of it?

A:

Observe that
$$ (4bxy+3x+5)y'+3x^2 + 8ax+2b^2+3y=0\iff$$
$$ (3x^2 + 8ax+2b^2+3y)dx+(4bxy+3x+5)dy=0$$
and the equation is exact iff
$$\frac{\partial}{\partial y}(3x^2 + 8ax+2b^2+3y)=\frac\partial{\partial x}(4bxy+3x+5)$$
Yet the left summand above is $\;3\;$, whereas the right one is $\;4by+3\;$, so the equation isn't exact or you miscopied the exercise.
Edit after the asker edited the question : The potential of the exact vector field is
$$F=\int(3x^2+8ax+2by^2+3y)dx=x^3+4ax^2+2by^2x+3yx+C(y)\implies$$
$$4bxy+3x+5=F'_y=4byx+3x+C'(y)\implies C'(y)=5\implies C(y)=5y+K\implies$$
the potential is $\;F=x^3+4ax^2+2by^2+3yx+5y+K\;,\;\;K=\,\text{a constant}\;$
and the solution to the dif. eq. is thus
$$x^3+4ax^2+2by^2+3yx+5y=C_1$$
This is a solution given in implicit form and I'm not sure why would you want to simplify it (in fact, to give explicitly $\;y\;$ as a function of $\;x\;$) or even if this is possible at all. When you try to apply the quadratic formula here you may be restricting the generality of the above, which is valid in the whole real line.

