Solving $y'(x)\left(4-3y(x)x^2\right)=4x$ 
Solve the differential equation
  $$y'(x)\left(4-3y(x)x^2\right)=4x$$

I would appreciate some help with this problem.
 A: i will write ode as $(4-3x^2y)\frac{dy}{dx} = 4x$ as $$4x \frac{dx}{dy} +3x^2 y = 4 $$ substituting $u = x^2,$ gives a linear equation $$2\frac{du}{dy} + 3uy = 4$$ for $u$.  you can take it from here. use integrating factor or variation of parameters.
A: $$\frac{dy}{dx}\left(4-3y(x)x^2\right)=4x$$
$$4-3yx^2=4x\frac{dx}{dy}$$
You can use the sustitution $u=x^2$, so $\frac{du}{dy}=2x\frac{dx}{dy}$
In the equation:
$$ 4-3yu=2\frac{du}{dy} $$
Rewriting:
$$ 2\frac{du}{dy}+(3y)u=4 $$
$$ \frac{du}{dy}+\frac{3}{2}yu=2   $$
That is a first order linear ordinary differential equation and its integrating factor is $ z(y)=e^{\int\frac{3y}{2}dy}=e^{3\frac{y^2}{4}} $.
Multyply both sides by z(y):
$$ e^{\frac{3y^2}{4}}\frac{du}{dy}+\frac{1}{2}(3e^{\frac{3y^2}{4}}y)u=2e^{\frac{3y^2}{4}} $$
Substitute $\frac{3}{2}e^{\frac{3y^2}{4}}y=\frac{d}{dy}(e^{\frac{3y^2}{4}})$.
Apply the reverse product rule $g\frac{df}{dy}+f\frac{dg}{dy}=\frac{d(fg)}{dy}$ to the left-hand side:
$$\frac{d}{dy}(e^\frac{3y^2}{4}u)=2e^{\frac{3y^2}{4}}$$
Integrate both sides with respecto to $y$:
$$\int\frac{d}{dy}(e^\frac{3y^2}{4}u)dy=\int2e^\frac{3y^2}{4}dy$$
Evaluate the integrals:
$$e^\frac{3y^2}{4}u=2\sqrt\frac{π}{3}f*e*(\frac{\sqrt3y}{2})+c_1$$
Divide both sides by $z(y)=e^\frac{3y^2}{4}$
$$u=\frac{{2\sqrt\frac{π}{3}}f*e*(\frac{\sqrt3y}{2})+c_1}{{e^\frac{3y^2}{4}}}$$
So:
$$x^2=\frac{{2\sqrt\frac{π}{3}}f*e*(\frac{\sqrt3y}{2})+c_1}{{e^\frac{3y^2}{4}}}$$
Where
$f*e*$: error function
$c_1$ : arbitrary constant
A: The closest approximation to an explicit formula for $y(x)$ might be that $z(x)=\frac34y(x)^2$ solves $$x^2=\frac2{\sqrt3}\mathrm e^{-z(x)}\int_{z(0)}^{z(x)}\frac{\mathrm e^t}{\sqrt{t}}\,\mathrm dt.$$
