Solve $3x(1-x^2)y^2\frac{dy}{dx}+(2x^2-1)y^3=ax^3$ I am solving this linear Differential equation which can be easily solve by using the formulas for the Bernoulli's Equations
I have solved till 
$$\frac{dy}{dx}+\frac{(2x^2-1)y^3}{3x(1-x^2)y^2}=\frac{ax^3}{3x(1-x^2)y^2}$$
$$y^2\frac{dy}{dx}+\frac{(2x^2-1)y^3}{3x(1-x^2)}=\frac{ax^3}{3x(1-x^2)}$$
Substituting $y^3=t$
so the equation will be
$$\frac{1}{3}\frac{dt}{dx}+\frac{(2x^2-1)t}{3x(1-x^2)}=\frac{ax^3}{3x(1-x^2)}$$
after this the integrating factor is
$$\frac{1}{x\sqrt{1-x^2}}$$
But I am unable to solve it forward.
 A: After substitution $t(x) = y^3(x)$
$$x(1-x^2) t'(x) + (2x^2-1) t(x) = ax^3.$$
Let's find a particular solution in form $t_0(x) = bx:$
$$bx - bx^3 + 2bx^3 - bx = ax^3 \Rightarrow b = a$$
So the full solution has form $t(x) = ax + z(x),$
$$x(1-x^2) z'(x) + (2x^2-1) z(x) = 0 \Rightarrow$$
$$\log z = - \int \frac{2x^2 -1}{x(1-x^2)} dx = 
\int \frac{1 - x^2}{x(1-x^2)} dx - \int \frac {x^2}{x(1-x^2)} dx =
\log x + \frac 1 2 \log\left(1-x^2\right) +C \Rightarrow
$$
$$t(x) = C x \sqrt{1 -x^2} + ax \Rightarrow y(x) = \left(C x \sqrt{1-x^2} + ax\right)^{\frac 1 3}$$
A: Notice, multiply the integration factor both the sides of the D.E. as follows $$\frac 13\frac{dt}{dx}\frac{1}{x\sqrt{1-x^2}}+\frac{(2x^2-1)t}{3x(1-x^2)}\frac{1}{x\sqrt{1-x^2}}=\frac{ax^3}{3x(1-x^2)}\frac{1}{x\sqrt{1-x^2}}$$
$$\frac{d}{dt}\left(\frac{t}{x\sqrt{1-x^2}}\right)=\frac{ax}{(1-x^2)^{3/2}}$$
$$\int d \left(\frac{t}{x\sqrt{1-x^2}}\right)=\int \frac{ax}{(1-x^2)^{3/2}}\ dx $$
$$\frac{t}{x\sqrt{1-x^2}}=-\frac a2\int \frac{d(1-x^2)}{(1-x^2)^{3/2}}$$
$$\frac{t}{x\sqrt{1-x^2}}=-\frac a2 \frac{(1-x^2)^{-1/2}}{-1/2}+C$$
$$\frac{y^3}{x\sqrt{1-x^2}}= \frac{a}{\sqrt{1-x^2}}+C$$
$$y=(ax+Cx\sqrt{1-x^2})^{1/3}$$
