Finding the particular solution of a differential equation using at least three different methods. Find the particular solution $(x^2+6y^2)dx-4xydy=0$; when $x=1$, $y=1$ using at least three different methods.
I have done the first two. Can somebody help me with the third method.
Method 1: Homogenous Equation
Let $y=vx; dy=vdx+xdv$
$(x^2+6x^2v^2)dx-4x(vx)(vdx+xdv)=0$
$(x^2+2x^2v^2)dx-4x^3vdv=0$
$(1+2v^2)dx-4xvdv=0$
$\int\frac{dx}{4x}-\int\frac{v}{1+2v^2}dv=0$
$\ln{x}-\ln{(2v^2+1)}=C$
$\ln{(\frac{x}{2v^2+1})}=\ln{C}$
$\frac{x}{2v^2+1}=\frac{1}{C}$
$C=3$
$3x=2v^2+1$
$3x^3=2y^2+x^2$
$2y^2=x^2(3x-1)$
The particular solution by method 1 is $2y^2=x^2(3x-1)$.
Method 2: Bernoulli Equation
$2y\frac{dy}{dx}-\frac{x^2+6y^2}{2x}=0$
$2y\frac{dy}{dx}-\frac{3y^2}{x}=\frac{x}{2}$
Let $v=y^2; dv=2ydy$
$\frac{dv}{dx}-\frac{3v}{x}=\frac{x}{2}$
$P(x)=-3x^{-1}$; I.F.$=e^{-3\int x^{-1}dx}=x^{-3}$
$vx^-3=\frac{1}{2}\int\frac{dx}{x^2}$
$2vx^{-3}=-x^{-1}+C^{-1}$
$2y^2x^{-3}+x^{-1}=C^{-1}$
$C=\frac{1}{3}$
$2y^2+x^2=3x^3$
$2y^2=x^2(3x-1)$
The particular solution by method 2 is also $2y^2=x^2(3x-1)$.

Compare
$$(x^2+6y^2)dx-4xydy=0~~~~(1)$$
with $$Mdx+Ndy=0 \implies \frac{\partial M}{\partial y}=12y,~~ \frac{\partial  N}{dx}=-4y $$ It's integrating factor is
$$I=\exp \left(\int \frac{12y +4y}{-4xy}\right)=x^{-4}$$
Multiplying (1) with $x^{-4}$,(1) becomes an exact ODE:
$$(x^{-2}+6x^{-4} y^2) dx- 4x^{-3} ydy==0$$
It's solution is
$$\int (x^{-2}+6x^{-4}y^2) dx~~~ \text{(treat y as constant)}=C$$
$$-x^{-1}-2x^{-3}y^{2}=C$$
By the condition $y(1)=1$, we get $C=-3$
Finally $$2y^2=3x^3-x^2.$$
