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 $2y^2=x^2(3x-1)$.

$$(x^2+6y^2)dx-4xydy=0$$
$$(x^2+6y^2)dx-2xdy^2=0$$
$$x^2dx+2(3y^2dx-xdy^2)=0$$
Multiply by $x^2$:
$$x^4dx+2(3x^2y^2dx-x^3dy^2)=0$$
$$x^4dx+2(y^2dx^3-x^3dy^2)=0$$
Divide by $x^6$:
$$\dfrac {dx}{x^2}-2d\left (\dfrac {y^2}{x^3}\right)=0$$
Integrate.
$$\dfrac {1}{x}+2\left (\dfrac {y^2}{x^3}\right)=C$$
$$y(1)=1 \implies C=3$$
Therefore:
$$x^2+2y^2=3x^3$$
