<s>[INST] 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 $2y^2=x^2(3x-1)$.
 [/INST] We can check if the ODE is exact:
$$\Delta=\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}=\frac{\partial}{\partial y}(x^2+6y^2)-\frac{\partial}{\partial x}(-4xy)=16y\ne0.$$
So it's not exact, but as
$$\frac{\Delta}{N}=\frac{16y}{-4xy}=-\frac{4}{x}=f(x),$$
we can get an integrating factor with the form
$$\mu(x)=\exp\left(\int f(x)\mathrm{d}x\right)=\exp\left(\int-\frac{4}{x}\mathrm{d} x\right)=e^{-4\log x}=x^{-4}.$$
Now, let's apply our IF, multiplying both sides by it:
$$x^{-4}(x^2+6y^2)\mathrm{d}x-4x^{-3}y\,\mathrm{d}y=0,$$
Let's check if the new ODE is exact (it must be, but just in case),
$$\Delta'=\frac{\partial M'}{\partial y}-\frac{\partial N'}{\partial x}=\frac{\partial}{\partial y}\bigl[x^{-4}(x^2+6y^2)\bigr]-\frac{\partial}{\partial x}(-4x^{-3}y)=0.$$
So, we are looking for a function $F(x,y)$ that satisfy
\begin{align}
\frac{\partial F}{\partial x} & =x^{-4}(x^2+6y^2);\\
\frac{\partial F}{\partial y} & =-4x^{-3}y.
\end{align}
From the second condition,
$$F(x,y)=\int-4x^{-3}y\,\mathrm{d}y+\varphi(x)=-2x^{-3}y^2+\varphi(x).$$
Now, we can plug $F$ in the first condition:
\begin{align*}
\frac{\partial}{\partial x}\bigl[-2x^{-3}y^2+\varphi(x)\bigr]=x^{-4}(x^2+6y^2) & \Longrightarrow6x^{-4}y^2+\varphi'(x)=x^{-4}(x^2+6y^2)\Longrightarrow\\
& \Longrightarrow\varphi'(x)=x^{-2}\Longrightarrow\varphi(x)=-\frac{1}{x}.
\end{align*}
So, the solution takes the form $F(x,y)=C$, and is
$$-2x^{-3}y^2-\frac{1}{x}=C\Longrightarrow2y^2=-Cx^3-x^2.$$
If $y=1$ when $x=1$,
$$2=-C-1\Longrightarrow C=-3,$$
and we we have
$$2y^2=3x^3-x^2.$$
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