Determine whether the following differential equations are exact or not exact. if exact, solve for...

Step 1
${\displaystyle (x+2y)dx-(2x+y)dy=0}$
The differential equation ${\displaystyle Mdx+Ndy=0}$ is a exact if
${\displaystyle \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}}$
Step 2
From the given differential equation
${\displaystyle M=x+2y}$
${\displaystyle N=-(2x+y)}$
${\displaystyle N=-(2x+y)}$
${\displaystyle \frac{\partial M}{\partial y}=2}$
${\displaystyle \frac{\partial N}{\partial x}=-2}$
Since, ${\displaystyle \frac{\partial M}{\partial y}\ne \frac{\partial N}{\partial x}}$
The given differential equation is not exact differential equation.

Simplifying
${\displaystyle (x+2y)\cdot dx+-1(2x+-1y)\cdot dy=0}$
Reorder the terms for easier multiplication:
${\displaystyle dx(x+2y)+-1(2x+-1y)\cdot dy=0}$
${\displaystyle (x\cdot dx+2y\cdot dx)+-1(2x+-1y)\cdot dy=0}$
Reorder the terms:
${\displaystyle (2dxy+{dx}^2)+-1(2x+-1y)\cdot dy=0}$
${\displaystyle (2dxy+{dx}^2)+-1(2x+-1y)\cdot dy=0}$
Reorder the terms for easier multiplication:
${\displaystyle 2dxy+{dx}^2+-1dy(2x+-1y)=0}$
${\displaystyle 2dxy+{dx}^2+(2x\cdot -1dy+-1y\cdot -1dy)=0}$
${\displaystyle 2dxy+{dx}^2+(-2dxy+1{dy}^2)=0}$
Reorder the terms:
${\displaystyle 2dxy+-2dxy+{dx}^2+1{dy}^2=0}$
Combine like terms: ${\displaystyle 2dxy+-2dxy=0}$
${\displaystyle 0+{dx}^2+1{dy}^2=0}$
${\displaystyle {dx}^2+1{dy}^2=0}$
Solving
${\displaystyle {dx}^2+1{dy}^2=0}$
Solving for variable d.
Move all terms containing d to the left, all other terms to the right.
Factor out the Greatest Common Factor (GCF), d.
${\displaystyle d(x^2+y^2)=0}$

It is a homogeneous equation
$\begin{array}{}\text{Put}y=vx\\ \text{So,}v+x\frac{dv}{dx}=\frac{x+2vx}{2x-vx}\\ v+x\frac{dv}{dx}=\frac{1+2v}{2-v}\\ x\frac{dv}{dx}=\frac{1+2v}{2-v}-v\\ x\frac{dv}{dx}=\frac{1+2v-2v+v^2}{2-v}\\ x\frac{dv}{dx}=\frac{1+v^2}{2-v}\\ \text{Integrating both sides we get,}\\ \int \frac{2-v}{1+v^2}dv=\int \frac{dx}{x}\\ \int \frac{2}{1+v^2}dv-\int \frac{v}{1+v^2}dv\\ =\int \frac{dx}{x}\\ 2{\tan }^{-1}v-\frac{1}{2}\log |1+v^2|\\ =\log |x|+\log c\\ 2{\tan }^{-1}v=\log |xc|+\log |1+v^2|\frac{1}{2}\\ e^{{\tan }^{-1}}v=\{1+v^2\}\frac{1}{2}xc\\ e^{{\tan }^{-1}}\frac{y}{x}=\{\frac{(y^2+x^2)}{x^2}{\}}^{\frac{1}{2}}xc\\ e^{{\tan }^{-1}}\frac{y}{x}=\{(y^2+x^2){\}}^{\frac{1}{2}}c\end{array}$