91Ó°ÊÓ

The given differential equation is$\left(y^2-xy\right)dx+\left(x^2+xy\right)dy=0$

A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself to its derivatives of various orders.

First we start by checking whether if this is an exact equation or not using eq. (4.4), that is

$\frac{∂}{∂y}\left(y^2-xy\right)=2y-x≠\frac{∂}{∂y}\left(x^2+xy\right)=2x+y$

Because, eq.(4.4) does not hold, and it's not simple to find an integration factor that will make it exact, so we will think about it as a homogeneous equation, therefore we can write

$$\begin{gathered} y\text{'}=\frac{xy-y^2}{x^2+xy} \\ y\text{'}=\frac{x^2\left({\displaystyle \frac{y}{x}}-{\displaystyle \frac{y^2}{x^2}}\right)}{x^2\left(1+{\displaystyle \frac{y}{x}}\right)} \end{gathered}$$

Assume v = y/x, so y' = xv' + v, therefore, the equation becomes

$$\begin{gathered} xv\text{'}+v=\frac{v-v^2}{1+v} \\ v\text{'}=\frac{1}{x}\left(\frac{52-2v^2}{1+v}\right) \end{gathered}$$

which is become a separable equation, so the solution is

$$\begin{gathered} -∫\frac{1+v}{2v^2}dv=∫\frac{dx}{x} \\ \frac{1}{2v}-\frac{1}{2}Inv=Inx+C \end{gathered}$$

substitute v = y/x, therefore,

$\frac{x}{y}-In\left(\frac{y}{x}\right)=2Inx+C_1$