y/x2dx–xy+1/xdy=0
P(x;y)dx+Q(x;y)dy=0
[m]\frac{y}{x^2}dx-(y+\frac{1}{x})dy=0[/m]
[m]P(x;y)= \frac{y}{x^2}[/m]
[m]Q(x;y)=-(y+\frac{1}{x})[/m]
[m]\frac{ ∂ P}{ ∂y}=\frac{ ∂ Q}{ ∂ x}=\frac{1}{x^2}[/m]
Уравнение в полных дифференциалах