Question

Is ((x/y)-2y)(dy/dx)+(y/x)=-1 an exact equation?

Answer 1

\[\left(\frac{x}{y}-2y\right)\frac{dy}{dx}+\frac{y}{x}=-1\\
\left(\frac{x}{y}-2y\right)\frac{dy}{dx}+\left(\frac{y}{x}+1\right)=0
\]
Here, \(\dfrac{\partial}{\partial x}\Psi(x,y)=M(x,y)=\dfrac{y}{x}+1\) and \(\dfrac{\partial}{\partial y}\Psi(x,y)=N(x,y)=\dfrac{x}{y}-2y\).
Check whether \(\dfrac{\partial}{\partial y}M(x,y)=\dfrac{\partial}{\partial x}N(x,y)\). If equality holds, you have an exact equation.

Answer 2

$$\left(\frac{x}y-2y\right)\frac{dy}{dx}+\frac{y}x+1=0\\\left(\frac{x}y-2y\right)dy+\left(\frac{y}x+1\right)dx=0$$If this is an exact equation, we have some potential function \(F\) s.t. \(\dfrac{\partial F}{\partial x}=\dfrac{x}y-2y,\dfrac{\partial F}{\partial y}=\dfrac{y}x+1\). Assuming \(F\) is continuous, we have symmetry of second derivatives \(\dfrac{\partial F}{\partial x\partial y}=\dfrac{\partial F}{\partial y\partial x}\). To test this, we take our derivatives:$$\frac{\partial F}{\partial x\partial y}=-\frac{x}{y^2}-2\\\frac{\partial F}{\partial y\partial x}=-\frac{y}x^2$$... therefore our equation is inexact.