Turning inexact solutions into exact solutions

[Yasmine Hemmati]

How do integrating factors turn an inexact solution into an exact solution.

[Victor Ivrii]

The question does not make any sense because there are no exact or non-exact solutions, but there are exact or non-exact equations.

The original equation and this equation multiplied by an integrating factor have the same solutions, but the first one is not exact, and the second is exact.

[Wei Cui]

If you ask how to turn the inexact equation into exact equation, then there are three cases:

Try 1: check $\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} =$ $f(x)$ function of $x$ only,
 
then let $\frac{u^{'}}{u} = f(x)$
           
           $\frac{du}{u} = f(x)dx$

            $u = e^{\int f(x)dx}$

Try 2: check $\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M} = g(y)$ function of $y$ only, and then same as the first one you let $\frac{u^{'}}{u} = g(y)$,
         
          $u = e^{\int g(y)dy}$

Try 3: if $\frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{Mx-Ny} = z(x, y)$, then
         
          $u = u(x, y)$ and $u = e^{\int z(x,y)}$

In each of these cases,  $u$ is the integrating factor, when you solve $u$ and you multiply the equation both sides with $u$ then you will turn the inexact equation into an exact equation.

[Victor Ivrii]

There are three cases for which on the lectures and in the textbook the recipe was provided. There are some other cases. And for some equations the simple guess will work.