Toronto Math Forum
MAT2442018F => MAT244Lectures & Home Assignments => Topic started by: Yasmine Hemmati on September 26, 2018, 03:15:55 PM

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

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

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 $\frac{\frac{\partial N}{\partial x}  \frac{\partial M}{\partial y}}{MxNy} = z(x, y)$, 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.

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.