### Author Topic: Turning inexact solutions into exact solutions  (Read 1402 times)

#### Yasmine Hemmati

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##### Turning inexact solutions into exact solutions
« on: September 26, 2018, 03:15:55 PM »
How do integrating factors turn an inexact solution into an exact solution.
« Last Edit: September 26, 2018, 04:30:57 PM by Victor Ivrii »

#### Victor Ivrii

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##### Re: Turning inexact solutions into exact solutions
« Reply #1 on: September 26, 2018, 04:34:27 PM »
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

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##### Re: Turning inexact solutions into exact solutions
« Reply #2 on: September 27, 2018, 01:43:07 AM »
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.
« Last Edit: September 27, 2018, 01:44:39 AM by Wei Cui »