Question

Differential Equations question...if the Integrating Factor method does not produce an Exact Differential Equation, do you use Integrating Factor again?

Answer 1

you mean if the integrating method does not produce a separable eqn?

Answer 2

integrating factor*

Answer 3

spearable eqn?

Answer 4

does integrating factor produce separable eqn?

Answer 5

cuszI don't think introducing an integrating factor can change its "exactness"

yes, usually using an integrating factor makes the equation separable

Answer 6

im talking about that \[\large \frac{1}{N} (\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})\] we\re talking about the same thing right?

Answer 7

no that is exact...
I don't see how integrating factors can change a DE's exactness, and have never combined the two methods, so I'm not sure how to answer your Q...

Answer 8

okay here's an example... \((x^2 + y^2 + 1)dx + x(x-2y)dy = 0\) is not exact because \(\frac{\partial M}{\partial y} \ne \frac{\partial N}{\partial x}\) right?

Answer 9

right, because of the minus sign

Answer 10

but if the integrating factor \(e^x\) is introduced then it becomes exact...my question is if the IF does not make the equation exact...do i find the IF of the new equation?

Answer 11

oops wait..i read the wrong example

Answer 12

that's not the IF

Answer 13

I have never tried to make an equation exact by introducing an integrating factor...
how did you get an integrating factor here at all? what formula did you use?
I only know of integrating factors for linear equations

Answer 14

it is in the form Mdx + Ndy = 0 right? \[*\frac{1}{N} (\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) = f(x)\] \[*\frac{1}{M} (\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) = g(y)\]

the IF = \(e^{\int f(x)dx}\) or \(e^{-\int g(y)dy}\)

Answer 15

therefore first i did \(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} = 2y - (2x + 2y) = 2y - 2x + 2y = -2x + 4y\)
\(= -2(x - 2y)\)

Answer 16

then i divided by N to make it a function of x \[\frac{-2(x-2y)}{x(x-2y)} = -\frac{2}{x}\] then i used the formula for I.F. \[e^{\int f(x)dx} = e^{\int -\frac{2}{x}dx} = e^{-2\ln x} = \frac{1}{x^2} \Longleftarrow \text{I.F.}\]

@TuringTest

Answer 17

news to me!
never seen this method befor

Answer 18

ahh i find it useful actually..that's why i want to know what to do when it fails because i'd love to abuse this method :D

Answer 19

@Zarkon DE help?