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Homework Help: Help in solving an inexact differential equation

  1. Nov 10, 2018 at 2:33 PM #1
    1. The problem statement, all variables and given/known data

    The question is to solve the inexact equation by turning it into exact.the equation is ##( x + y + 4 ) d x + ( - x + y + 6 ) d y = 0##
    Where "x" and "y" are variable.

    2. Relevant equations

    1.(x+y+4)=m and (-x+y+6)=n
    2.Integrating Factor =##\frac { 1 } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y }##
    3.a=[(Integrating factor)*m]=$$\frac { ( x + y + 4 ) } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y }$$ and b=[(Integrating factor)*m]=$$\frac { ( x + y + 4 ) } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y }$$
    4.##t = (x ^ { 2 } + y ^ { 2 } + 4 x + 6 y)##
    3. The attempt at a solution
    The given equation is a non exact homogeneous equation and is in the form
    mdx+ndy = 0
    now let's take
    ( x + y + 4 ) = m and ( - x + y + 6 ) = n
    Here ##m x+n y##
    =## x ^ { 2 } + x y + 4 x - x y + y ^ { 2 } + 6 y##
    = ##x ^ { 2 } + y ^ { 2 } + 4 x + 6 y \neq 0##
    So the integrating factor of the equation should be
    I.F = ##\frac { 1 } { m x + n y } = \frac { 1 } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y }##
    So now let's convert the inexact equation to exact by multiplying each term with integrating factor.so the new exact equation should be
    ##\frac { ( x + y + 4 ) } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y } + \frac { ( - x + y + 6 ) } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y } = 0##
    =##a x + b y =0##
    Where a=$$\frac { ( x + y + 4 ) } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y }$$ and b= $$\frac { ( x + y + 4 ) } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y }$$
    Now the solution should be ##\int a d x + \int b d y = constant##
    Where [in term "a" y is constant and the term "b" is free from x]
    =##\int \frac { ( x + y + 4 ) d x } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y } = C##
    (As there is no term in "b" free from x)
    (##C##=constant)
    So$$ \int \frac { ( x + y + 4 ) d x } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y } = $$
    =$$\int \frac { x + 2 } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y } + \int \frac { ( y + 2 ) d x } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y }$$
    [Now let's put ##(x ^ { 2 } + y ^ { 2 } + 4 x + 6 y) = t##
    So ##( x + 2 ) d x = \frac { d t } { 2 }##]
    =$$\frac { 1 } { 2 } \int \frac { d t } { t } + (\int \frac { ( y + 2 ) d x } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y } )$$
    =$$\frac { 1 } { 2 } \log t + ( \int \frac { ( y + 2 ) d x } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y }) $$
    =$$\log t ^ { 2 } + ( \int \frac { ( y + 2 ) d x } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y })$$
    =$$\log ( x ^ { 2 } + ( y ^ { 2 } + 4 x + 6 y ) ^ { 2 } + (\int \frac { ( y + 2 ) d x } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y }) $$
    So now ##\log ( x ^ { 2 } + y ^ { 2 } + 4 x + 6 y ) ^ { 2 } + ( \int \frac { ( y + 2 ) d x } { x ^ { 2 } + y ^ { 2 } + 4 x + 6 y }) ##= constant

    Now I can't figure out what to do next.please help.
     
  2. jcsd
  3. Nov 10, 2018 at 11:56 PM #2

    haruspex

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    I don't think you have the right way of obtaining the integration factor.
    From what I read online, if the factor is 1/P(x,y) then it needs to satisfy
    ∂(M/P)/∂y=∂(N/P)/∂x

    where the inexact equation is M.dx+N.dy=0
    Your factor does not seem to have that property.
    This is not an area I have ever worked in, so I don't know of a better way; I assumed P was a quadratic in x and y and obtained a solution by sheer algebra.

    As a check, if the x2 coefficient is 1 then the constant term in the quadratic is 26.
     
  4. Nov 11, 2018 at 12:20 AM #3
    IMG-20181111-114757-HDR.jpg
    Here you can see the formula and I have done according to this.
     
  5. Nov 11, 2018 at 2:16 AM #4

    haruspex

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    I see it is followed by an example. Would you post that too, please?
     
  6. Nov 11, 2018 at 3:28 AM #5

    Delta²

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    I certainly can see a mistake in the OP. You say at some point that the term b is free of x, but if I am not mistaken it is

    ##b=\frac{-x+y+6}{x^2+y^2+4x+6y}## (another mistake i see is that you seem to say that ##b=\frac{x+y+4}{x^2+y^2+4x+6y}## which is equal to a, I reckon this is a cut and paste mistake ), it clearly is NOT free of x.
     
  7. Nov 11, 2018 at 3:30 AM #6
    Yes I know that, but even if I corrected that mistake,I can't get my results.and I said that because of the method which is to integrate the part of the "b", which is free of x,and no part of "b" is free of x,so here that value is 0.
     
  8. Nov 11, 2018 at 3:46 AM #7
    IMG-20181111-150640558.jpg
    IMG-20181111-150653461.jpg
    Here are they.
     
  9. Nov 11, 2018 at 3:56 AM #8

    Delta²

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    Ok as i understand you need some hint finding the integral ##\int \frac{y+2}{x^2+y^2+4x+6y}dx##. y is a constant for this. so essentially you have the integral

    ##(y+2)\int\frac{1}{x^2+4x+L}dx## where ##L=y^2+6y##. Try to solve this with the help of the integral ##\int\frac{1}{t^2+1}dt=\arctan t##
     
  10. Nov 11, 2018 at 4:36 AM #9

    haruspex

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    Interesting...
    What I see online is that the integrating factor being a function of x or y only are easy special cases. Indeed, they are the only examples I have found. I note also that the IF in the example satisfies the equation
    where 1/P is the IF. But the IF you obtained using the formula in the text does not satisfy this, so I am suspicious that the formula is not general. Certainly it does not yield the IF I got that satisfies the equation above.
     
  11. Nov 11, 2018 at 4:52 AM #10
    I have just now realized that my equation is not a homogeneous equation so that's why the formula isn't working.I can't believe I missed such a tiny thing. But can you still tell me now how to find the I.F of this non exact,non homogeneous differential equation?
     
  12. Nov 11, 2018 at 5:18 AM #11

    haruspex

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    Try my method in post #2. It's a bit of a slog, but it gets there.
    Assume the IF is 1/P where P is some quadratic. You can take the x2 coefficient as 1. Find a solution to ∂(M/P)/∂y=∂(N/P)/∂x.
    What to do after that I am not sure.
     
  13. Nov 11, 2018 at 5:23 AM #12
    Thanks for trying but I can't still reach the answer.
     
  14. Nov 11, 2018 at 5:26 AM #13

    Delta²

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    What is the answer key?

    BTW I don't understand why the equation is non homogeneous. Non homogeneous would be if it was ##Mdx+Ndy=x^2+y^2## for example but it is ##Mdx+Ndy=0##
     
  15. Nov 11, 2018 at 5:27 AM #14

    haruspex

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    Where do you get stuck? It is just a matter of substituting P=x2+axy+by2+ etc. in the PDE I quoted and collecting up terms to find the coefficients a, b, ...
    One tip: there is no xy term.
     
  16. Nov 11, 2018 at 5:29 AM #15

    haruspex

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    M and N are not homogeneous functions.
     
  17. Nov 11, 2018 at 4:38 PM #16

    LCKurtz

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    I think you can eliminate the 4 and 6 with the substitutions ##r=x+y+4,~s = -x+y+6## and make a homogeneous equation in ##r## and ##s## out if it. (Homogeneous in the ##f(\lambda x,\lambda y) = \lambda^n f(x,y)## sense).
     
  18. Nov 13, 2018 at 11:43 PM #17

    haruspex

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    I'm sure there's a better way, but I embarked on solving the general problem of finding an integration factor for the general first order case: M=Ax+By+C, N=Dy+Ex+F. The IF must be 1/P where P satisfies ∂(M/P)/∂y=∂(N/P)/∂x. I assumed P to be a quadratic in x and y.

    I got these coefficients:
    x2: (AD-BE)A
    xy: (AD-BE)(B+E)
    y2: (AD-BE)D
    x: AEF-BCE-CEE+2ACD-ABF
    y: BCD-BBF-BEF+2ADF-CDE
    1: AFF+CCD-BCF-CEF
     
  19. Nov 14, 2018 at 1:36 PM #18

    SammyS

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    Consider the following (from Wikipedia: Homogeneous Diff. Eq. (Special Case) )
    ##( x + y + 5 - 1 ) d x + ( - x + y + 5 + 1 ) d y = 0##

    ##( x -1 + y + 5 ) d x + ( - x +1 + y + 5 ) d y = 0##

    ##( (x -1) + (y + 5) ) d x + ( - (x -1) + (y + 5) ) d y = 0##

    Make the appropriate change of variables.


    Alternatively: Solve for y and get d'Alembert's equation

    This is messy, but should work.
     
  20. Nov 14, 2018 at 5:54 PM #19

    LCKurtz

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    Edit: corrected copy/paste.

    If anyone cares, what I got using my idea in post #16 is:
    (1/2)*ln(2*x^2-4*x+2*y^2+20*y+52)-arctan((-x+y+6)/(x+y+4)) = C
    I haven't checked it though.
     
    Last edited: Nov 14, 2018 at 6:15 PM
  21. Nov 14, 2018 at 7:07 PM #20

    haruspex

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    I tried differentiating that and could not get back to the original equation.

    Edit:
    I get ##\frac 12\ln((x-1)^2+(y+5)^2)+\arctan(\frac{x-1}{y+5})##.

    The easiest way to make it homogeneous is to adjust each of x and y by a constant: x=X+1, y=Y-5.
    (X+Y).dX+(-X+Y).dY=0
    Now we can apply the formula in the text to get the IF ##\frac 1{X^2+Y^2}##. This leads easily to the solution ##\frac 12\ln(X^2+Y^2)+\arctan(\frac{X}{Y})=C##.
     
    Last edited: Nov 14, 2018 at 7:45 PM
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