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Differential equation, need help

  1. Nov 1, 2008 #1

    rock.freak667

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    1. The problem statement, all variables and given/known data

    Solve the following

    (x+2y-1)dx +3(x+2y)dy=0

    2. Relevant equations

    3. The attempt at a solution

    I said that it was in form M(x,y) dx + N(x,y) dy=0

    then found [itex]\frac{\partial{N}}{\partial{y}}[/itex] and similarly [itex]\frac{\partial{M}}{\partial{x}}[/itex]

    Meaning that it wasn't exact, but then tried to find the integrating factors, given that the equation wasn't exact. But in doing so, The formulas for the integrating factors didn't work out. Did I start this question wrong?
     
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  3. Nov 1, 2008 #2

    gabbagabbahey

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    Your approach looks correct, why don't you show us your work so that we can see where you may have erred?
     
  4. Nov 1, 2008 #3

    HallsofIvy

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    What "formulas for the integrating factors" do you mean?
     
  5. Nov 1, 2008 #4

    rock.freak667

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    This is from my notes:

    If [itex]\frac{\partial{M}}{\partial{y}} \neq \frac{\partial{N}}{\partial{x}}[/itex] i.e. not exact equation

    Then an integrating factor is found as follows:

    If (i)

    [tex]\frac{\frac{\partial{M}}{\partial{y}}- \frac{\partial{N}}{\partial{x}}}{N}=f(x)[/tex]


    Where f(x) is a function in x-only , then ef(x) is an integrating factor.

    (ii)
    [tex]\frac{\frac{\partial{M}}{\partial{y}}- \frac{\partial{N}}{\partial{x}}}{M}= - g(y)[/tex]

    then e+g(y) is an integrating factor

    ______________________________________________________________________________
    For (x+2y-1)dx +3(x+2y)dy=0

    M= x+2y-1 so that [itex]\frac{\partial{M}}{\partial{y}}=2[/itex]
    N=3x+6y => [itex]\frac{\partial{N}}{\partial{x}}=3[/itex] Clearly [itex] 2 \neq 3[/itex]

    But if I put it into either of the two previously stated formulas, I do not get a function in x only for (i) OR a function in y only for equation (ii)
     
  6. Nov 1, 2008 #5

    gabbagabbahey

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    There is an easier way to find the integrating factor in this case: In general, whenever [itex]M(x,y)[/itex] and [itex]N(x,y)[/itex] are homogeneous functions of the same degree, then [itex]\frac{1}{xM+yN}[/itex] is an integrating factor. So, are [itex]M(x,y)[/itex] and [itex]N(x,y)[/itex] homogeneous functions of the same degree in this case?
     
  7. Nov 1, 2008 #6

    rock.freak667

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    M(x,y) isn't homogeneous since it doesn't involve terms in x and y only...
    N(x,y) is homogeneous of degree 1
     
  8. Nov 1, 2008 #7

    gabbagabbahey

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    Hmmm.. yes, M is inhomogeneous so that won't work....What other methods have you been taught for finding integrating factors?
     
  9. Nov 1, 2008 #8

    rock.freak667

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    Those two are the only two equations which I know of when it is a non-exact differential equation.
     
  10. Nov 1, 2008 #9

    rock.freak667

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    I think I got it out. All I needed to do was to put t=x+2y-1
     
  11. Nov 1, 2008 #10

    gabbagabbahey

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    This may be above the scope of your course; but let's look at the definition of integrating factors:

    [itex]\mu(x,y)[/itex] is an integrating factor of the ODE [itex] M(x,y)dx+N(x,y)dy[/itex] if the ODE [itex] \mu(x,y)M(x,y)dx+\mu(x,y)N(x,y)dy[/itex] is exact. This means that [itex]\mu(x,y)[/itex]
    must satisfy the condition:

    [tex]\frac{\partial}{\partial y}(\mu(x,y)M(x,y))=\frac{\partial}{\partial x}(\mu(x,y)N(x,y))[/tex]

    Or, equivalently;

    [tex]M(x,y)\frac{\partial \mu}{\partial y}-N(x,y)\frac{\partial \mu}{\partial x}=\mu(x,y) \left( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)[/tex]

    In this case, we have:

    [tex](x+2y-1)\frac{\partial \mu}{\partial y}-3(x+2y)\frac{\partial \mu}{\partial x}=\mu(x,y) \left( (3)-(2) \right)=\mu (x,y)[/tex]

    Do you have any tools at your disposal that you could use to solve this PDE for [itex]\mu(x,y)[/itex]?
     
  12. Nov 1, 2008 #11

    gabbagabbahey

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    hmmm... yes, that should work!
     
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