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Homework Help: Not understanding integrating factorsplease clarify/explain

  1. Apr 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Specifically what i dont understand is a situation where you cant get the equation in the standard form of y' + p(x)y = q(x)

    the integrating factor is e^integ p(x), but what if y is part of a sin or cos, as in this equation:
    (x+2)sin y dx + xcos y dy = 0

    if i rearrange so that it is: y' + (x+2)sin y / x cos y = 0, it still isnt in standard form.

    So at that point how am i supposed to proceed? What step do i take to get the integrating factor?

    So, if equation isnt separable, and isnt exact, and cant be put into standard form, how do you solve it?
    thanks for any help
    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 18, 2010 #2

    Mark44

    Staff: Mentor

    You are confusing two different techniques. A differential equation of the form y' + p(x)y = q(x) is a linear, first-order, nonhomogeneous equation. It's linear because y and its derivative occur to the first power. There's a technique that involves finding an integrating factor that can be used to solve this type of equation.

    The equation (x + 2)sin y dx + xcosy dy = 0 can be written as y' + (1 + 2/x)tany = 0, so this is also first-order, but it's nonlinear because y appears as part of a trig function. The standard approach for equations written as (x + 2)sin y dx + xcosy dy = 0 is to determine whether the equation is exact. If so, then this equation represents the total derivative of some as-yet-undetermined function F(x, y). As you have already found out, this equation is not exact, so that approach isn't applicable.

    What other techniques do you know?
     
  4. Apr 18, 2010 #3
    Well the problem asks me to solve by finding an integrating factor, mu(x), so thats why im stuck on getting an integrating factor. I also know about 1st order homogeneous equations but i dont think that applies here because I cant manipulate it so that y/x. ( v = y/x)
    thank you
     
  5. Apr 18, 2010 #4
    multiply both sides by mu(x)??
     
  6. Apr 18, 2010 #5

    Mark44

    Staff: Mentor

    Is this the technique where you assume there is an integrating factor mu(x) and work it into the equation so as to make the equation exact? It's been awhile for me since I thought about this.
     
  7. Apr 18, 2010 #6

    Mark44

    Staff: Mentor

    Unfortunately two different mathematicians came up with categorizations for d.e.s and both settled on the same term for completely different things - homogeneous. The meaning that seems to me to be used most often is when you have a differential equation of the form f(y, y', y'', ...) = 0. E.g., the one in another thread you started - y'' + 3y' + 2y = 0.

    The other meaning - that you can write the diff. equation as f(y/x) or something seems much more obscure to me, and doesn't seem to appear as much.
     
  8. Apr 18, 2010 #7
    Mark could you please tell me what you think best technique to use on this? ,--the thing is, i could never really grasp the professors notes so im learning most of this off internet and I have exam tomorrow, so what do you personally think is best method (easiest) to solve it? --if you wont show me the steps can you tell me most recommended method so i can learn it, then apply it here? ..thanks for any more help.
     
  9. Apr 18, 2010 #8
    The specific question asks:
    Solve (x + 2) sin y dx + x cos y dy = 0 by finding an integrating factor, mu(x).
    (from homework probs)
    thanks
     
  10. Apr 18, 2010 #9

    Mark44

    Staff: Mentor

    Your equation doesn't appear to be homogeneous in the sense of being y' = f(y/x), but it is solvable by applying the right integrating factor [itex]\mu[/itex], where [itex]\mu[/itex] can be a function of x ([itex]\mu[/itex] = [itex]\mu[/itex](x)) or of y ([itex]\mu[/itex] = [itex]\mu[/itex](y)).


    Let's start with [itex]\mu[/itex](x).
    If we multiply both sides of your original equation by [itex]\mu[/itex](x), we get
    [[itex]\mu[/itex](x) * (x + 2) * sin y] dx + [[itex]\mu[/itex](x) * x * cosy] dy = 0

    The expression in the first pair of brackets is M, and the expression in the second pair of brackets in N.

    For exactness, we must have My = Nx. Can you continue from here?

    It's possible in certain cases to get a differential equation for [itex]\mu[/itex](x) that involves both x and y, which wouldn't be solvable. In that case, we take exactly the same approach, but with [itex]\mu[/itex] as a function of y.
     
  11. Apr 18, 2010 #10
    thanks.

    [mu(x) * (x + 2) * sin y] dx + [mu(x) * x * cosy] dy = 0
    ......M.....................................N

    M_y(x,y) = cos y

    N_x(x,y) = cos y

    so now it is exact.

    so now i integrate M wrt x, and then take derivative of that result and equate it to N and solve for h'(y). Then integrate that to get h(y) and write as final equation.

    But how am i supposed to view mu(x) ?? Is it a constant?
    I mean, when integrating wrt x the M term:

    mu(x)(x+2)siny there are three terms, so how does integ by parts apply or even u-sub apply?

    thanks

    EDIT: ok if im integ wrt x there are two terms multiplied, mu(x) * (x+2) and the sin y is treated as constant.
    but im still confused how to treat mu(x). thanks
     
    Last edited: Apr 18, 2010
  12. Apr 18, 2010 #11

    Mark44

    Staff: Mentor

    Not so fast. In differentiation with respect to y, x is seen as a constant, so My = mu(x) * (x + 2) * cos y. For Nx, you have to use the product rule.
     
  13. Apr 18, 2010 #12
    My = mu(x) * (x + 2) * cos y

    Nx = (mu'(x)*x + mu(x) ) cos y

    So not exact afterall ?
     
  14. Apr 19, 2010 #13

    Mark44

    Staff: Mentor

    It depends on what mu(x) is. What does it take for the d.e. to be exact? If it's exact, then My = Nx, right?

    Can you solve for mu(x) so that the equation is exact?
     
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