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Linear DE

  1. Dec 11, 2014 #1
    1. The problem statement, all variables and given/known data

    This problem appears as one of exercise, of a linear differential equations chapter in my DE book.

    ##y \ dx - 4(x + y^6) \ dy = 0##

    How to change above equation in a standard form of a linear differential equation?

    The standard form of a linear differential equation is:

    ##\frac{dy}{dx} + P(x) y = f(x)##

    It appears that the above equation is not linear because the power of ##y## is not 1 instead of 6.

    Also, what is its integrating factor ##(e^{\int P(x) \ dx})##?

    2. Relevant equations


    3. The attempt at a solution

    ##y \ dx - 4(x + y^6) \ dy = 0##
    ##y - 4(x + y^6) \frac{dy}{dx} = 0##
    ##4(x + y^6) \frac{dy}{dx} - y = 0##
    ##\frac{dy}{dx} - \frac{1}{4(x + y^6) }y = 0##
     
  2. jcsd
  3. Dec 11, 2014 #2

    LCKurtz

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    You are correct. It is not linear.

    That's the wrong question to ask. Your equation is in the form ##Pdx + Qdy=f(x)##. Look in your text about exact DE's and what condition on ##P## and ##Q## makes the equation exact.

    [Edit, added] And if the equation isn't exact is it possible to find an integrating factor that makes it exact?
     
    Last edited: Dec 11, 2014
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