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Question on linearity of a D.E.

  1. Sep 7, 2014 #1
    Hi, I was just wondering if :

    (y^2 - 1)*dx/dy + x = 0

    In this case, x is the dependent variable.

    Is linear? I know it is but I want to understand why. My question is with the coefficients.
    The first coefficient has a y to the power of 2 to it and a constant. It is also a function of y (independent var in this case) for obvious reasons. I'm wondering if the power in the coefficient matters ? For example , instead of getting (y^2 - 1) as a coeff. , would simply y^5 as the coefficient still make the eq. linear? Also , for an equation to be non-linear , must the coefficients involve the dependent var. instead of the independent one?

    Thank you very much!
     
  2. jcsd
  3. Sep 7, 2014 #2
    supposing y is "space" then, this is a linear, spatially-variant ODE. It wouldn't be considered nonlinear unless you couldn't write this ode as

    [tex] ...a_1(y) \frac{dx}{dy} + a_0(y) x = 0 [/tex]

    If any of the "a" coefficients had an x or any of its derivatives in it, it would be nonlinear.

    For example the following cannot be written in the above form (and are non linear)

    [tex] ...(y^2 - x)\frac{dx}{dy} + a_0(y) x = 0 [/tex]

    [tex] ...(y^2 - 1)(\frac{dx}{dy})^2 + a_0(y) x = 0 [/tex]

    [tex] ...(x^2 - 1)\frac{dx}{dy} + x^2 = 0 [/tex]

    But the last example is at least "spatially invariant".
     
    Last edited: Sep 7, 2014
  4. Sep 8, 2014 #3

    HallsofIvy

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    The equation you give is linear because It is "linear" in the dependent variable x. Specifically, if we had two functions, [itex]x_1(y)[/itex] and [itex]x_2(y)[/itex], that satisfy that equation, and any two numbers, [itex]\alpha[/itex] and [itex]\beta[/itex], then the "linear combination", [itex]\alpha x_1(y)+ \beta x_2(y)[/itex] also satisfies it:
    [tex]a_1(y)\frac{d(\alpha x_1+ \beta x_2}{dy}+ a_0(y)(\alpha x_1+ \beta x_2)[/tex]
    [tex]= \alpha\left(a_1(y)\frac{d x_1}{dy}+ a_2(y)x_1\right)+ \beta \left(a_1(y)\frac{dx_2}{dy}+ a_2(y)x_2\right)= \alpha(0)+ \beta(0)= 0[/tex]
     
    Last edited by a moderator: Sep 8, 2014
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