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Question about linearity

  1. Jan 27, 2008 #1
    I am reading through my Diff Eqs Text and I follow most of the lingo. However I am just a tad confused by the statement:

    An nth order ODE is said to be linear if F is linear in y,y',...y^(n)

    Then it gives the example:

    [tex]a_n(x)\frac{d^ny}{dx^n}+a_{n-1}(x)..+a_0(x)y=g(x)[/tex]

    It then says: 'On the left-hand side of the above equation the dependent variable y and all of its derivatives, y,y',y'',...y[itex]^n[/itex] are of the first degree.

    Clearly I missed something in Calc. If n=2, I have: [tex]\frac{d^2y}{dx^2}[/tex]

    Why is this linear if n=2?

    Thanks,
    Casey
     
  2. jcsd
  3. Jan 27, 2008 #2

    cepheid

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    You're confusing the *order* of the derivative with its *degree* in the equation.

    y'' + y = 0

    is a second-order linear equation. Second-order because the highest order derivative in the equation is a second dervative. Linear because the equation itself is linear in both y'' and y.

    In contrast:

    (y'')^2 + ay^3 = 0

    is a second-order NON-linear D.E. Second-order for the same reason. NON-linear because it is QUADRATIC in y'' and CUBIC in y. I hope this clears things up.
     
    Last edited: Jan 27, 2008
  4. Jan 27, 2008 #3
    I think it does. y' just means "the 1st derivative" and similarly for y" however if either one y' or y" or y for that matter were raised to any power above 1, the DE would no longer be linear.

    Thanks!
     
  5. Jan 27, 2008 #4

    HallsofIvy

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    Yes, that is correct. Also note that other "non-linear" functions of the dependent variable, y or its derivatives, such as sin(y) or exp(y"), would make the equation non-linear.
     
  6. Jan 27, 2008 #5
    Great, thanks Halls and cepheid. Hey also, I know that the right-hand side can be equal to 0 or a function of the independent variable; what about a constant?

    Like [itex]\frac{d^2y}{dx^2}-\frac{dy}{dx}+6y=7[/itex] ?
     
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