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How to find if eq is linear or non linear

  1. Jan 29, 2007 #1
    Please let me know if this is true or not

    Equation is linear if dependent variable is not in any other power than 1 e,g y

    Equation is linear if dependent variable y is only being mulitiplied by constant.

    Equation is linear if dependent variable y is not being multiplied by variable x

    sin(y) and cos(y) make equation non linear.

    Now i went to a site and they said following equation is linear
    x^2y''+ xy' + (x^2-v^2)y = 0

    y'' is being multiplied by x^2. and also y' being multiplied by x
    Is there any easy way of steps to go thru to see if eq is linear or not besides the above mentioned...I am trying my best but i will be need lots of help with DE and I hope this forum will help me out.

    Thnk you all
  2. jcsd
  3. Jan 29, 2007 #2


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    The form of an ODE that is linear is:

    [tex] a_{n}(x) y^{(n)} + a_{n-1}(x) y^{(n-1)} + .... + a_{1}(x) y' + a_{0}(x) y = g(x) [/tex]

    if you look at it real close you'll notice all derivatives and the dependent variable are of first degree, and that there are no coefficients of dependent variables y.

    For example:

    This is linear:

    [tex] y'' = \sin (x) [/tex]

    This is nonlinear

    [tex] (y-1)y' = x [/tex]

    This is nonlinear

    [tex] y' = y^2 [/tex]

    This is linear

    [tex] x^2y''+ xy' + (x^2-v^2)y = 0 [/tex]

    There are only independent variables as coefficients, and all derivatives plus dependent variables are of degree 1.
  4. Jan 29, 2007 #3
    so in second equation (y-1)y'=x becomes non linear becuase of y.y'?

    secondly does 2y or 2y' make equation non linear becuase they are cofffcient of y
    or is it only if its y.y' in this case both are being multiplied together.
  5. Jan 29, 2007 #4
    i am confused about
    x y' = 1 is non-linear because y' is not multiplied by a constant (http://www.myphysicslab.com/classify_diff_eq.html)

    x^(2)y' term is in last equation which is linear but xy'=1 is non linear
    Can you help me figure this out?
  6. Jan 30, 2007 #5


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    xy' =1 will be linear if you were differentiating by x. Above says multiple variable problems, in the equation you provided i assume v as a constant.
  7. Jan 30, 2007 #6
    oh ok got it. I appreciate it cyclovenom...i am sure i will be asking you a lot more in future...lol..thanks again
  8. Jan 30, 2007 #7


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    There is nothing in the site you give that say "x y'= 1 is non-linear". The closest thing I can find is "x x'= 1" is non-linear. That's true because x is the dependent variable being differentiated with respect to some independent variable (perhaps t).
    I would interpret xy'= 1 as meaning that y is the dependent variable (since it is differentiated) and x is the independent variable (since it is the only other variable mentioned). In that case, xy'= 1 is a "linear differential equation with constant coefficient".
  9. Jan 31, 2007 #8
    what about sin(x)y=1?
    or sin(x)y'=1

  10. Feb 1, 2007 #9


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    What part of this is not getting through to you? A differential equation is "linear" if and only if it contains only linear functions of the dependent variable. What functions there are of the independent variable is irrelevant.

    sin(x)y= 1 is not a differential equation at all! There is no derivative. It is equivalen to y= cosec(x).

    sin(x)y'= 1 is a linear differential equation. The only occurence of the dependent variable is the y' and it is to first degree mutliplied by a function of x only- which doesn't effect linearity.
  11. Feb 2, 2007 #10
    I got it now. Thanks. I was confused about the independent and dependent variables.
  12. Feb 3, 2007 #11


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    In differential equations, at least, the "dependent" variable is the one that is being differentiated and the independent variable is the one you are differenitating with respect to.

    In y'= f(x,y), y is the dependent variable, x is the independent variable. In
    y'= y2, y is the dependent variable, the independent variable is not explicitely given and you are welcome to use whatever symbol you want.

    In what is sometimes called "symmetric" form, f(x,y)dx+ g(x,y)dy= 0, there is no way (or reason) to distinguish "dependent" and "independent" variable. But that only occurs in first order equations where the distinction between "linear" and "nonlinear" equations is much less important.
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