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Why is this a linear differential equation?

  1. Jan 19, 2012 #1
    Why is y'-2xy=x a linear differential equation? I thought it would be nonlinear due to the 2xy...?
     
  2. jcsd
  3. Jan 19, 2012 #2

    Ray Vickson

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    It is linear *in y*; it would be a nonlinear DE if it contained things like y^2, exp(y), 1/(1+y), etc. Another way to see it is: if y1 and y2 are two solutions and a, b are constants, then the linear combination a*y1 + b*y2 is also a solution. That would generally fail for a nonlinear DE.

    RGV
     
  4. Jan 20, 2012 #3

    Ray Vickson

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    Sorry: I mean that the above linear combination property is true of the _homogeneous_ equation, with 0 on the right-hand-side; it may, of course, fail for a nonzero right-hand-side. Also: I should have mentioned that a DE is nonlinear as well if it has terms nonlinear in y and/or y', such as y*y' or (y')^2, etc.

    RGV
     
  5. Jan 20, 2012 #4
    Another way of looking at it is to consider the operator [itex]L=\frac{d}{dx}-2x[/itex], so that the differential equation becomes [itex]Ly=x[/itex]. Then we say the differential equation is linear if that operator [itex]L[/itex] is linear, i.e. [tex]L(f+g)=L(f)+L(g)[/tex] and [tex]L(cf)=cL(f)[/tex] for all (suitably smooth) functions [itex]f[/itex] and [itex]g[/itex] and constants [itex]c[/itex] (where addition of functions and multiplication of a function by a constant are defined point-wise as usual). This is equivalent to what Ray Vickson just said: if you consider the equation [itex]Ly=0[/itex], then (as a result of the linearity of [itex]L[/itex]) any linear combination of solutions is also a solution.
     
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