anonymous
  • anonymous
Consider the function f(x) ={ 1, if x<= 0 2, if x>0} Show that the differential equation x dot = f(x) has no solution on any open interval about t knot =0. Could someone help me out? I am not sure how to do this. I tried integrating for f(x)=1 and f(x)=2 to find x(t), but I don't know where to go from there.
Differential Equations
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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sirm3d
  • sirm3d
continuity on an closed interval is the first requirement for differentiability in the open interval. The function is discontinuous on any closed interval containing 0, so it is not differentiable there (at zero). take a look at the graph of \(f(x)\) |dw:1359845550730:dw|
anonymous
  • anonymous
Is there a way to tell without knowing what the graph looks like though?
sirm3d
  • sirm3d
the graph is not needed. just check the value of the function at x=0.

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sirm3d
  • sirm3d
for example, if f(x) = { 2x+5, x<0, x^2+3 for x>=0 at 2(0)+5=5, (0)^2+3=3 since the two values are different, the function is discontinuous at x=0, so it's not differentiable there.
anonymous
  • anonymous
is it like saying the limit going to 0 is different, so the limit does not exist?
sirm3d
  • sirm3d
yes, that's right. the correct argument is to use limits.
anonymous
  • anonymous
okay, that makes sense. thanks!

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