anonymous
  • anonymous
does the IVP dy/dt=6y^(2/3), y(1)=0 have a unique solution on an interval containing t=1?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
first step, i have y=(2t+c)^3 is it right?
JamesJ
  • JamesJ
Yes.
JamesJ
  • JamesJ
and now use the initial condition to find c.

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anonymous
  • anonymous
that's makes it much easier, thank u
anonymous
  • anonymous
is y=0 a solution to the IVP? yes isn't it, because when t=1, y=0 or is there another way to determine the solution?
JamesJ
  • JamesJ
Yes, the zero function is the solution of your ODE. But it's a very uninteresting solution and we don't count it as a solution for the purposes of describing uniqueness.
anonymous
  • anonymous
ok thx
anonymous
  • anonymous
i have another problem: show that y=tant satisfies the IVP y'=1/(1+t^2), y(0)=0. i am confused because the y'(tant) is not 1/(1+t^2). they both do equal 0 with initial condition set to 0 but what do they have to do with each other?
anonymous
  • anonymous
i don't know where to go now because i'm trying to look for the limits of (t,y) to find the largest open interval but tan and arctan limits are not the same, at least for t

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