does the IVP dy/dt=6y^(2/3), y(1)=0 have a unique solution on an interval containing t=1?
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that's makes it much easier, thank u
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?
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.
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?
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