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- anonymous

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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- anonymous

does the IVP dy/dt=6y^(2/3), y(1)=0 have a unique solution on an interval containing t=1?

- schrodinger

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- anonymous

first step, i have y=(2t+c)^3 is it right?

- JamesJ

Yes.

- JamesJ

and now use the initial condition to find c.

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- anonymous

that's makes it much easier, thank u

- 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

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

ok thx

- 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

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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