do initial conditions have to be where t=0,

Question

anonymous · Answer

No.

"Initial conditions" is an arbitrary marker. So is t=0.

anonymous · Answer

boundary is when they give t=a, t=b. rather than y(a)=k1 , y'(a)= k2 and so on.

anonymous · Answer

If those are the only boundaries they give you, I can't tell you anything about "initial conditions". I would guess that they mean for t=a to be the initial condition, but it's not necessarily 0 unless there's other information.

turingtest · Answer

There's not really enough info for me to understand what you're doing, but it looks to me like you have boundary values, not initial values.

anonymous · Answer

Show that the problem y'=2x, y(0)=0, y(1)=100, has no solution. Is this an initial value problem?

turingtest · Answer

It looks like a boundary value problem, but I don't think it really is. It just demonstrates that the two initial values contradict$$y'=2x$$$$y=x^2+C$$$$y(0)=C=0$$$$y(1)=1+C=100\to C=99$$so C is trying to be two different numbers at once, which is not possible.
Not sure what you call this type of 'impossible initial values' set-up...

turingtest · Answer

I think BVP's are at least second order

anonymous · Answer

if it is an initial value problem then there is no solution since c cant be two different numbers at once, where as if this was a boundary problem we just identified the boundary values for the boundary conditions.

turingtest · Answer

Yes, I think it is an IVP with two contradictory values. A first-order BVP doesn't really make any sense since you only need one condition to find C, and boundary conditions come in pairs or more.

anonymous · Answer

thanks