anonymous 4 years ago do initial conditions have to be where t=0,

1. anonymous

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

2. anonymous

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

3. anonymous

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.

4. TuringTest

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.

5. anonymous

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

6. TuringTest

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...

7. TuringTest

I think BVP's are at least second order

8. anonymous

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.

9. TuringTest

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.

10. anonymous

thanks

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