A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
do initial conditions have to be where t=0,
anonymous
 4 years ago
do initial conditions have to be where t=0,

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No. "Initial conditions" is an arbitrary marker. So is t=0.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0boundary is when they give t=a, t=b. rather than y(a)=k1 , y'(a)= k2 and so on.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If 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
 4 years ago
Best ResponseYou've already chosen the best response.1There'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
 4 years ago
Best ResponseYou've already chosen the best response.0Show that the problem y'=2x, y(0)=0, y(1)=100, has no solution. Is this an initial value problem?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1It 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' setup...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1I think BVP's are at least second order

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if 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
 4 years ago
Best ResponseYou've already chosen the best response.1Yes, I think it is an IVP with two contradictory values. A firstorder BVP doesn't really make any sense since you only need one condition to find C, and boundary conditions come in pairs or more.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.