I Initial conditions for an ODE

SeM

Hi, I am trying to solve an ODE, however, the initial conditions are not known. From PDE examples, which are quite different, I see that some examples have initial conditions given by functions, and not by constants, i.e::

y(0) = x^2

I may have not modeled the problem correctly yet, however, I see that the inhomogenous ODE should have a condition which is as such:

y(0) = cos(x)

MATLAB can't do it. Is it something that can't be done at all?

Thanks!
 

BvU

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Do we have to guess what ODE ? Why don't you post it ?
 

SeM

y'' + iy' = 0
 

BvU

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In other words, y(0) = 1 ?

(your ODE doesn't look inhomogeneous to me at all ?!)
 

SeM

In other words, y(0) = 1 ?

(your ODE doesn't look inhomogeneous to me at all ?!)
Thanks, I think that explains it. I see this is a PDE not an ODE I want to solve. Cheers (y(0)=cos(theta)
 

BvU

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Don't see no partial differential either. Is it me ?
 

SeM

Don't see no partial differential either. Is it me ?
It's not you, don't worry. It's me who is horribly wrong here.
 
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y'' + iy' = 0
I see this is a PDE not an ODE I want to solve.
y appears to be a function of a single variable, so the above is an ODE. BTW, the solution I get, without considering an initial condition, is ##y = -\frac i C e^{-it} + D = -\frac i C (\cos(t) - i\sin(t)) + D##.
 
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SeM

y appears to be a function of a single variable, so the above is an ODE. BTW, the solution I get, without considering an initial condition, is ##y = -\frac i C e^{-it} + D = -\frac i C (\cos(t) - i\sin(t)) + D##.
Thanks Mark!

Cheers
 
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As C is an arbitrary constant anyway, this can be written shorter as ##y=Ce^{-it} + D##.
 
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As C is an arbitrary constant anyway, this can be written shorter as ##y=Ce^{-it} + D##.
Right. I thought of this, but didn't let i be absorbed into the constant, so as to make things a bit clearer.
 
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