Can initial conditions for an ODE be given by functions instead of constants?

[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]

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

 

[SeM]

y'' + iy' = 0

 

[BvU]

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]

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.

 

[Mark44]

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 $y = -\frac i C e^{-it} + D = -\frac i C (\cos(t) - i\sin(t)) + D$.

 

[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

 

[mfb]

As C is an arbitrary constant anyway, this can be written shorter as $y=Ce^{-it} + D$.

 

[Mark44]

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.