Refresh my calculus for general solution of Aexp(-ikx)

[Chaste]

Homework Statement

If \psi(x) = Ae^-ikx
then the general solution \psi(x) = Asin(kx) + Bsin(kx)

May I know if this is ODE? or? sorry, I need a refresher on my calculus.

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Char. Limit]

This is mainly done by rewriting e^(i u) as cos(u) + i sin(u), I think.

 

[Chaste]

This is mainly done by rewriting e^(i u) as cos(u) + i sin(u), I think.

so is it a first order ODE? and may I ask how does rewriting derive at the general solution? Please advise.

 

[HallsofIvy]

There was NO differential equation in anything you wrote!

It is true that "Asin(kx)+ Bcos(kx)" is the general solution of the second order differential equation y''+ k^2y= 0. Other than that, I can't see what you are asking.

 

[Chaste]

There was NO differential equation in anything you wrote!

It is true that "Asin(kx)+ Bcos(kx)" is the general solution of the second order differential equation y''+ k^2y= 0. Other than that, I can't see what you are asking.

alright, let me rephrase my question more clearly this time.
How does Ae^-ikx transform into Asin(kx)+ Bcos(kx)?

 

[Char. Limit]

alright, let me rephrase my question more clearly this time.
How does Ae^-ikx transform into Asin(kx)+ Bcos(kx)?

Let e^-ikx = cos(kx) - i sin(kx). Then Ae^-ikx = Acos(kx) - i Asin(kx). Now let B=-iA. Then Ae^-ikx = Acos(kx)+Bsin(kx).

And the first statement is Euler's Theorem, or something like that. It's easily proven using the Taylor series for the two functions.

 

[Chaste]

Let e^-ikx = cos(kx) - i sin(kx). Then Ae^-ikx = Acos(kx) - i Asin(kx). Now let B=-iA. Then Ae^-ikx = Acos(kx)+Bsin(kx).

And the first statement is Euler's Theorem, or something like that. It's easily proven using the Taylor series for the two functions.

Thanks! I see it now!

 

[Char. Limit]

No problem, glad I could help!