1d free particle. How do I find the solution to the DE?

[richyw]

Homework Statement

Sorry if this should be in intro section. I'm not sure the math required. So I have
$$\frac{d^2\psi}{dx^2}=-\frac{2mE}{\hbar^2}\psi$$

I would like to see how I can get the solution that my book gives me which I know is
$$\psi(x)=A\sin kx+B\cos kx$$

where A and B are constant and k is 2mE/h^2

Homework Equations

shown above

The Attempt at a Solution

Is this easy? I am rusty on my differential equations and keep getting the wrong answer. Would it be better to split it into a system of differential equations?

 

[Mute]

Homework Statement

Sorry if this should be in intro section. I'm not sure the math required. So I have
$$\frac{d^2\psi}{dx^2}=-\frac{2mE}{\hbar^2}\psi$$

I would like to see how I can get the solution that my book gives me which I know is
$$\psi(x)=A\sin kx+B\cos kx$$

where A and B are constant and k is 2mE/h^2

Homework Equations

shown above

The Attempt at a Solution

Is this easy? I am rusty on my differential equations and keep getting the wrong answer. Would it be better to split it into a system of differential equations?

How are you trying to solve it? It would help if you show us some work so we can get an idea of where you're going wrong.

To give you some hints, you probably don't need to split it up into two first order differential equations. Does the term 'characteristic polynomial' ring any bells? That's the typical method used to solve an autonomous ODE like this. (You'll have to be familiar with complex numbers to use this method)

 

[richyw]

I just searched my textbook from my intro ODE course and characteristic polynomial is mentioned once in the systems of ODE section. I remember what a characteristic equation is from linear algebra...

 

[richyw]

ok I think I know where I went wrong. I accidentaly wrote down y''=-ky'. Ill try it again real quick.

 

[richyw]

aha, that was my problem HAHA.

just out of curiosity in this case would you call the "characteristic polynomial" the part that you have to solve for zero? like in this case $(\lambda^2+k)$

 

[richyw]

oh my god :O, If I write it as a system it IS the characteristic polynomial from linear algebra!

 

[richyw]

thank you very much for your help!