1. Mar 15, 2006

### NINHARDCOREFAN

$$\frac{\partial^2 \phi}{\partial x^2} + \lambda \phi = 0$$

We have to analyze eigenvalues.

My prof. gave me this answer for $$\lambda>0$$ it is
$$c*sin(\sqrt(\lambda))x+d*cos(\sqrt(\lambda))x$$

Shouldn't it be -$$\lambda$$ inside the squareroot? If not can someone explain how he got this?

Last edited: Mar 15, 2006
2. Mar 15, 2006

### greytomato

The problem you'er presenting is homogenic.
sin and cos are natural solutions for this kind of equations.
pay attention that if you use diff on sin twice, you get -sin.
same goes for cos.
c,d = constants
you need 2 inputs to get the private solution. otherwisw it's a general solution.

3. Mar 15, 2006

### NINHARDCOREFAN

I know its a general solution but I don't understand how he got the positive $$\lambda$$ inside the squareroot.

4. Mar 15, 2006

### Hammie

Looks right to me.. for lambda > 0, the eigenvalue is imaginary.

Yes, it is $$\sqrt(\lambda)$$ or $$sqrt(/lambda))i$$

should be square root of neg lambda in the first.. give me a while to get latex right..

Last edited: Mar 15, 2006
5. Mar 15, 2006

### CarlB

If you put a minus sign in the square root, you would have to change the sin and cosine to exponential functions. I.e. $$C \exp(\sqrt{-\lambda}x)+D\exp(\sqrt{-\lambda}x).$$