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

Your prof was right.

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

Carl