Why Does the Eigenvalue Equation Use Positive Lambda in the Solution?

[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?

 

[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.

 

[NINHARDCOREFAN]

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

 

[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..

 

[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