The classic second order differential equation

[M. next]

If we have \frac{1}{X(x)} \frac{d^2 X}{dx^2}=-κ^2, the literature is saying that the solution must be: e^(±iκx), but am always getting e^(±k^2x).

Isn't the approach is to decently integrate twice and then raise the ln by the exponential? Where am I going wrong? Thanks

 

[arildno]

What do you mean with "integrating twice"?
Show your steps.

 

[jambaugh]

The devil is in the details... could you show your integration work?

Keep in mind that to integrate you need a differential instead of derivative format. letting Y=\frac{dX}{dx} your equation is of the form:
\frac{1}{X}\frac{dY}{dx} = -k^2 so
dY = -k^2 X dx
Integrating doesn't work out directly:
Y(x)=k^2 \int X(x)dx
You can try different integration by parts and substitutions... but eventually you come back to the problem of finding an integration factor for the original equation:

YdY = -k^2 XYdx = -k^2 XdX
Y^2/2 = -k^2 X^2/2
Y = -ik X
dX = -ikX dx, dX/X = -ik dx
X = e^{-ikx}
(here I'm sloppy with the integration constants, you need to work them in.)

 

[M. next]

Jambaugh, thank you for the reply. But that was derivation with second order. You worked it out with first order, right?

 

[M. next]

One more question, if for x=0, X=0 then X=sinkx according to your result? Or would it be isinkx? Am just confused with the imaginary number..

 

[arildno]

Show your work, and we'll tell you where you are mistaken.

 

[LCKurtz]

If we have \frac{1}{X(x)} \frac{d^2 X}{dx^2}=-κ^2, the literature is saying that the solution must be: e^(±iκx), but am always getting e^(±k^2x).

Isn't the approach is to decently integrate twice and then raise the ln by the exponential? Where am I going wrong? Thanks

It is just a constant coefficient equation $X''+k^2X=0$ with characteristic equation $r^2+k^2=0$. So $r=\pm ik$.