# The classic second order differential equation

1. Sep 28, 2013

### 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

Last edited: Sep 28, 2013
2. Sep 28, 2013

### arildno

What do you mean with "integrating twice"?

3. Sep 28, 2013

### 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.)

4. Sep 28, 2013

### M. next

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

5. Sep 28, 2013

### 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..

6. Sep 28, 2013

### arildno

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

7. Sep 29, 2013

### LCKurtz

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