# The classic second order differential equation

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

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## Answers and Replies

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

jambaugh
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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.)

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

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

LCKurtz
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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##.