Could anyone explain to me how do we solve this second order linear differential equation ($\kappa = \text{constant}$):

$$\frac{d^2 f(x)}{dx^2}=\kappa^2 f(x)$$

It is said that general solution is:

$$f(x) = Ae^{\kappa\, x} + Be^{-\kappa\, x}$$

But how do we get it?

  • $\begingroup$ Actually this differential equation is linear. $\endgroup$ – roger Mar 29 '13 at 20:04
  • $\begingroup$ do you know about separation of variables? $\endgroup$ – Lost1 Mar 29 '13 at 20:06
  • $\begingroup$ But it doesnt have 1st derivative? $\endgroup$ – 71GA Mar 29 '13 at 20:16
  • $\begingroup$ @71GA Saying that a differential equation is linear is not saying that it only involves first order derivatives. It's saying that if you have two (or more) solutions then their addition is also a solution. That's why you need initial conditions to specify which solution is the solution for your (specific) problem, i.e., in the case of your question, specify the constants $A$ and $B$. $\endgroup$ – PML Mar 29 '13 at 20:28
  • $\begingroup$ @roger: This is probably better as a comment as opposed to an answer. Regards $\endgroup$ – Amzoti Mar 29 '13 at 20:29

Assume that our solution is in the form of $f(x)=e^{c x}$ after you put into first equation you get

$c^2 e^{c x}-\kappa^2 e^{c x}=0$

If you take $e^{c x}$ out of bracket, you get

$e^{c x} (c^2-\kappa^2)=0$

or $c^2=\kappa^2$

Now you should express $c$ as $\pm\kappa$ and insert into first equation, and of course multiply each variant by some constant $A_1$, $A_2$ and add.

  • $\begingroup$ @71GA could you read this solution? i mean symbols $\endgroup$ – dato datuashvili Mar 29 '13 at 20:26
  • $\begingroup$ I am not quite sure where your first equation originates from. Did you use $k$ instead of $\kappa$? $\endgroup$ – 71GA Mar 29 '13 at 21:13

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