how to solve
(d^(2) y/dx^2) + k^(2) y = 0??

Question

how to solve
(d^(2) y/dx^2) + k^(2) y = 0??

anonymous · Answer

let y=Acosx+Bsinx

isabel☺ · Answer

DE?

anonymous · Answer

sorry I responded but somehow posting is missing

anonymous · Answer

you can start by writing down the characteristics equation

anonymous · Answer

m^2+k^2=0
so m=+-ki
therefore y=Acoskx+Bsinkx
I left out the k in the initial response

isabel☺ · Answer

okay..

isabel☺ · Answer

i just don't know how :(

ganeshie8 · Answer

which part is confusing ?

isabel☺ · Answer

i don't know how to solve DE like those.

ganeshie8 · Answer

So basically you want to solve a second order linear homogeneous DE of form :

$\large    y'' + Ay' + By = 0$

isabel☺ · Answer

yes.. please.. :)

ganeshie8 · Answer

Asume the required solution looks like below :
$\large y = c_1 y_1 + c_2 y_2$

ganeshie8 · Answer

where $c_1$  and $c_2$ are arbitrary constants,
$y_1$ and $y_2$ are two different solutions to the DE.

ganeshie8 · Answer

So our goal is to simply find two solutions $y_1$ and $y_2$
then plug them in above ^^

ganeshie8 · Answer

Okay, so far ? :)

ganeshie8 · Answer

Notice that two arbitrary constants occur because it is a second order DE.

isabel☺ · Answer

may you show me your solution please>?

ganeshie8 · Answer

solution is easy :

$\large y = c_1  \sin (kt) +  c_2 \cos (kt)   $

isabel☺ · Answer

and then?

ganeshie8 · Answer

thats the solution ^^

isabel☺ · Answer

ohh!! hihi thanks!
wait what's the difference bet. ORDER and DEGREE of a DE?

ganeshie8 · Answer

order  =   highest `derivative` in the DE

ganeshie8 · Answer

degree =  power of the highest derivative

ganeshie8 · Answer

for example :

$$\large \left(\dfrac{d^3y}{dx^3}\right)^2 +   \dfrac{dy}{dx} =  1$$

ganeshie8 · Answer

order = 3
degree = 2

isabel☺ · Answer

ahhhhhh!!!! THANKS!! ☺