Help with this solving this equation !!!!!

Discussion in 'Math' started by kingmort, Mar 20, 2011.

1. kingmort Thread Starter New Member

Mar 4, 2011
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X" + ( a^2 )X = 0 where X= Asin(ax) + Bcos(ax)

Cause I got a different expression with an exponential and when I use maclaurin to approximate my expression and the supposed function , I found them to be different .
Help would be greatly appreciated !!!!

2. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
This might help

$X=Asin(ax)+Bcos(ax)$

$X'=aAcos(ax)-aBsin(ax)$

$X"=-a^2Asin(ax)-a^2Bcos(ax)$

$a^2X=a^2Asin(ax)+a^2Bcos(ax)$

$X"+a^2X=?$

SaraSa likes this.
3. kingmort Thread Starter New Member

Mar 4, 2011
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0
hi , thanks for the reply , however would you mind trying to solve the differential equation above ? i m trying to see if i did the solved it wrongly ... thanks !!!

4. kingmort Thread Starter New Member

Mar 4, 2011
4
0
i mean solved it wrongly

Mar 4, 2011
4
0