Help with this solving this equation !!!!!

kingmort

Joined Mar 4, 2011
4

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 !!!!

t_n_k

Joined Mar 6, 2009
5,455
This might help

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

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

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

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

$$X"+a^2X=?$$

kingmort

Joined Mar 4, 2011
4
This might help

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

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

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

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

$$X"+a^2X=?$$
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 !!!

kingmort

Joined Mar 4, 2011
4
i mean solved it wrongly