Help with this solving this equation !!!!!

Thread Starter

kingmort

Joined Mar 4, 2011
4
Please help me to prove that

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


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

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

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

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

\(X"+a^2X=?\)
 

Thread Starter

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 !!!
 
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