Re: Linear Dependence problem

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Posted by Brad Paul on October 11, 2002 at 15:48:08:

In Reply to: Re: Linear Dependence problem posted by colin on October 11, 2002 at 12:59:46:

: : You should check my work but on my first try I got:

: : W(f,g)=B e^{2 A t}

: Brad, could you possibly show me how you got that result?

: Also, how did you markup your post to get the exponents displayed in superscript? That's cool!
f= e^Atcos(Bt)
g=e^Atsin(Bt)

f'=A e^Atcos(Bt)-B e^Atsin(Bt)
g'=A e^Atsin(Bt)+B e^Atcos(Bt)

W=f g'-g f'
W=e^2At{Asin(Bt)cos(Bt)+B cos²(Bt)-Asin(Bt)cos(Bt)+B sin²(Bt)}

W=B e^{2 AT}

To see how I get the neat formating look at the comments section where
it is just text. But be warned it can go real bad if there is a
typo. See some of my older posts to see some bad formating.

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Comments:
: : : You should check my work but on my first try I got: : : : W(f,g)=B e2 A t : : Brad, could you possibly show me how you got that result? : : Also, how did you markup your post to get the exponents displayed in superscript? That's cool! : f= eAtcos(Bt) : g=eAtsin(Bt) : f'=A eAtcos(Bt)-B eAtsin(Bt) : g'=A eAtsin(Bt)+B eAtcos(Bt) : W=f g'-g f' : W=e2At{Asin(Bt)cos(Bt)+B cos2(Bt)-Asin(Bt)cos(Bt)+B sin2(Bt)} : W=B e2 AT : : To see how I get the neat formating look at the comments section where : it is just text. But be warned it can go real bad if there is a : typo. See some of my older posts to see some bad formating.

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