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Is anyone awake who could calculate the second derivative...

of y=Ae^(-3x)+Bxe^(-3x)+x-1

I've got an answer but my friend has something slightly different....


(A,B are arbitrary constants)
Reply 1
Avatar for bubba.ok
bubba.ok
OP
15
oops forgot to post what i got

9Ae^(-3x)-9Bxe^(-3x)-3Be^(3x)


She got the same but with a -6Be^(3x) at the end instead...
Yeah ermmmmmm. im awake but ermmmmmm. i don't know anything about mathematics. sorry.
Reply 3
Avatar for cj_134
cj_134
9
Original post by bubba.ok
of y=Ae^(-3x)+Bxe^(-3x)+x-1

I've got an answer but my friend has something slightly different....


(A,B are arbitrary constants)


Wolfram Alpha, my freind.
Reply 4
Avatar for jimjimmeny
jimjimmeny
I think in my incredibly tired state that your friend is correct.

dy/dx= -3Ae^(-3x) + Be^(-3x)-3Bxe^(-3x)+1
d2y/dx2=9Ae^(-3x) -3Be^(-3x)-3Be^(-3x)+9Bxe^(-3x)

=9Ae^(-3x) + 9Bxe^(-3x) - 6Be^(-3x)

Hope that helps/is correct

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