anonymous
  • anonymous
Solve the differential equation(antiderivatives calculus 2) f(x)=  12x^3+ 6x^2-7 with initial condition f(1)=  4
Mathematics
jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
Hi Saif, you here?
anonymous
  • anonymous
yea i am here
anonymous
  • anonymous
its college calculus 2, antiderivatives are killin me. SO confusing

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anonymous
  • anonymous
Confused. There is no differential as written. Does "dx" multiply the polynomial?
anonymous
  • anonymous
what do you mean?
anonymous
  • anonymous
Are you asking us to just integrate/ find the antiderivative of the function f(x)?
amistre64
  • amistre64
You want to the integral and its anchor at (1,4) right?
anonymous
  • anonymous
yea i think so I thought it was just asking the antiderivative with that condition
amistre64
  • amistre64
F(x)= (12/4)x^4+ (6/3)x^3 -7x + C F(x) = 3x^4 +2x^3 -7x +C solve for F(1) = 4 :)
anonymous
  • anonymous
you are using big F(x), its giving you f'(x), wouldnt the notation change?
anonymous
  • anonymous
its not the same convention as a general antiderivative
amistre64
  • amistre64
naming is naming, I just did it to be flamboyant :)
anonymous
  • anonymous
ok so the totation would be f(x) correct and then you find f(1)
anonymous
  • anonymous
i mean you find C
amistre64
  • amistre64
We find the "antiderivative" of the function to find a family of curves that we can adopt... then we apply (1,4) into it to see what our anchoring constant turns out to be, right?
anonymous
  • anonymous
yea correct
amistre64
  • amistre64
then "suit up" and find our constant :)
anonymous
  • anonymous
i got 6
amistre64
  • amistre64
4 = 3 +2 -7 +C 4 = 5-7 + C 4 = -2 + C 6 = C F(x) = 3x^4 +2x^3 -7x +6 yay!!!
anonymous
  • anonymous
thanks so much. Can you take a look at the other antiderivative I posted. I am getting confused, he took like 2 antiderivatives?
amistre64
  • amistre64
Didja notice me new "title"..... funny, i never was a sandwich before :)
anonymous
  • anonymous
hahaahahaah
amistre64
  • amistre64
ill check em out...
anonymous
  • anonymous
its 2 below the this question

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