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Integrating a polynomial

  • Thread starter snowJT
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I Just want to know if this is how this should be done..

[tex]\frac{dy}{dx} = \frac{3y^2}{x}[/tex]

[tex]dy = \frac{3y^2}{x}dx[/tex]

[tex]dy = 3^2 x^-^1dx[/tex]

[tex]\frac {1}{3} \int \frac {dy}{y^2} = \int x^-^1dx[/tex]

[tex]\frac {1}{3} ln| y^2| = ln | x | + C[/tex]

because I forget how to integrate the polynomial...
 

Answers and Replies

cristo
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The right hand side is correct. Look again at [tex]\int y^{-2}dy[/tex] How do you integrate this?
 
AlephZero
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[tex]\frac {1}{3} \int \frac {dy}{y^2} = \int x^-^1dx[/tex]

That's OK, but the integral of 1/y^2 isn't ln(y^2).

Look up how to integrate polynomials, if you forgot. Or if you know how to differentiate polynominals, work it out from the fact that integration is "anti-differentiation".
 
118
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its not [tex]\frac{1}{3y}[/tex] is it?
 
cristo
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its not [tex]\frac{1}{3y}[/tex] is it?
Close; it's -1/(3y), since you must divide by the power of y(which is -1)
 
118
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oops.. thanks I was thinking that was how you did it originally, I was just confused and did it that other way and forgot ket things... thanks

EDIT: Hmmm I also need to then solve for Y

[tex]y^-^1 = \frac {-ln|x|}{3} - \frac {C}{3}[/tex]

[tex]y = \frac{-3}{ln|x|} - \frac{3}{C}[/tex]
 
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that looks wrong to me
 
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not even close i dont think
 
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[tex]y=\frac{ln x + C}^-1{3}[/tex]
 
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cristo
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[tex]y=\frac{ln x + C}^-1{3}[/tex]
thats what i got but im dumb so dont trust me
 
20
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oops that -13 is supposed to be an exponent of the numerator (-1) and 3 in the denominator
 
cristo
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[tex]y=\frac{(ln x + C)^{-1}}{3}[/tex]
thats what i got but im dumb so dont trust me
[tex]\frac{(ln x + C)^{-1}}{3}=\frac{1}{3(ln x + C)}[/tex] The 3 should definitely be in the numerator, and we need a - sign too. See my post:

cristo said:
[tex]y=\frac{-3}{(ln|x|+C)}[/tex]
 
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118
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[tex]y^{-1} = \frac {-ln|x|}{3} - \frac{C}{3}=\frac{-(ln|x|+C)}{3}[/tex]

[tex]y=\frac{-3}{(ln|x|+C)}[/tex]

Can you follow what I've done?
no I can't figure it out, common denominator?
 
118
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oh.. yes I see

why is C positive and not negative
 
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cristo
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no I can't figure it out, common denominator?
Yea you just write ln|x| and C as one fraction with denominator 3. Then take the reciprocal of both sides (since the LHS is 1/y) to obtain an expression in terms of y.

snowJT said:
oh! you simplified it before you swapped the fractions! is that a rule?
Well, yes. In order to take the reciprocal of both sides, there must only be one fraction on each side.
 
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looking back to the origonal i don't see how there is a negative
 
cristo
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looking back to the origonal i don't see how there is a negative
The minus sign comes about when integrating; integral(y-2)=-y-1 (+C)
 
763
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snowJT what does C mean?

it means any constant...it can be a minus but since we don't know we just write + C
 
118
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snowJT what does C mean?

it means any constant...it can be a minus but since we don't know we just write + C
figured so, there were some examples I just found in my book
 
cristo
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snowJT said:
why is C positive and not negative
Because you defined it as positive!! When you integrated you wrote +c on the RHS, then multiplied through with the minus sign.
 
763
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figured so, there were some examples I just found in my book
you could find the answer in your books if you just read with an open mind, but it feels more real, less boring and it is less likely that you'd forget the anwswer when you ask some one else
 
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cristo
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I see you've got that homework helper recognition, is that new?
 

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