# Integrating a polynomial

I Just want to know if this is how this should be done..

$$\frac{dy}{dx} = \frac{3y^2}{x}$$

$$dy = \frac{3y^2}{x}dx$$

$$dy = 3^2 x^-^1dx$$

$$\frac {1}{3} \int \frac {dy}{y^2} = \int x^-^1dx$$

$$\frac {1}{3} ln| y^2| = ln | x | + C$$

because I forget how to integrate the polynomial...

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cristo
Staff Emeritus
The right hand side is correct. Look again at $$\int y^{-2}dy$$ How do you integrate this?

AlephZero
Homework Helper
$$\frac {1}{3} \int \frac {dy}{y^2} = \int x^-^1dx$$

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".

its not $$\frac{1}{3y}$$ is it?

cristo
Staff Emeritus
its not $$\frac{1}{3y}$$ is it?
Close; it's -1/(3y), since you must divide by the power of y(which is -1)

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

$$y^-^1 = \frac {-ln|x|}{3} - \frac {C}{3}$$

$$y = \frac{-3}{ln|x|} - \frac{3}{C}$$

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that looks wrong to me

not even close i dont think

$$y=\frac{ln x + C}^-1{3}$$

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cristo
Staff Emeritus
EDIT: Hmmm I also need to then solve for Y
$$y^{-1} = \frac {-ln|x|}{3} - \frac{C}{3}=\frac{-(ln|x|+C)}{3}$$

$$y=\frac{-3}{(ln|x|+C)}$$

Can you follow what I've done?

that looks wrong to me
not even close i dont think
JJ420; we are here to help, not criticise!

$$y=\frac{ln x + C}^-1{3}$$
thats what i got but im dumb so dont trust me

oops that -13 is supposed to be an exponent of the numerator (-1) and 3 in the denominator

cristo
Staff Emeritus
$$y=\frac{(ln x + C)^{-1}}{3}$$
thats what i got but im dumb so dont trust me
$$\frac{(ln x + C)^{-1}}{3}=\frac{1}{3(ln x + C)}$$ The 3 should definitely be in the numerator, and we need a - sign too. See my post:

cristo said:
$$y=\frac{-3}{(ln|x|+C)}$$

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$$y^{-1} = \frac {-ln|x|}{3} - \frac{C}{3}=\frac{-(ln|x|+C)}{3}$$

$$y=\frac{-3}{(ln|x|+C)}$$

Can you follow what I've done?
no I can't figure it out, common denominator?

oh.. yes I see

why is C positive and not negative

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cristo
Staff Emeritus
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.

looking back to the origonal i don't see how there is a negative

cristo
Staff Emeritus
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)

snowJT what does C mean?

it means any constant...it can be a minus but since we don't know we just write + C

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
Staff Emeritus
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.

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

Because you defined it as positive!!
:rofl: you don't have to shout

cristo
Staff Emeritus