How do I integrate a polynomial with a variable in the denominator?

[snowJT]

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...

 

[cristo]

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

 

[AlephZero]

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

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".

 

[snowJT]

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

 

[cristo]

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

Close; it's -1/(3y), since you must divide by the power of y(which is -1)

 

[snowJT]

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

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

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

 

[JJ420]

that looks wrong to me

 

[JJ420]

not even close i don't think

 

[JJ420]

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

 

[cristo]

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 don't think

JJ420; we are here to help, not criticise!

 

[JJ420]

$$y=\frac{ln x + C}^-1{3}$$
thats what i got but I am dumb so don't trust me

 

[JJ420]

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

 

[cristo]

$$y=\frac{(ln x + C)^{-1}}{3}$$
thats what i got but I am dumb so don't 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:

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

 

[snowJT]

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

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

Can you follow what I've done?

no I can't figure it out, common denominator?

 

[snowJT]

oh.. yes I see

why is C positive and not negative

 

[cristo]

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.

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.

 

[JJ420]

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

 

[cristo]

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)

 

[sara_87]

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]

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]

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.

 

[sara_87]

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

 

[sara_87]

Because you defined it as positive!

you don't have to shout

 

[cristo]

you don't have to shout

Haha, I didn't. You'd know when I WAS SHOUTING!

 

[sara_87]

I see you've got that homework helper recognition, is that new?

 

[cristo]

I see you've got that homework helper recognition, is that new?

Yea, it just popped up yesterday.

 

[sara_87]

congratulations! I'm jelous but unlike yourself i don't deserve one
(and i won't post up anything unrelated to this thread anymore, appologies snowJT)

 

[cristo]

Thanks. Yea, this thread has got very off topic. Sorry, although I think the question is answered now!

 

[Gib Z]

C is always positive and never has a coefficient is the final result because any multiple of some constant is just another constant, and A negative value of C is allowed, we write it Positive because its easier to work with.

Edit: I always do this..didnt see this page, just the first one..omg kill me

 

[matt grime]

It doesn't matter what sign you use for C, or what letter. C is just an undetermined constant. The standard practice whenever you have undetermined constants is to write them in as simple form as possible. Thus you would choose to write Cx, not -Cx if you had the chance. You can also add them together: C+D can be rewritten as a single new constant E. Also, to save using an unnecessary number of letters being used, we often replace C+D with just another C. This is equivalent to replacing C+D with E, and then, now we've freed up the C we can reuse that symbol by using C instead of E. Note that you cannot simplify Cx+D.