Why is the constant positive in y = 1/(6x-x²+13)?

[UrbanXrisis]

f(3)=1/4

$$\int\frac{dy}{y}=\int (6-2x)dx$$
$$-\frac{1}{y}=6x-x^2+C$$
$$-4=18-9+C$$
$$C=-13$$
$$-\frac{1}{y}=6x-x^2-13$$
$$y=-\frac{1}{6x-x^2-13}$$
the answer is...
$$y=\frac{1}{6x-x^2+13}$$

why?

 

[James R]

Looks like the answer is wrong.

 

[UrbanXrisis]

I think that the negative was distributed before the Constant was added.. I just don't know why?

 

[whozum]

$$\int \frac{dy}{y}$$ is not $$\frac{-1}{y}$$

 

[UrbanXrisis]

sorry, it's y^2

f(3)=1/4

$$\int\frac{dy}{y^2}=\int (6-2x)dx$$
$$-\frac{1}{y}=6x-x^2+C$$
$$-4=18-9+C$$
$$C=-13$$
$$-\frac{1}{y}=6x-x^2-13$$
$$y=-\frac{1}{6x-x^2-13}$$
the answer is...
$$y=\frac{1}{6x-x^2+13}$$

why?

 

[whozum]

$$-\frac{1}{y}=6x-x^2+C$$

$$y = -\frac{1}{6x-x^2+c}$$

$$-\frac{1}{4} = \frac{1}{9+C}$$

C = 13

 

[UrbanXrisis]

sub that back in... you get the same equation as I do...

 

[whozum]

I'm sorry, C is -13, and yeah, the one you got is correct.

 

[UrbanXrisis]

the book's answer is http://home.earthlink.net/~urban-xrisis/phy001.jpg

I don't see how they got that answer...

 

[whozum]

You wrote the answer in correct in your first post, and unless there's a new math system where

C + 9 = 4, and C = 18 theyre wrong. C is -13

 

[UrbanXrisis]

I took the answers right off of the college-board website! wow...

 

[Hippo]

Nobody's infall-yable.