Solve for y: F(3) = 1/4; Integration: f(3) = 1/4; y = 1/(6x-x^2+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.