Question

What is the integration of dy/dt = (y/8) (6-y)

Answer 1

first separate the variables

dy/dt = (y/8)(6-y)

dy = (y/8)(6-y) dt

dy/(y/8)(6-y) = dt

int( dy/(6y/8 - y^2/8) = int(dt)

can you solve this?

Answer 2

yeah I got up to that but I don't know how to solve after would that be ln of that value on the left side ?

Answer 3

For this one I think you have to integrate by parts like so:

\[\int\limits_{}^{}1/(y/8)(6-y)dy = \int\limits_{}^{}(A/(y/8) + B/(6-y))dy\]

Have you learned this technique of integration yet?

Answer 4

Oh I was doing it like this \[\ln \left| 6y-y^2 \right| = e^{x/8+C}\]
And that's where I'm having trouble going on to the next step.

Answer 5

e^ ln on the left sorry. And then I have 6y-y^2 = e^(x/8+C)

Answer 6

I do not think that you integrated the left-hand term correctly. Try integrating by parts on the y term

Answer 7

How do I integrate (6y-y^2)^-1 dy

Answer 8

Here's how you integrate by parts:
1/(y/8)(6-y) = A/(y/8) + B/(6-y), where you solve for A and B

First, find a common denomenator
1/(y/8)(6-y) = A(6-y)/(y/8)(6-y) + B(y/8)/(6-y)(y/8)

Then get rid of the denominator by multiplying both sides by the denominator

1 = A(6-y) + B(y/8)

Then set values for y so that you can solve for A and B individually

let y = 0
1 = A(6-0) + B(0)
A = 1/6

let y = 6
1 = A(0) + B(6/8)
B = 8/6 or 4/3

plug in A and B

int( (1/6)/(y/8) + (4/3)/(6-y))dy)

this should be an integral you can solve. If you follow a textbook, they probably go into detail on how to solve these problems

Answer 9

\[
\text{we have:}\quad \int{8\over y(6-y)}dy=\int dt\\
{\rm let}\quad {A\over y}+{B\over 6-y}={8\over y(6-y)}\\
\implies A(6-y)+By=8\\
{\rm let}\quad y=0\implies 6A=8\implies \boxed{A={4\over3}}\\
{\rm let}\quad y=6\implies 6B=8\implies \boxed{B={4\over3}}\\
\text{the integrand becomes:}\\
\int \left[{4/3\over y}+{4/3\over 6-y}\right]dy+C=t\\
\]this is now readily solvable

Answer 10

thank you (:

Answer 11

yw

Answer 12

this is called integration by parts

Answer 13

Uh I think this way is called something else. Integration by parts would use the formula u*v-integral v du.