Question

Challenge

Answer 1

C = 250
and
k = 0.001

Answer 2

find "t" when y=100

Answer 3

"8" were introduced.
hence,
when t = 0, y = 8

Answer 4

\[
\int\frac{dy}{ky(C-y)}=\int dt
\]

Answer 5

use partial fractions on the left integral

Answer 6

partial fractions.. separate the left part to sum of two fractions.
\[\frac{A}{ky}+\frac{B}{C-y}=\frac{1}{ky(y-C)}\\
A(C-y)+B(ky)=1\\
A={1\over C}\qquad B={1\over kC}\\\;\\
\int\left[{1/C\over ky}+{1/kC\over y-C}\right]dy=\int dt
\]

Answer 7

dont worry about it. it is just a const number.

Answer 8

\[{1\over kC}\int\left[{1\over y}+{1\over y-C}\right]dy=\int dt\]

Answer 9

now, it will be just a couple of logs