Question

What is M? dM/dt=(100-kM)
k is a constant

Answer 1

\[\Large \frac{dM}{dt} = 100 - kM\]

\[\Large dM = (100 - kM)*dt\]

\[\Large \frac{dM}{100 - kM} = dt\]

\[\Large \int\frac{dM}{100 - kM} = \int 1*dt\]

I'll let you finish

Answer 2

I got to that part. I don't know how to integrate the left side. How do I take care of the kM. I would have to multiply by -1/M right?

Answer 3

let u = 100 - kM and then derive both sides with respect to M


u = 100 - kM

du/dM = -k


and then isolate dM


du/dM = -k

du = -k*dM

-du/k = dM

dM = -du/k

Answer 4

\[\Large \int\frac{dM}{100 - kM} = \int 1dt\]

turns into

\[\Large \int \frac{-\frac{du}{k}}{u} = \int 1dt \]

\[\Large -\frac{1}{k}\int \frac{du}{u} = \int 1dt \]

\[\Large -\frac{1}{k}\int \frac{1}{u}du = \int 1dt \]

Answer 5

so is the answer -1/k*ln|100-kM|=t

Answer 6

oh + C

Answer 7

yeah

\[\Large -\frac{1}{k}ln(|100-kM|) = t+C\]

from here you solve for M

Answer 8

How do I take care of the absolute value sign?

Answer 9

if |x| = k, then x = k or x = -k for some positive number k

put another way, if |x| = k, then x = +-k

Answer 10

so I have to put +- in front of the right side?

Answer 11

Like this

\[\Large -\frac{1}{k}ln(|100-kM|) = t+C\]

\[\Large ln(|100-kM|) = -k(t+C)\]

\[\Large |100-kM| = e^{-k(t+C)}\]

\[\Large 100-kM = \pm e^{-k(t+C)}\]

Answer 12

Okay I think I understand now. What a struggle. Thanks