Question

Write and solve the differential equation that models the verbal statement. (Use k for the constant of proportionality. Use C for the constant of integration.)
The rate of change of Q with respect to u is inversely proportional to the square of u.
dQ/du =
Q =

Answer 1

dq/du (is proportional to) 1/u^2
\[\frac{dQ}{dU} \alpha 1/u^2\]

Thus:

dq/du = k*(1/u^2)
dq/du = k(1/u^2)

Answer 2

oic so we just integrated 1/u...

Answer 3

err nmvm I guess that's what we do now...

Answer 4

it's separable... separate the variables...

Answer 5

Not quite, you don't integrate just yet. You make separate the variables, then it can be integrated.

All the U's must be on one side of the equation, all the q's on the other.

Answer 6

@chaise latex note:

\propto\[\propto\]

Answer 7

dq = k(1/u^2)du?

Answer 8

q = k(-1/u) + C?

Answer 9

So is this correect? :)

Answer 10

yup guess so :P

Answer 11

I'm almost certain this is correct.

Answer 12

if my initial is 250g how would 250108 work? The practice version is like this too.... It says 147000 as the answer, but then says y=97....?

Answer 13

(ln(125/250))/1599=k

Try using this for your k value, my bad. I think. :x

Answer 14

Because the initial quantity is 150 grams,
y = 150ekt.
Because the half-life is 1599 years,
10 = 20ek(1599)
k = 1/1599 ln(1/2)

.
So, y = 150e[ln(1/2)/1599]t.
When t = 1000, y = 150e[ln(1/2)/1599](1000) ≈ 97.24 g.
When t = 3000, y ≈ 40.86 g.

Answer 15

this is 149976 lool

Answer 16

150e[ln(1/2)/1599](1000)

Answer 17

I don't get it..

Answer 18

http://www.wolframalpha.com/input/?i=250e%5E%28%28ln%281%2F2%29%2F1599%5D%281000%29%29

After 1000 years there will be 162 grams left.
This makes sense, because after 1599 years there will be 125 grams left.

Answer 19

ah everything was e^ huh LOL :(

Answer 20

Stupid crap :(.