Question

Find the solution of \[y'=5 y(3-y)\] with \[y(0)=9\].

Answer 1

\[y'=5 y(3-y)\] with \[y(0)=9\]

Answer 2

so, how to antidifferentiate \[1/(3y-y^2)\]

Answer 3

partial fractions

Answer 4

\[\frac{1}{y(3-y)}=\frac{1}{3y}+\frac{1}{3(3-y)}\]

Answer 5

aha! partial fractions? that was fast, dude.

Answer 6

so \[\int\limits[1/3y+1/(3(3−y))] = \ln(3y)/3 - \ln(9-3y)/3\] is what i got

Answer 7

\frac{}{}

Answer 8

hey how do you get the eq editor to show top/bottom fractions like that?

Answer 9

oh cool, ty.

Answer 10

could i say \[ln(3y) - ln(9-3y) = ln(3y/(9-3y)\]?

Answer 11

yes

Answer 12

you can make it look nice by doing this

\[\ln\left(\frac{3y}{9-3y}\right)\]

\ln\left(\frac{3y}{9-3y}\right)

Answer 13

so i got \[\frac{y}{3-y} = -\frac{3}{2}*e^{15t}\]

Answer 14

looks like it will be implicit

Answer 15

that cant be right, it requires an answer in the form of y(t) = ...

Answer 16

Here is a different method. the answer you got above can be solved for y, and i think it will produce what I got. Im checking it out right now.

Answer 17

yeah, what you have above my solution produces the same solution.

Answer 18

\[\frac{y}{3-y} = -\frac{3}{2}*e^{15t}\]

\[\frac{y}{3-y} = a\]

\[y=(3-y)a\]
\[y=3a-ya\]

\[y+ya=3a\]

\[y(1+a)=3a\]

\[y=\frac{3a}{1+a}\]

Answer 19

from there, divide the numerator and denominator by 3a, and you'll get what i got.

Answer 20

WOOOOOWWWW!!!! where do you guys come up with this?!

Answer 21

black magic lol. Diff equations just seems to be a bunch of rules you have to memorize -.-
"if it looks like this, do this"
"if it looks like that, do that"

Answer 22

lol. On a related note, Im surprised the prof expects me to reach that conclusion. He never taught us at least half a dozen tricks you two just used.

Answer 23

very cute!
i wish i could have stayed up for this problem
i enjoy SOME differential problems