can anyone help me solve for x(t)? (dx/dt)=kx(n-x)

Question

anonymous · Answer

i am really not sure on what to do

anonymous · Answer

ok i have an answer but i am unsure if it is right

anonymous · Answer

i got $$x(t)= (e^t-n^{1/t})/(k-1^{1/k})$$

amistre64 · Answer

separate it up right?

anonymous · Answer

correct

amistre64 · Answer

$$dx/dt=kx(n-x)$$
$$x(n-x)^{-1}dx =kdt$$
$$\int x(n-x)^{-1}dx =kt +c$$
gotta see what we can do abt the left over there

amistre64 · Answer

int by parts might be best, or you got something else in mind?

anonymous · Answer

int by parts is fine. we have just learned that but was unsure how to get it to work out

anonymous · Answer

the follow up question is suppose t is in days, k = 1/250, and suppose 2 people start a rumor at t = to 0 in a population of n=1000

amistre64 · Answer

u = x       v=ln(n-x)
du = dx    dv = (n-x)^-1

$$\int udv=uv-\int vdu$$
$$\int x(x-1)^{-1}dx=xln(n-x)-\int ln(n-x)dx $$

anonymous · Answer

when will havlf the population will hear the rumor. i should be able to do that part but idk if the numbers help at all

amistre64 · Answer

say; r = n-x
      dr =  -dx

$$-\int -ln(r)dr=rln(r)-r+c$$

$$(n-x)ln(n-x)-(n-x)+c\to (n-x)(ln(n-x)-1)+c$$

amistre64 · Answer

$$\int x(x-1)^{-1}dx=xln(n-x)+(n-x)\left(ln(n-x)-1 \right)+C$$

amistre64 · Answer

gotta dbl chk meself ...

anonymous · Answer

so you are saying you think that x(t)=xln(n−x)+(n−x)(ln(n−x)−1)+C? right?

amistre64 · Answer

thats what im thinking but i tend to miss up on the simple stuff ... the wolf gives a different answer i think

amistre64 · Answer

http://www.wolframalpha.com/input/?i=integrate+x%28n-x%29%5E%28-1%29+dx im not sure if its just a simplified version, or if i did it wring :)

anonymous · Answer

lol ok should there be a k?

amistre64 · Answer

this is just the left side; the k is over there onthe right with the t http://www.wolframalpha.com/input/?i=xln%28n-x%29-%28integrate+ln%28n-x%29dx%29

anonymous · Answer

oh ok i see

amistre64 · Answer

http://www.wolframalpha.com/input/?i=xln%28n%E2%88%92x%29%2B%28n%E2%88%92x%29%28ln%28n%E2%88%92x%29%E2%88%921%29 yeah, i think it says that simplifies in the end to what they got ;)

anonymous · Answer

ahhh i cant solve for c in the answer wolfram gives

amistre64 · Answer

soo lets just use what they got :)

-n ln(n-x) -x = kt +c

$$ln(n-x)^{-n} -x = kt +c$$

why yes, the wolf will take the original and give you a solution to solve :)

amistre64 · Answer

http://www.wolframalpha.com/input/?i=+%28dx%2Fdt%29%3Dkx%28n-x%29 ewww thats ugly

anonymous · Answer

haha ok i can see that :) thank you. any guess on how to solve it?

anonymous · Answer

ya but we know when t= 0 so i think it simplifies to 2=((1000e^(1000c1))/(e^(1000c1)-1)

amistre64 · Answer

i see a place i went wrong at; i forgot to under the x ...

$$\int x(n-x)^{-1}dx =kt +c$$
spose to be
$$\int (x(n-x))^{-1}dx =kt +c$$

amistre64 · Answer

partial fraction decomp goes:
$$\frac{1}{x(n-x)}=\frac{A}{x}+\frac{B}{n-x}$$
$$1=A(n-x)+Bx$$$$when~x=0; A=\frac1n$$$$when~x=n; B=\frac1n$$
$$\int \frac{1}{x(n-x)}dx=\int \frac{1/n}{x}+\frac{1/n}{n-x} dx=\frac{ln(x)}{n}-\frac{ln(n-x)}{n-x}$$

anonymous · Answer

ohhh i got something close to that lol

amistre64 · Answer

finding x(t) explicitly looks like a challenge now

anonymous · Answer

a little bit of one lol

amistre64 · Answer

$$\frac{ln(x)}{n}-\frac{ln(n-x)}{n-x}=kt+C$$
$$ln(x^{n^{-1}}) -ln(n-x)^{(n-x)^{-1}}=kt+C$$
$$ln(\frac{x^{n^{-1}}}{(n-x)^{(n-x)^{-1}}}) =kt+C$$
$$exp\left(ln(\frac{x^{n^{-1}}}{(n-x)^{(n-x)^{-1}}}) =kt+C\right)$$
$$\frac{x^{n^{-1}}}{(n-x)^{(n-x)^{-1}}} =exp(kt+C)$$
$$\frac{x^{n^{-1}}}{(n-x)^{(n-x)^{-1}}} =exp(kt)C$$
$$\frac{x^{n^{-1}}}{e^{kt}(n-x)^{(n-x)^{-1}}} =C$$
now what did i mess up?

amistre64 · Answer

i solved for C and not x(t) lol

anonymous · Answer

lol wow i hate this problem so much i have to go in like 10 mins to turn this in lol

amistre64 · Answer

nothing like waiting for the last possible moment  :)

you sure this is calc2? intro to diffy qs?

anonymous · Answer

nope its calc 2 lol my friend said the same thing. i had another test this week so i have been doing that lol. umm he gave a hint that said log identities are your friend

amistre64 · Answer

they are, and thats what ive been trying to use but i think my partial decomp might be off

anonymous · Answer

haha its ok its no big deal. im sure we will go over it in class becasue we havent dont anyting close this this

anonymous · Answer

thank your for the help though

amistre64 · Answer

$$\int \frac{1}{x(n-x)}dx=\int \frac{1/n}{x}+\frac{1/n}{n-x} dx=\frac{ln(x)}{n}-\frac{ln(n-x)}{n-x}$$

that n-x on the end is wrong :)


$$\int \frac{1}{x(n-x)}dx=\int \frac{1/n}{x}+\frac{1/n}{n-x} dx=\frac{ln(x)}{n}-\frac{ln(n-x)}{n}$$
is better

amistre64 · Answer

now it simplifies nicer

anonymous · Answer

but then the x's cancel out dont they?

amistre64 · Answer

typos persist :)

sooo

$$\frac{ln(x)}{n}-\frac{ln(n-x)}{n}=kt+C$$
$$\frac{ln(\frac{x}{n-x})}{n} =kt+C$$
$$\frac1n {ln(\frac{x}{n-x})} =kt+C$$
$${ln(\frac{x}{n-x})} =knt+C$$
$$\frac{x}{n-x} =Ce^{knt}$$
$$x=Ce^{knt}(n-x)$$
$$x=Cne^{knt}-xCe^{knt}$$
$$x+xCe^{knt}=Cne^{knt}$$
$$x(1+Ce^{knt})=Cne^{knt}$$
$$x=\frac{Cne^{knt}}{1+Ce^{knt} }$$

anonymous · Answer

that looks good to me :)

amistre64 · Answer

HIT PRINT AND RUN WITH IT LOL .... pinky hit a caps  :)

anonymous · Answer

lol thank you verrryyyy much