Question

Solve the initial value problem: dx/dt = 10x-x^2 , x(0) =1

Answer 1

use separable variable method

dx/ (10x-x^2)=dt

integrate and find constant C

Answer 2

i should say constant C(t)

Answer 3

Before integrating I got \[dt = \frac{ 1 }{ 10x } + \frac{ 1 }{ 10(1-x) }\] Is that right?

Answer 4

method of partial fractions or try to make it a perefect square, if thats easy for you, many methods are available,its your call :)

Answer 5

Answer in the book is \[x(t) = \frac{ 10 }{ 1+9e ^{-10t} }\]

I don't known how they got that. My answer is t(something) not x(something), and the ln's make 0 invalid so I'm completely lost.

Answer 6

how did you intergrate 10x-x^2

Answer 7

Partial fraction, see above.

Answer 8

you should split that up as x(10-x), use partial fractions
example

1/(10x-x^2)= (a/x)+(b/10-x)

Answer 9

You see the post above where I have the result of my partial fraction right?

Answer 10

no you do not, you taken out 10 as a common factor which is not the case,

Answer 11

Ok how about this: \[dt = \frac{ 1 }{ x } + \frac{ 1 }{ 10-x }\]
\[t = \ln|x| - \ln|10-x| +c\]

Answer 12

No, wait, the 10 in the denom came as a result of 10A=1 from the partial fraction process not a common factoring error.

Answer 13

b is also 1/10

Answer 14

i checked your work, and values of a and b are fine

Answer 15

k so now how do I use that to get the answer in the book posted earlier.

Answer 16

what are the values of integrals?

Answer 17

you are almost there, the book uses log properties like say ln (ab)= ln(a)+ln(b) and also ln(a)-ln(b)=ln(a/b)

Answer 18

\[t = \frac{ \ln|x| }{ 10 } - \frac{ \ln|x-10| }{ 10 }\]

Answer 19

remember you need to find value of constant c(t), put t=0 and find value of contant c(0), as you already know that x(0)=1

Answer 20

So t = 0 and x = 1?

Answer 21

tell me what do you understand by x(0)=1, ? x is a function of t, so when t=0, x(0)= 1, does that help

you will need to use this concept to find value of constant C, without that your answer is incorrect, or couple of points will be docked off on your test : )

Answer 22

i hope that makes sense : )

Answer 23

So I get: \[c(t) = \frac{ \ln|9| }{ 10 }\] Is that correct?

Answer 24

If I put that into the general soln, I get a big mess, and it looks nothing like the book answer :(

Answer 25

And if x is a function of t, why did I get t being a function of x after integrating. Where is x(t) even coming from?

Answer 26

Three hours and still on the first question, with tons of other HW to do, awesome.

Answer 27

ok, you have it up to here:\[t = \frac{ \ln|x| }{ 10 } - \frac{ \ln|10-x| }{ 10 }\](you switched the -10 and the x for some reason, but since its absolute value I guess it doesn't matter.
I would simplify this using\[\ln a-\ln b=\ln\frac ab\]\[t=\frac1{10}\ln|\frac x{10-x}|\]

Answer 28

Yes, I got that far, but the answer is in terms of x's, and the book soln is in terms of t's...

Answer 29

sorry forgot the thing we were looking for, the integration constant C\[t=\frac1{10}\ln|\frac x{10-x}|+C\]\[x(0)=1\]so plug in t=0 and x=1\[0=\frac1{10}\ln|\frac19|+C\implies C=\frac{\ln 9}{10}\]so the problem must just be the form of the answer, since you seem right about C

Answer 30

I made my calc solve for x, it gave:\[x = \frac{ 10e ^{10t} }{ e ^{10t}+9 }\]

Close, but not it..

Answer 31

I get it if I divide top and bottom by 1/e^10t, is that a valid maneuver?

Answer 32

\[t=\frac1{10}\ln|\frac x{10-x}|+\frac{\ln9}{10}=\frac1{10}\ln|\frac{9x}{10-x}|\]\[\exp(10t)=\exp\left(\ln|\frac{9x}{10-x}|\right)\]\[e^{10t}=\frac{9x}{10-x}\]\[10e^{10t}-e^{10t}x=9x\]yes it is!

Answer 33

so that's the only problem? lol
cuz I was gonna say, the answer is right

Answer 34

\[t=\frac1{10}\ln|\frac x{10-x}|+\frac{\ln9}{10}=\frac1{10}\ln|\frac{9x}{10-x}|\]\[\exp(10t)=\exp\left(\ln|\frac{9x}{10-x}|\right)\]\[e^{10t}=\frac{9x}{10-x}\]\[10e^{10t}-e^{10t}x=9x\]\[9x+e^{10t}x=10e^{10t}\]\[x=\frac{10e^{10t}}{9+e^{10t}}=\frac{10}{9e^{-10t}+1}\]

Answer 35

is that what you were looking for?

Answer 36

Your terms in the bottom are backwards to the answer, but I'm sure due to some arbitrary math rule its equivalent.

I guess it's solved, I'm not sure I know what I just did or what any of this means, but that's not your problem.

Answer 37

but the 9 is next to the e^(-10) right?

Answer 38

Yup.

Answer 39

so it's just
x=10/(1+9e^(-10t))
yes you should definitely realize that a+b=b+a, so those two are the same

Answer 40

:) I was imaging a subtraction between them.

Answer 41

ah, yeah algebra never gets to be very fun because of things like that ;)

Answer 42

K, thanks for coming to the rescue, on the the next problem.

Answer 43

Post it if you are having trouble I'll do what I can, but I am learning my own stuff too :)

Answer 44

post separately of course ...

Answer 45

Roger that.