Diff.equation homework

416
0
The following diff.equation determines the number of people with a spesific illness:
[tex]
\frac{{dx}}{{dt}} = k\left( {150000 - x\left( t \right)} \right)
[/tex]
At t=0, 30000 people are infected, and at t=15, 60000. How long will it take for 120000 are infected? Here is my work:
[tex]
\begin{array}{l}
x\left( t \right) = 30000 \cdot e^{k \cdot t} \\
x\left( {15} \right) = 30000 \cdot e^{k \cdot 15} = 60000 \\
\Rightarrow k = 0,046 \\
x\left( t \right) = 30000 \cdot e^{0,046 \cdot t} = 120000 \\
\end{array}
[/tex]
Problem is that this leads to no good, the answer is supposed to be about t=72 (days), but I'm not sure how to implement the 150000 in the beginning (at least I think that's the prob)...anyone?
 
185
3
dx/dt = k*(p - x) has solutions of the form x = Aexp(-kt) + p,
not just x = A*exp(-k*t).
 

TD

Homework Helper
1,020
0
Solving your DE will give an extra constant and you already had the constant k. You can determine the values of these two constants with x(0) = 30000 and x(15) = 60000. Are you sure of your solution of the DE?
 
416
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You mean am I sure it's supposed to be 72? That's just what it says in the back of the book that's all...
I just want to know where in my attempt I turned wrong...
 

TD

Homework Helper
1,020
0
No, I'm talking about the solution of the differential equation.
 
416
0
No, I'm not at all sure about my solution becasue I'm not taking the 150000 into effect since I don't know where or how to do so. How would you go about solving it?
 

saltydog

Science Advisor
Homework Helper
1,579
2
TSN79 said:
No, I'm not at all sure about my solution becasue I'm not taking the 150000 into effect since I don't know where or how to do so. How would you go about solving it?
TSN, as Qbert stated, the solution to the DE is:

[tex]x(t)=Ae^{-kt}+150000[/tex]

right?

You arrange it into the form:

[tex]x^{'}+kx=150000k[/tex]

obtain an integrating factor, integrate, done deal.

So you have two unknowns in the solution above, k and A, but you have two conditions:

x(0)=30000

x(15)=60000

Substituting the first condition into the equation yields:

[tex]30000=Ae^{-0k}+150000[/tex]

you can solve for A right?

now that you have A, substitute the second condition:

[tex]60000=Ae^{-15k}+150000[/tex]

That allows you to find k then. You know, rearrange, take the ln of both sides, solve for k.

Then once you have these values, you can easily solve for what t yields 120000:

[tex]120000=Ae^{-kt}+150000[/tex]

I bet you can do this.:smile:
 

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