Partial fractions to solve ODE's

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Homework Help Overview

The discussion revolves around using partial fractions to solve a first-order ordinary differential equation (ODE) of the form (dx/dt) = 3x(x-5) with the initial condition x(0) = 2. Participants are exploring the application of partial fraction decomposition and the subsequent steps in solving the ODE.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss their attempts to apply partial fractions, with varying results for constants A and B. There are questions regarding the correctness of their values and the steps leading to the solution. Some express confusion about the relationship between their derived expressions and the book's answer.

Discussion Status

The discussion is active, with participants sharing their calculations and questioning each other's results. There is no clear consensus on the values of A and B, and multiple interpretations of the steps involved are being explored. Some participants have provided alternative forms of the equations, suggesting different approaches to the problem.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. There is an ongoing discussion about the assumptions made in their calculations and the potential for errors in sign or constant values.

cue928
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So I am being asked to use partial fractions to solve ODE's. In particular, I have (dx/dt) = 3x(x-5) with x(0) = 2.

I came up with A = -1/5 and B = 15.

Then, I had (1/15)*[(x-5)/(3x)] = Be^t. But this is where things break down. I thought B = -1/2. Yet the book's answer is 10/(2+3e^15t). What am I missing?
 
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cue928 said:
So I am being asked to use partial fractions to solve ODE's. In particular, I have (dx/dt) = 3x(x-5) with x(0) = 2.

I came up with A = -1/5 and B = 15.
I get A = -1/5 and B = +1/5.

When you separate the diff. eqn., it's a little more convenient to write it as
dx/(x(x - 5)) = 3dt
rather than having a denominator of 3x(x - 5) on the left side.

I also get x(t) = 10/(2 + 3e^(15t))
cue928 said:
Then, I had (1/15)*[(x-5)/(3x)] = Be^t. But this is where things break down. I thought B = -1/2. Yet the book's answer is 10/(2+3e^15t). What am I missing?
 
Okay so I tried that and after using PF's and exponentiating, I get the following:
(5-x)/x = Ce^15t.

When I solve that for x, I get x = 5/(1+Ce^15t). Yikes, where am I going wrong? I ran thru it twice and still don't see where I'm missing the 3 and 10?
 
cue928 said:
Okay so I tried that and after using PF's and exponentiating, I get the following:
(5-x)/x = Ce^15t.
You have switched a sign somewhere. I get (x - 5)/x = Ae^(15t), which is a little different from what you got. I also found that A = -3/2.

cue928 said:
When I solve that for x, I get x = 5/(1+Ce^15t). Yikes, where am I going wrong? I ran thru it twice and still don't see where I'm missing the 3 and 10?
 
So I'm confused, is A=3/2 or the 1/5? I'm getting 1/5?
 
Sorry, I meant to write (x - 5)/x = Ce^(15t). C = - 3/2.

1/x + 1/(x - 5) = (-1/5)/x + (1/5)/(x - 5)
 

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