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Partial fractions to solve ODE's

  • Thread starter cue928
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  • #1
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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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  • #2
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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.
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))
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?
 
  • #3
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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?
 
  • #4
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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.

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?
 
  • #5
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So I'm confused, is A=3/2 or the 1/5? I'm getting 1/5?
 
  • #6
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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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