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Bernoulli ODE question

  • Thread starter cue928
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  • #1
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So I have the following Bernoulli ODE:
x^2*y' + 2xy = 5y^3
I use an integrating factor of x^-4, my n value is 3. I am okay until I get to the very end, where I get y^-2 = (2+Cx^5)/x; the book shows y^2 = x/(2+Cx^5). Am I must missing an algebra step or did I make a mistake somewhere within?
 

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  • #2
dextercioby
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Well, y^(-2) = 1/y^2.
 
  • #3
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I see. I was multiplying by -1/2 and wondering why no sq rt on the right hand side.
 
  • #4
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Let me ask you about a logistics equation problem: I got the equation dp/dt = kp^.5 for the following problem: Time rate of change of a population is proportional to sq root of P. At time t = 0, the population numbers 100 rabbits and is increasing at rate of 20 rabbits per month. How many will there be after one year? I got a k value of 2 but I don't see where to sub in the rate of change of 20 rabbits per month?
 

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