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frank0520

  • one year ago

Use separations of variables to solve the Differential Equation: K dN/dt = -r(N-K)(N-A) After doing partial fractions and the integration I get: (N-A)/(N-K) = C_1 e^((A-K)(-rt)/K) I am stuck solving for N

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  1. frank0520
    • one year ago
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    \[\frac{ N-A }{N-K }=C_1e^{\frac{ (A-K)(-rt) }{ K }}\] I am stuck in this part, solving for N

  2. zepdrix
    • one year ago
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    \[\large\rm \frac{ N-A }{N-K }=c_1e^{stuff}\]Multiply both sides by (N-K),\[\large\rm N-A=(N-K)c_1 e^{stuff}\]

  3. zepdrix
    • one year ago
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    Distribute,\[\large\rm N-A=N c_1 e^{stuff}-K c_1 e^{stuff}\]Let's subtract N to move it to the right side, and add Kc_1 e^(stuff) to the other side by adding,\[\large\rm K c_1 e^{stuff}-A=N c_1 e^{stuff}-N\]Then factor an N out of each term on the right side,\[\large\rm K c_1 e^{stuff}-A=N (c_1 e^{stuff}-1)\]And divide to isolate your N,\[\large\rm \frac{K c_1 e^{stuff}-A}{c_1 e^{stuff}-1}=N\]

  4. zepdrix
    • one year ago
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    If your K is just a constant, you can probably just absorb it into the c.\[\large\rm \frac{c_2 e^{stuff}-A}{c_1 e^{stuff}-1}=N\]

  5. frank0520
    • one year ago
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    Thanks for the help.

  6. zepdrix
    • one year ago
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    :D

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