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anonymous

  • one year ago

please help Solve the differential equation dN/dt= 0.02 N (1−0.01 N) , obtaining an expression for t in terms of N.. when N=20, t=0

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  1. hartnn
    • one year ago
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    you can easily separate the variables in that differential equation. tried that?

  2. anonymous
    • one year ago
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    having some doubt

  3. anonymous
    • one year ago
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    could you help pls ?

  4. hartnn
    • one year ago
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    sure, whats the doubt?

  5. anonymous
    • one year ago
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    can u try it and show me what u've got ?

  6. hartnn
    • one year ago
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    i got this integral, \(\Large \int \dfrac{dN}{0.02N \times (1-0.01)N} = \int dt\) you got the same?

  7. anonymous
    • one year ago
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    yes but when doing it further

  8. hartnn
    • one year ago
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    (1-0.01)*0.02 is a constant ...you can bring it out of integral

  9. anonymous
    • one year ago
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    how's that ?

  10. hartnn
    • one year ago
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    \(\Large \int \dfrac{dN}{0.02N \times (1-0.01)N} = \int dt\) \(\dfrac{dN}{0.02N \times (1-0.01)N} = \dfrac{dN}{0.0198N^2} = \dfrac{50.5dN}{N^2}\) \(\Large 50.5 \int \dfrac{dN}{N^2} = \int dt \)

  11. anonymous
    • one year ago
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    @phi please help me out...

  12. anonymous
    • one year ago
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    @hartnn not understood

  13. hartnn
    • one year ago
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    you got those integrals right ? so you didn't get how \(0.02N \times (1-0.01)N = 0.098N^2\) ? or what exactly did u not understand?

  14. hartnn
    • one year ago
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    ah, just noticed i took (1-0.01)N instead of given (1-0.01N)

  15. hartnn
    • one year ago
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    \(\Large \int \dfrac{dN}{0.02N \times (1-0.01N)} = \int dt\) partial fractions for left integral!

  16. phi
    • one year ago
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    yes, the N was in the wrong spot

  17. hartnn
    • one year ago
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    1/0.02 = 50 to avoid decimals, we can multiply numerator and denominator by 100 \(\Large 50 \times 100 \int \dfrac{dN}{N \times (100-N)} = \int dt\)

  18. phi
    • one year ago
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    If you don't know how to do partial fraction expansion, when you have time, see https://www.khanacademy.org/math/algebra2/polynomial_and_rational/partial-fraction-expansion/v/partial-fraction-expansion-1

  19. hartnn
    • one year ago
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    @yajna , can you please try some steps on your own... and we can help you with each steps..

  20. Michele_Laino
    • one year ago
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    hint: following the good suggestion of @hartnn we have to determine 2 constants, say K and H, such that the subsequent decomposition holds: \[\Large \frac{1}{{N\left( {100 - N} \right)}} = \frac{K}{N} + \frac{H}{{100 - N}}\]

  21. anonymous
    • one year ago
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    okay

  22. Michele_Laino
    • one year ago
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    now, developing the right side of that expression, we get: \[\Large \frac{1}{{N\left( {100 - N} \right)}} = \frac{{K\left( {100 - N} \right) + HN}}{{N\left( {100 - N} \right)}}\] please simplify

  23. Michele_Laino
    • one year ago
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    please simplify the right side, what do you get?

  24. Michele_Laino
    • one year ago
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    hint: what is: K(100-N)=...?

  25. Michele_Laino
    • one year ago
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    apply the distributive property of multiplication over addition: K(100-N)=...?

  26. anonymous
    • one year ago
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    i dont understand the way you are doing it.. i would rather prefer the partial.. i am stuck here \[\frac{ 1 }{ 0.02N( 1-0.01N) } = \frac{ A }{ 0.02N } +\frac{ B }{ 1-0.01N }\]

  27. Michele_Laino
    • one year ago
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    ok! It is the same: here we can write this: \[\large \begin{gathered} \frac{1}{{0.02N\left( {1 - 0.01N} \right)}} = \frac{1}{{0.02}} \times \frac{1}{{N\left( {1 - 0.01N} \right)}} = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{A}{N} + \frac{B}{{1 - 0.01N}}} \right) \hfill \\ \end{gathered} \]

  28. anonymous
    • one year ago
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    okay.. but can i do it without factoring out the 1/0.02 ?

  29. Michele_Laino
    • one year ago
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    Yes! since it is a factor which multiplies the entire fraction. now we have to find the constants A, and B so we have to write this: \[\large \begin{gathered} \frac{1}{{0.02N\left( {1 - 0.01N} \right)}} = \frac{1}{{0.02}} \times \frac{1}{{N\left( {1 - 0.01N} \right)}} = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{A}{N} + \frac{B}{{1 - 0.01N}}} \right) = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{{A\left( {1 - 0.01N} \right) + BN}}{{N\left( {1 - 0.01N} \right)}}} \right) \hfill \\ \end{gathered} \]

  30. anonymous
    • one year ago
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    what are the values are A and B ?

  31. Michele_Laino
    • one year ago
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    the constants A and B are the numerators of the decomposition, as you can see

  32. anonymous
    • one year ago
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    yes but did u get the value of the constants ? just to check my answer

  33. Michele_Laino
    • one year ago
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    we have to develop the numerator, as below: \[\begin{gathered} \frac{1}{{0.02N\left( {1 - 0.01N} \right)}} = \frac{1}{{0.02}} \times \frac{1}{{N\left( {1 - 0.01N} \right)}} = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{A}{N} + \frac{B}{{1 - 0.01N}}} \right) = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{{A\left( {1 - 0.01N} \right) + BN}}{{N\left( {1 - 0.01N} \right)}}} \right) = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{{A - 0.01AN + BN}}{{N\left( {1 - 0.01N} \right)}}} \right) \hfill \\ \end{gathered} \]

  34. anonymous
    • one year ago
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    okay

  35. Michele_Laino
    • one year ago
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    next step is: \[\begin{gathered} \frac{1}{{0.02N\left( {1 - 0.01N} \right)}} = \frac{1}{{0.02}} \times \frac{1}{{N\left( {1 - 0.01N} \right)}} = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{A}{N} + \frac{B}{{1 - 0.01N}}} \right) = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{{A\left( {1 - 0.01N} \right) + BN}}{{N\left( {1 - 0.01N} \right)}}} \right) = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{{A - 0.01AN + BN}}{{N\left( {1 - 0.01N} \right)}}} \right) \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \frac{{A + N\left( {B - 0.01A} \right)}}{{N\left( {1 - 0.01N} \right)}} \hfill \\ \end{gathered} \]

  36. Michele_Laino
    • one year ago
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    am I right?

  37. anonymous
    • one year ago
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    yes

  38. Michele_Laino
    • one year ago
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    ok! So we can write: \[\frac{1}{{0.02}} \times \frac{1}{{N\left( {1 - 0.01N} \right)}} = \frac{1}{{0.02}} \times \frac{{A + N\left( {B - 0.01A} \right)}}{{N\left( {1 - 0.01N} \right)}}\]

  39. Michele_Laino
    • one year ago
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    or: \[\frac{1}{{N\left( {1 - 0.01N} \right)}} = \frac{{A + N\left( {B - 0.01A} \right)}}{{N\left( {1 - 0.01N} \right)}}\]

  40. Michele_Laino
    • one year ago
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    am I right?

  41. Michele_Laino
    • one year ago
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    Now applying the identity principle of polynomials we can write the subsequent algebraic system: \[\left\{ \begin{gathered} A = 1 \hfill \\ B - 0.01A = 0 \hfill \\ \end{gathered} \right.\] please solve for A and B, what do you get?

  42. Michele_Laino
    • one year ago
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    hint: A=1, so substituting A=1 into the second equation, we get: B-0.01*1=0 what is B?

  43. anonymous
    • one year ago
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    0.01

  44. Loser66
    • one year ago
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    If you have partial fraction as \[\dfrac{1}{0.02N(1-0.01N}=\dfrac{A}{0.02N}+\dfrac{B}{1-0.01N}\] then, by multiple the first term by (1-0.01N) and the second term by 0.02N, you have \[A(1-0.01N) +B(0.02N) =1\] If N=0, then B(0.02N) =0, that gives us \(A (1-0.01*0) =1\), hence A =1 If N = 100, then \(A(1-0.01N) = 0\) , hence \(B(0.02*100) = 2B =1\), that gives us B = 1/2 Now, plug back

  45. Loser66
    • one year ago
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    I believe you can handle the rest from here, right?

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