anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
hartnn
  • hartnn
you can easily separate the variables in that differential equation. tried that?
anonymous
  • anonymous
having some doubt
anonymous
  • anonymous
could you help pls ?

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hartnn
  • hartnn
sure, whats the doubt?
anonymous
  • anonymous
can u try it and show me what u've got ?
hartnn
  • hartnn
i got this integral, \(\Large \int \dfrac{dN}{0.02N \times (1-0.01)N} = \int dt\) you got the same?
anonymous
  • anonymous
yes but when doing it further
hartnn
  • hartnn
(1-0.01)*0.02 is a constant ...you can bring it out of integral
anonymous
  • anonymous
how's that ?
hartnn
  • hartnn
\(\Large \int \dfrac{dN}{0.02N \times (1-0.01)N} = \int dt\) \(\dfrac{dN}{0.02N \times (1-0.01)N} = \dfrac{dN}{0.0198N^2} = \dfrac{50.5dN}{N^2}\) \(\Large 50.5 \int \dfrac{dN}{N^2} = \int dt \)
anonymous
  • anonymous
@phi please help me out...
anonymous
  • anonymous
@hartnn not understood
hartnn
  • hartnn
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?
hartnn
  • hartnn
ah, just noticed i took (1-0.01)N instead of given (1-0.01N)
hartnn
  • hartnn
\(\Large \int \dfrac{dN}{0.02N \times (1-0.01N)} = \int dt\) partial fractions for left integral!
phi
  • phi
yes, the N was in the wrong spot
hartnn
  • hartnn
1/0.02 = 50 to avoid decimals, we can multiply numerator and denominator by 100 \(\Large 50 \times 100 \int \dfrac{dN}{N \times (100-N)} = \int dt\)
phi
  • phi
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
hartnn
  • hartnn
@yajna , can you please try some steps on your own... and we can help you with each steps..
Michele_Laino
  • Michele_Laino
hint: following the good suggestion of @hartnn we have to determine 2 constants, say K and H, such that the subsequent decomposition holds: \[\Large \frac{1}{{N\left( {100 - N} \right)}} = \frac{K}{N} + \frac{H}{{100 - N}}\]
anonymous
  • anonymous
okay
Michele_Laino
  • Michele_Laino
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
Michele_Laino
  • Michele_Laino
please simplify the right side, what do you get?
Michele_Laino
  • Michele_Laino
hint: what is: K(100-N)=...?
Michele_Laino
  • Michele_Laino
apply the distributive property of multiplication over addition: K(100-N)=...?
anonymous
  • anonymous
i dont understand the way you are doing it.. i would rather prefer the partial.. i am stuck here \[\frac{ 1 }{ 0.02N( 1-0.01N) } = \frac{ A }{ 0.02N } +\frac{ B }{ 1-0.01N }\]
Michele_Laino
  • Michele_Laino
ok! It is the same: here we can write this: \[\large \begin{gathered} \frac{1}{{0.02N\left( {1 - 0.01N} \right)}} = \frac{1}{{0.02}} \times \frac{1}{{N\left( {1 - 0.01N} \right)}} = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{A}{N} + \frac{B}{{1 - 0.01N}}} \right) \hfill \\ \end{gathered} \]
anonymous
  • anonymous
okay.. but can i do it without factoring out the 1/0.02 ?
Michele_Laino
  • Michele_Laino
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: \[\large \begin{gathered} \frac{1}{{0.02N\left( {1 - 0.01N} \right)}} = \frac{1}{{0.02}} \times \frac{1}{{N\left( {1 - 0.01N} \right)}} = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{A}{N} + \frac{B}{{1 - 0.01N}}} \right) = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{{A\left( {1 - 0.01N} \right) + BN}}{{N\left( {1 - 0.01N} \right)}}} \right) \hfill \\ \end{gathered} \]
anonymous
  • anonymous
what are the values are A and B ?
Michele_Laino
  • Michele_Laino
the constants A and B are the numerators of the decomposition, as you can see
anonymous
  • anonymous
yes but did u get the value of the constants ? just to check my answer
Michele_Laino
  • Michele_Laino
we have to develop the numerator, as below: \[\begin{gathered} \frac{1}{{0.02N\left( {1 - 0.01N} \right)}} = \frac{1}{{0.02}} \times \frac{1}{{N\left( {1 - 0.01N} \right)}} = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{A}{N} + \frac{B}{{1 - 0.01N}}} \right) = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{{A\left( {1 - 0.01N} \right) + BN}}{{N\left( {1 - 0.01N} \right)}}} \right) = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{{A - 0.01AN + BN}}{{N\left( {1 - 0.01N} \right)}}} \right) \hfill \\ \end{gathered} \]
anonymous
  • anonymous
okay
Michele_Laino
  • Michele_Laino
next step is: \[\begin{gathered} \frac{1}{{0.02N\left( {1 - 0.01N} \right)}} = \frac{1}{{0.02}} \times \frac{1}{{N\left( {1 - 0.01N} \right)}} = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{A}{N} + \frac{B}{{1 - 0.01N}}} \right) = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{{A\left( {1 - 0.01N} \right) + BN}}{{N\left( {1 - 0.01N} \right)}}} \right) = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{{A - 0.01AN + BN}}{{N\left( {1 - 0.01N} \right)}}} \right) \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \frac{{A + N\left( {B - 0.01A} \right)}}{{N\left( {1 - 0.01N} \right)}} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
am I right?
anonymous
  • anonymous
yes
Michele_Laino
  • Michele_Laino
ok! So we can write: \[\frac{1}{{0.02}} \times \frac{1}{{N\left( {1 - 0.01N} \right)}} = \frac{1}{{0.02}} \times \frac{{A + N\left( {B - 0.01A} \right)}}{{N\left( {1 - 0.01N} \right)}}\]
Michele_Laino
  • Michele_Laino
or: \[\frac{1}{{N\left( {1 - 0.01N} \right)}} = \frac{{A + N\left( {B - 0.01A} \right)}}{{N\left( {1 - 0.01N} \right)}}\]
Michele_Laino
  • Michele_Laino
am I right?
Michele_Laino
  • Michele_Laino
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?
Michele_Laino
  • Michele_Laino
hint: A=1, so substituting A=1 into the second equation, we get: B-0.01*1=0 what is B?
anonymous
  • anonymous
0.01
Loser66
  • Loser66
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
Loser66
  • Loser66
I believe you can handle the rest from here, right?

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