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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
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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hartnn
 one year ago
Best ResponseYou've already chosen the best response.2you can easily separate the variables in that differential equation. tried that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0could you help pls ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can u try it and show me what u've got ?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2i got this integral, \(\Large \int \dfrac{dN}{0.02N \times (10.01)N} = \int dt\) you got the same?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes but when doing it further

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2(10.01)*0.02 is a constant ...you can bring it out of integral

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2\(\Large \int \dfrac{dN}{0.02N \times (10.01)N} = \int dt\) \(\dfrac{dN}{0.02N \times (10.01)N} = \dfrac{dN}{0.0198N^2} = \dfrac{50.5dN}{N^2}\) \(\Large 50.5 \int \dfrac{dN}{N^2} = \int dt \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@phi please help me out...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@hartnn not understood

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2you got those integrals right ? so you didn't get how \(0.02N \times (10.01)N = 0.098N^2\) ? or what exactly did u not understand?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2ah, just noticed i took (10.01)N instead of given (10.01N)

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2\(\Large \int \dfrac{dN}{0.02N \times (10.01N)} = \int dt\) partial fractions for left integral!

phi
 one year ago
Best ResponseYou've already chosen the best response.0yes, the N was in the wrong spot

hartnn
 one year ago
Best ResponseYou've already chosen the best response.21/0.02 = 50 to avoid decimals, we can multiply numerator and denominator by 100 \(\Large 50 \times 100 \int \dfrac{dN}{N \times (100N)} = \int dt\)

phi
 one year ago
Best ResponseYou've already chosen the best response.0If you don't know how to do partial fraction expansion, when you have time, see https://www.khanacademy.org/math/algebra2/polynomial_and_rational/partialfractionexpansion/v/partialfractionexpansion1

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2@yajna , can you please try some steps on your own... and we can help you with each steps..

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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2now, 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
 one year ago
Best ResponseYou've already chosen the best response.2please simplify the right side, what do you get?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2hint: what is: K(100N)=...?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2apply the distributive property of multiplication over addition: K(100N)=...?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont understand the way you are doing it.. i would rather prefer the partial.. i am stuck here \[\frac{ 1 }{ 0.02N( 10.01N) } = \frac{ A }{ 0.02N } +\frac{ B }{ 10.01N }\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2ok! 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
 one year ago
Best ResponseYou've already chosen the best response.0okay.. but can i do it without factoring out the 1/0.02 ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2Yes! 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
 one year ago
Best ResponseYou've already chosen the best response.0what are the values are A and B ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the constants A and B are the numerators of the decomposition, as you can see

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes but did u get the value of the constants ? just to check my answer

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2we have to develop the numerator, as below: \[\begin{gathered} \frac{1}{{0.02N\left( {1  0.01N} \right)}} = \frac{1}{{0.02}} \times \frac{1}{{N\left( {1  0.01N} \right)}} = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{A}{N} + \frac{B}{{1  0.01N}}} \right) = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{{A\left( {1  0.01N} \right) + BN}}{{N\left( {1  0.01N} \right)}}} \right) = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{{A  0.01AN + BN}}{{N\left( {1  0.01N} \right)}}} \right) \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2next 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
 one year ago
Best ResponseYou've already chosen the best response.2ok! 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
 one year ago
Best ResponseYou've already chosen the best response.2or: \[\frac{1}{{N\left( {1  0.01N} \right)}} = \frac{{A + N\left( {B  0.01A} \right)}}{{N\left( {1  0.01N} \right)}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2Now 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
 one year ago
Best ResponseYou've already chosen the best response.2hint: A=1, so substituting A=1 into the second equation, we get: B0.01*1=0 what is B?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0If you have partial fraction as \[\dfrac{1}{0.02N(10.01N}=\dfrac{A}{0.02N}+\dfrac{B}{10.01N}\] then, by multiple the first term by (10.01N) and the second term by 0.02N, you have \[A(10.01N) +B(0.02N) =1\] If N=0, then B(0.02N) =0, that gives us \(A (10.01*0) =1\), hence A =1 If N = 100, then \(A(10.01N) = 0\) , hence \(B(0.02*100) = 2B =1\), that gives us B = 1/2 Now, plug back

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I believe you can handle the rest from here, right?
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