## 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

1. hartnn

you can easily separate the variables in that differential equation. tried that?

2. anonymous

having some doubt

3. anonymous

could you help pls ?

4. hartnn

sure, whats the doubt?

5. anonymous

can u try it and show me what u've got ?

6. hartnn

i got this integral, $$\Large \int \dfrac{dN}{0.02N \times (1-0.01)N} = \int dt$$ you got the same?

7. anonymous

yes but when doing it further

8. hartnn

(1-0.01)*0.02 is a constant ...you can bring it out of integral

9. anonymous

how's that ?

10. hartnn

$$\Large \int \dfrac{dN}{0.02N \times (1-0.01)N} = \int dt$$ $$\dfrac{dN}{0.02N \times (1-0.01)N} = \dfrac{dN}{0.0198N^2} = \dfrac{50.5dN}{N^2}$$ $$\Large 50.5 \int \dfrac{dN}{N^2} = \int dt$$

11. anonymous

12. anonymous

@hartnn not understood

13. 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?

14. hartnn

ah, just noticed i took (1-0.01)N instead of given (1-0.01N)

15. hartnn

$$\Large \int \dfrac{dN}{0.02N \times (1-0.01N)} = \int dt$$ partial fractions for left integral!

16. phi

yes, the N was in the wrong spot

17. hartnn

1/0.02 = 50 to avoid decimals, we can multiply numerator and denominator by 100 $$\Large 50 \times 100 \int \dfrac{dN}{N \times (100-N)} = \int dt$$

18. 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

19. hartnn

20. 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: $\Large \frac{1}{{N\left( {100 - N} \right)}} = \frac{K}{N} + \frac{H}{{100 - N}}$

21. anonymous

okay

22. 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

23. Michele_Laino

please simplify the right side, what do you get?

24. Michele_Laino

hint: what is: K(100-N)=...?

25. Michele_Laino

apply the distributive property of multiplication over addition: K(100-N)=...?

26. anonymous

i dont understand the way you are doing it.. i would rather prefer the partial.. i am stuck here $\frac{ 1 }{ 0.02N( 1-0.01N) } = \frac{ A }{ 0.02N } +\frac{ B }{ 1-0.01N }$

27. Michele_Laino

ok! It is the same: here we can write this: $\large \begin{gathered} \frac{1}{{0.02N\left( {1 - 0.01N} \right)}} = \frac{1}{{0.02}} \times \frac{1}{{N\left( {1 - 0.01N} \right)}} = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{A}{N} + \frac{B}{{1 - 0.01N}}} \right) \hfill \\ \end{gathered}$

28. anonymous

okay.. but can i do it without factoring out the 1/0.02 ?

29. 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: $\large \begin{gathered} \frac{1}{{0.02N\left( {1 - 0.01N} \right)}} = \frac{1}{{0.02}} \times \frac{1}{{N\left( {1 - 0.01N} \right)}} = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{A}{N} + \frac{B}{{1 - 0.01N}}} \right) = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{{A\left( {1 - 0.01N} \right) + BN}}{{N\left( {1 - 0.01N} \right)}}} \right) \hfill \\ \end{gathered}$

30. anonymous

what are the values are A and B ?

31. Michele_Laino

the constants A and B are the numerators of the decomposition, as you can see

32. anonymous

yes but did u get the value of the constants ? just to check my answer

33. Michele_Laino

we have to develop the numerator, as below: $\begin{gathered} \frac{1}{{0.02N\left( {1 - 0.01N} \right)}} = \frac{1}{{0.02}} \times \frac{1}{{N\left( {1 - 0.01N} \right)}} = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{A}{N} + \frac{B}{{1 - 0.01N}}} \right) = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{{A\left( {1 - 0.01N} \right) + BN}}{{N\left( {1 - 0.01N} \right)}}} \right) = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{{A - 0.01AN + BN}}{{N\left( {1 - 0.01N} \right)}}} \right) \hfill \\ \end{gathered}$

34. anonymous

okay

35. Michele_Laino

next step is: $\begin{gathered} \frac{1}{{0.02N\left( {1 - 0.01N} \right)}} = \frac{1}{{0.02}} \times \frac{1}{{N\left( {1 - 0.01N} \right)}} = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{A}{N} + \frac{B}{{1 - 0.01N}}} \right) = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{{A\left( {1 - 0.01N} \right) + BN}}{{N\left( {1 - 0.01N} \right)}}} \right) = \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \left( {\frac{{A - 0.01AN + BN}}{{N\left( {1 - 0.01N} \right)}}} \right) \hfill \\ \hfill \\ = \frac{1}{{0.02}} \times \frac{{A + N\left( {B - 0.01A} \right)}}{{N\left( {1 - 0.01N} \right)}} \hfill \\ \end{gathered}$

36. Michele_Laino

am I right?

37. anonymous

yes

38. Michele_Laino

ok! So we can write: $\frac{1}{{0.02}} \times \frac{1}{{N\left( {1 - 0.01N} \right)}} = \frac{1}{{0.02}} \times \frac{{A + N\left( {B - 0.01A} \right)}}{{N\left( {1 - 0.01N} \right)}}$

39. Michele_Laino

or: $\frac{1}{{N\left( {1 - 0.01N} \right)}} = \frac{{A + N\left( {B - 0.01A} \right)}}{{N\left( {1 - 0.01N} \right)}}$

40. Michele_Laino

am I right?

41. 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?

42. Michele_Laino

hint: A=1, so substituting A=1 into the second equation, we get: B-0.01*1=0 what is B?

43. anonymous

0.01

44. 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

45. Loser66

I believe you can handle the rest from here, right?