Question

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

Answer 1

you can easily separate the variables in that differential equation.

tried that?

Answer 2

having some doubt

Answer 3

could you help pls ?

Answer 4

sure, whats the doubt?

Answer 5

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

Answer 6

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

you got the same?

Answer 7

yes but when doing it further

Answer 8

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

Answer 9

how's that ?

Answer 10

\(\Large \int \dfrac{dN}{0.02N \times (1-0.01)N} = \int dt\)

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

\(\Large 50.5 \int \dfrac{dN}{N^2} = \int dt \)

Answer 11

@phi please help me out...

Answer 12

@hartnn not understood

Answer 13

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?

Answer 14

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

Answer 15

\(\Large \int \dfrac{dN}{0.02N \times (1-0.01N)} = \int dt\)

partial fractions for left integral!

Answer 16

yes, the N was in the wrong spot

Answer 17

1/0.02 = 50

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

Answer 18

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

Answer 19

@yajna , can you please try some steps on your own...
and we can help you with each steps..

Answer 20

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

Answer 21

okay

Answer 22

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

Answer 23

please simplify the right side, what do you get?

Answer 24

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

Answer 25

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

Answer 26

i dont understand the way you are doing it.. i would rather prefer the partial.. i am stuck here

\[\frac{ 1 }{ 0.02N( 1-0.01N) } = \frac{ A }{ 0.02N } +\frac{ B }{ 1-0.01N }\]

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

what are the values are A and B ?

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

okay

Answer 35

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

Answer 36

am I right?

Answer 37

yes

Answer 38

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

Answer 39

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

Answer 40

am I right?

Answer 41

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?

Answer 42

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

Answer 43

0.01

Answer 44

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

Answer 45

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