What is the solution for (1/N+(0.0004)/(1-0.0004N))dN=dt ?

The solution for (1/N+(0.0004)/(1-0.0004N))dN=dt is t=ln(N)-ln(0.0004N-1)+c_{1}

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(1/N+(0.0004)/(1-0.0004N))dN=dt

(\frac{1}{N} + \frac{0.0004}{1 - 0.0004 N}) d N = d t

t = ln (N) - ln (0.0004 N - 1) + c_{1}

Solution steps

(\frac{1}{N} + \frac{0.0004}{1 - 0.0004 N}) d N = d t

Let

t

be the dependent variable. Divide by

d N

\frac{1}{N} + \frac{0.0004}{1 - 0.0004 N} = \frac{d t}{d N}

Substitute

\frac{d t}{d N}

with

t^{'} (N)

\frac{1}{N} + \frac{0.0004}{1 - 0.0004 N} = t^{^{'}} (N)

Isolate

t^{'} (N) : t^{'} (N) = \frac{1}{N ( - 0.0004 N + 1 )}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

t = \int \frac{1}{N ( - 0.0004 N + 1 )} d N

\int \frac{1}{N ( - 0.0004 N + 1 )} d N = ln (N) - ln (0.0004 N - 1) + c_{1}

t = ln (N) - ln (0.0004 N - 1) + c_{1}

x^{2} - 10 x + 24 > 0

\frac{5 x - 4}{4} = \frac{x - 5}{5}

f a c t or 5 a^{6} - 16 a^{5} - 45 a^{4}

\frac{2}{5} (5 x - 1) + \frac{3}{10} (10 x - 3) = - \frac{1}{2} (x - 2) - \frac{6}{5}

r^{2} - r - 30 = 0

What is the solution for (1/N+(0.0004)/(1-0.0004N))dN=dt ?
The solution for (1/N+(0.0004)/(1-0.0004N))dN=dt is t=ln(N)-ln(0.0004N-1)+c_{1}