What is the solution for (dy)/(y(A-y))= k/A dx ?

The solution for (dy)/(y(A-y))= k/A dx is y=-(Ae^{c_{1}k})/(e^{kx)-e^{c_{1}k}}

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(dy)/(y(A-y))= k/A dx

\frac{d y}{y ( A - y )} = \frac{k}{A} d x

y = - \frac{A e ^{c_{1} k}}{e ^{k x} - e ^{c_{1} k}}

Solution steps

\frac{d y}{y ( A - y )} = \frac{k}{A} d x

Let

x

be the dependent variable. Divide by

d y

\frac{1}{y ( A - y )} = \frac{k}{A} \cdot \frac{d x}{d y}

Substitute

\frac{d x}{d y}

with

x^{^{'}}

\frac{1}{y ( A - y )} = \frac{k}{A} x^{^{^{'}}}

Isolate

x^{^{'}} : x^{^{'}} = \frac{A}{k y ( x ) ( A - y )}

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

then

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

x = \int \frac{A}{k y ( x ) ( A - y )} d y

\int \frac{A}{k y ( x ) ( A - y )} d y = - \frac{1}{k} (- ln (\frac{y}{A}) + ln (\frac{y}{A} - 1)) + c_{1}

x = - \frac{1}{k} (- ln (\frac{y}{A}) + ln (\frac{y}{A} - 1)) + c_{1}

Simplify

x = - \frac{- ln ( \frac{y}{A} ) + ln ( \frac{y}{A} - 1 )}{k} + c_{1}

Isolate

y : y = - \frac{A e ^{c_{1} k}}{e ^{k x} - e ^{c_{1} k}}

y = - \frac{A e ^{c_{1} k}}{e ^{k x} - e ^{c_{1} k}}

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What is the solution for (dy)/(y(A-y))= k/A dx ?
The solution for (dy)/(y(A-y))= k/A dx is y=-(Ae^{c_{1}k})/(e^{kx)-e^{c_{1}k}}