What is the solution for (dy)/(dx)=y(10-y)-9 ?

The solution for (dy)/(dx)=y(10-y)-9 is y=-(9e^{8x-8c_{1}})/(1-e^{8(x-c_{1))}}+1/(1-e^{8(x-c_{1))}}

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(dy)/(dx)=y(10-y)-9

\frac{d y}{d x} = y (10 - y) - 9

y = - \frac{9 e ^{8 x - 8 c_{1}}}{1 - e ^{8 (x - c_{1})}} + \frac{1}{1 - e ^{8 (x - c_{1})}}

Solution steps

\frac{d y}{d x} = y (10 - y) - 9

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = y (10 - y) - 9

Solve for

x, y^{'} (x) = \frac{d y}{d x} = \frac{1}{\frac{d x}{d y}} = \frac{1}{x ^{^{'}}}

(\frac{1}{x ^{^{^{'}}}}) = y (10 - y) - 9

Isolate

x^{^{'}} : x^{^{'}} = \frac{1}{10 y - y ^{2} - 9}

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

then

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

x = \int \frac{1}{10 y - y ^{2} - 9} d y

\int \frac{1}{10 y - y ^{2} - 9} d y = \frac{1}{8} (ln (\frac{y - 5}{4} + 1) - ln (\frac{y - 5}{4} - 1)) + c_{1}

x = \frac{1}{8} (ln (\frac{y - 5}{4} + 1) - ln (\frac{y - 5}{4} - 1)) + c_{1}

Simplify

x = \frac{1}{8} ln (\frac{y - 1}{y - 9}) + c_{1}

Isolate

y : y = - \frac{9 e ^{8 x - 8 c_{1}}}{1 - e ^{8 (x - c_{1})}} + \frac{1}{1 - e ^{8 (x - c_{1})}}

y = - \frac{9 e ^{8 x - 8 c_{1}}}{1 - e ^{8 (x - c_{1})}} + \frac{1}{1 - e ^{8 (x - c_{1})}}

\frac{d}{d x} (\frac{e ^{- x}}{1 + e ^{- x}})

y = ln (x^{2} + 1)

\frac{d r}{d s} = e^{r - 2 s}

\int \frac{16}{x ^{3} + 4 x} d x

\int 6 x^{5} y d x

What is the solution for (dy)/(dx)=y(10-y)-9 ?
The solution for (dy)/(dx)=y(10-y)-9 is y=-(9e^{8x-8c_{1}})/(1-e^{8(x-c_{1))}}+1/(1-e^{8(x-c_{1))}}