What is the solution for y^'=0.0012y(1700-y) ?

The solution for y^'=0.0012y(1700-y) is y=(2.04)/(-e^{-2.03999…t-c_{1)}+0.0012}

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y^'=0.0012y(1700-y)

y^{^{'}} = 0.0012 y (1700 - y)

y = \frac{2.04}{- e ^{- 2.03999 \dots t - c_{1}} + 0.0012}

Solution steps

y^{'} (t) = 0.0012 y (1700 - y)

Solve Bernoulli ODE:

y = \frac{2.04}{- e ^{- 2.03999 \dots t - c_{1}} + 0.0012}

y = \frac{2.04}{- e ^{- 2.03999 \dots t - c_{1}} + 0.0012}

y^{^{'}} = \frac{( 2 x ^{2} + 3 y ^{2} )}{2 x y}

\frac{\partial}{\partial v} (tan (\frac{u}{v}))

n = 3 \sum \infty \frac{2}{n ( n + 2 )}

\int - 5 csc^{2} (x) d x

x \to - 3 + lim (\frac{x + 4}{x + 3})

What is the solution for y^'=0.0012y(1700-y) ?
The solution for y^'=0.0012y(1700-y) is y=(2.04)/(-e^{-2.03999…t-c_{1)}+0.0012}