What is the solution for y^'=ky(4000-y),y(0)=150 ?

The solution for y^'=ky(4000-y),y(0)=150 is y=(12000e^{4000kt})/(77+3e^{4000kt)}

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y^'=ky(4000-y),y(0)=150

y^{^{'}} = k y (4000 - y), y (0) = 150

y = \frac{12000 e ^{4000 k t}}{77 + 3 e ^{4000 k t}}

Solution steps

y^{'} (t) = k y (4000 - y)

Solve Bernoulli ODE:

y = \frac{12000 e ^{4000 k t}}{77 + 3 e ^{4000 k t}}

y = \frac{12000 e ^{4000 k t}}{77 + 3 e ^{4000 k t}}

What is the solution for y^'=ky(4000-y),y(0)=150 ?
The solution for y^'=ky(4000-y),y(0)=150 is y=(12000e^{4000kt})/(77+3e^{4000kt)}