What is the solution for y^'=y(3-y),y(0)=5 ?

The solution for y^'=y(3-y),y(0)=5 is y=(15e^{3t})/(-2+5e^{3t)}

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y^'=y(3-y),y(0)=5

y^{^{'}} = y (3 - y), y (0) = 5

y = \frac{15 e ^{3 t}}{- 2 + 5 e ^{3 t}}

Solution steps

y^{'} (t) = y (3 - y)

Solve Bernoulli ODE:

y = \frac{15 e ^{3 t}}{- 2 + 5 e ^{3 t}}

y = \frac{15 e ^{3 t}}{- 2 + 5 e ^{3 t}}

f (x) = 3 sin (x) + cos (x)

x \to - \infty lim (x, - 2)

\int \frac{- 4}{( 3 x ) ^{3}} d x

y^{^{'}} = \frac{x ^{2} + y ^{2}}{x y}

\frac{\partial}{\partial x} (x + 3 y)

What is the solution for y^'=y(3-y),y(0)=5 ?
The solution for y^'=y(3-y),y(0)=5 is y=(15e^{3t})/(-2+5e^{3t)}