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

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

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

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

y = \frac{3 a}{- a e ^{3 t} + 3 e ^{3 t} + a}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{3 a}{- a e ^{3 t} + 3 e ^{3 t} + a}

y = \frac{3 a}{- a e ^{3 t} + 3 e ^{3 t} + a}

n = 0 \sum \infty \frac{n x ^{n}}{n + 2}

f (x) = 5^{x^{2}}

\frac{\partial}{\partial y} (\frac{1}{x + y})

\frac{d y}{d x} = - \frac{( x + 6 )}{7 y ^{2}}

\frac{d θ}{d t} = 40 - \frac{2}{5} (θ - 15)

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