What is the solution for y^'=y(3-yt) ?

The solution for y^'=y(3-yt) is y=(9e^{3t})/(3e^{3t)t-e^{3t}+c_{1}}

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y^'=y(3-yt)

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

y = \frac{9 e ^{3 t}}{3 e ^{3 t} t - e ^{3 t} + c _{1}}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{9 e ^{3 t}}{3 e ^{3 t} t - e ^{3 t} + c _{1}}

y = \frac{9 e ^{3 t}}{3 e ^{3 t} t - e ^{3 t} + c _{1}}

\frac{d}{d x} (3 \cdot sin (x))

2 x^{2} y^{^{''}} + 5 x y^{^{'}} + y = 18 x + 4

y^{^{'}} + y^{2} sin (x) = 0

x \to \infty lim (sin (1))

\frac{d y}{d x} = \frac{2 y}{2 x + 1}

What is the solution for y^'=y(3-yt) ?
The solution for y^'=y(3-yt) is y=(9e^{3t})/(3e^{3t)t-e^{3t}+c_{1}}