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

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

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y^'=y(3-ty)

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

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

Solution steps

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

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}}

\int \frac{3 x ^{2} - x + 5}{x ^{4}} d x

y^{^{'}} = - \frac{x}{y}

x \to - 3 lim (\frac{x + 2}{x - 2})

\frac{\partial}{\partial x} (cos^{2} (x) y)

\frac{d}{d a} (sec (a) \cdot tan (a) \cdot cos (a) \cdot csc (a))

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