What is the solution for y^'=y(3-ty),y(0)=1 ?

The solution for y^'=y(3-ty),y(0)=1 is y=(9e^{3t})/(3e^{3t)t-e^{3t}+10}

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y^'=y(3-ty),y(0)=1

y^{^{'}} = y (3 - t y), y (0) = 1

y = \frac{9 e ^{3 t}}{3 e ^{3 t} t - e ^{3 t} + 10}

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} + 10}

y = \frac{9 e ^{3 t}}{3 e ^{3 t} t - e ^{3 t} + 10}

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

x \to - 100 lim (3 x^{2})

\frac{d}{d x} (2 x^{2} - 3 x)

x \to 7 lim (\frac{1}{( x - 7 ) ^{2}})

\frac{d}{d y} (\frac{1}{y} e^{x y} - ln (y^{2}))

What is the solution for y^'=y(3-ty),y(0)=1 ?
The solution for y^'=y(3-ty),y(0)=1 is y=(9e^{3t})/(3e^{3t)t-e^{3t}+10}