What is the solution for 18y^'=y(y^2-9) ?

The solution for 18y^'=y(y^2-9) is y= 3/(sqrt(1-e^{t+c_{1))}},y=-3/(sqrt(1-e^{t+c_{1))}}

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18y^'=y(y^2-9)

18 y^{^{'}} = y (y^{2} - 9)

y = \frac{3}{1 - e ^{t + c_{1}}}, y = - \frac{3}{1 - e ^{t + c_{1}}}

Solution steps

18 y^{'} (t) = y (y^{2} - 9)

Solve separable ODE:

y = \frac{3}{1 - e ^{t + c_{1}}}, y = - \frac{3}{1 - e ^{t + c_{1}}}

y = \frac{3}{1 - e ^{t + c_{1}}}, y = - \frac{3}{1 - e ^{t + c_{1}}}

x \to 1 lim (\frac{1}{( 1.5 - 1 ) ^{2}})

\frac{\partial}{\partial x} (tan (4 x y^{5}))

\frac{d}{d x} (\frac{2 x ^{2} + x - 1}{x ^{2} - 1})

x^{2} - 2 x + 3, x = - 2, x = 1

\frac{d}{d x} ((2 x^{2} - 6)^{- 10})

What is the solution for 18y^'=y(y^2-9) ?
The solution for 18y^'=y(y^2-9) is y= 3/(sqrt(1-e^{t+c_{1))}},y=-3/(sqrt(1-e^{t+c_{1))}}