What is the solution for 18y^'=y^3-9y ?

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

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18y^'=y^3-9y

18 y^{^{'}} = y^{3} - 9 y

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

Solution steps

18 y^{'} (t) = y^{3} - 9 y

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 2 lim (1 - x^{2})

y = 4 x, y = \frac{4 x}{3}

y^{^{'}} = 1 - t + y - t y

\int \frac{ln ^{3} ( x ^{2} )}{x} d x

\frac{6 x}{4 + x ^{2}}

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