What is the solution for y^'+9y^3=0,y(0)=y^0 ?

The solution for y^'+9y^3=0,y(0)=y^0 is y=sqrt(-1/(2(-9t+c_{1)))},y=-sqrt(-1/(2(-9t+c_{1)))}

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y^'+9y^3=0,y(0)=y^0

y^{^{'}} + 9 y^{3} = 0, y (0) = y^{0}

y = - \frac{1}{2 ( - 9 t + c _{1} )}, y = - - \frac{1}{2 ( - 9 t + c _{1} )}

Solution steps

y^{'} (t) + 9 y^{3} = 0

Solve separable ODE:

y = - \frac{1}{2 ( - 9 t + c _{1} )}, y = - - \frac{1}{2 ( - 9 t + c _{1} )}

y = - \frac{1}{2 ( - 9 t + c _{1} )}, y = - - \frac{1}{2 ( - 9 t + c _{1} )}

\frac{d y}{d x} = \frac{( 3 x ^{2} - e ^{x} )}{( 2 y - 5 )}

2 x y (\frac{d y}{d x}) = 1

+ y^{^{'}} + y = 0

(x y + x) d x + (y x + y) d y = 0

(e^{2 x} + 1) \frac{d y}{d x} = y

What is the solution for y^'+9y^3=0,y(0)=y^0 ?
The solution for y^'+9y^3=0,y(0)=y^0 is y=sqrt(-1/(2(-9t+c_{1)))},y=-sqrt(-1/(2(-9t+c_{1)))}