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

The solution for y^'=y^3,y(0)=a is y=sqrt(-(a^2)/(2ta^2-1)),y=-sqrt(-(a^2)/(2ta^2-1))

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y^'=y^3,y(0)=a

y^{^{'}} = y^{3}, y (0) = a

y = - \frac{a ^{2}}{2 t a ^{2} - 1}, y = - - \frac{a ^{2}}{2 t a ^{2} - 1}

Solution steps

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

Solve separable ODE:

y = - \frac{a ^{2}}{2 t a ^{2} - 1}, y = - - \frac{a ^{2}}{2 t a ^{2} - 1}

y = - \frac{a ^{2}}{2 t a ^{2} - 1}, y = - - \frac{a ^{2}}{2 t a ^{2} - 1}

What is the solution for y^'=y^3,y(0)=a ?
The solution for y^'=y^3,y(0)=a is y=sqrt(-(a^2)/(2ta^2-1)),y=-sqrt(-(a^2)/(2ta^2-1))