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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

1 + x^{2}

\frac{d}{d x} (\frac{e ^{x}}{9 x ^{2}})

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

n = 0 \sum \infty \frac{( - 1 ) ^{n + 1} ( x - 1 ) ^{n + 1}}{n + 1}

\frac{sin ( t )}{4 + 4 cos ( t )}

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