What is the solution for (dy)/(dt)=t^2y^3 ?

The solution for (dy)/(dt)=t^2y^3 is y=sqrt(-3/(2(t^3+c_{1)))},y=-sqrt(-3/(2(t^3+c_{1)))}

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(dy)/(dt)=t^2y^3

\frac{d y}{d t} = t^{2} y^{3}

y = - \frac{3}{2 ( t ^{3} + c _{1} )}, y = - - \frac{3}{2 ( t ^{3} + c _{1} )}

Solution steps

\frac{d y}{d t} = t^{2} y^{3}

Solve separable ODE:

y = - \frac{3}{2 ( t ^{3} + c _{1} )}, y = - - \frac{3}{2 ( t ^{3} + c _{1} )}

y = - \frac{3}{2 ( t ^{3} + c _{1} )}, y = - - \frac{3}{2 ( t ^{3} + c _{1} )}

\frac{d}{d x} (x^{3} (3 x - 8))

\int_{6}^{11} 10 arccot (x) d x

f (x) = sin (sin (8 e^{x}))

\int \frac{1}{u ^{2} - a ^{2}} d u

x^{8} (x - 4)^{7}

What is the solution for (dy)/(dt)=t^2y^3 ?
The solution for (dy)/(dt)=t^2y^3 is y=sqrt(-3/(2(t^3+c_{1)))},y=-sqrt(-3/(2(t^3+c_{1)))}