What is the solution for y^'=t^2y^3 ?

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

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y^'=t^2y^3

y^{^{'}} = t^{2} y^{3}

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

Solution steps

y^{'} (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} )}

y^{^{'''}} - 3 y^{^{''}} - 9 y^{^{'}} + 27 y = 0

\frac{\partial}{\partial x} (2 x yz)

4 x + \frac{1}{x}

\int \frac{6 + 6 x}{1 + x ^{2}} d x

- \frac{13}{s ^{5}}

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