What is the solution for y^'=(t^2)/y ?

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

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y^'=(t^2)/y

y^{^{'}} = \frac{t ^{2}}{y}

y = \frac{2 t ^{3}}{3} + c_{1}, y = - \frac{2 t ^{3}}{3} + c_{1}

Solution steps

y^{'} (t) = \frac{t ^{2}}{y}

Solve separable ODE:

y = \frac{2 t ^{3}}{3} + c_{1}, y = - \frac{2 t ^{3}}{3} + c_{1}

y = \frac{2 t ^{3}}{3} + c_{1}, y = - \frac{2 t ^{3}}{3} + c_{1}

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