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}

7 x^{- 3}

f (x) = 4 x, (6, 9.798)

\int \frac{x ^{3}}{x ^{2} + 49} d x

\frac{d}{d x} (\frac{x ^{4}}{ln ( x )})

\int (\frac{1}{2} x) d x

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)}