What is the solution for u^'=((\H(t))/t)^{1/3} ?

The solution for u^'=((\H(t))/t)^{1/3} is u(t)=integral of ((\H(t))/t)^{1/3}

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u^'=((\H(t))/t)^{1/3}

u^{^{'}} = (\frac{H ( t )}{t})^{\frac{1}{3}}

u (t) = \int (\frac{H ( t )}{t})^{\frac{1}{3}} d t

Solution steps

u^{'} (t) = (\frac{H ( t )}{t})^{\frac{1}{3}}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

u (t) = \int (\frac{H ( t )}{t})^{\frac{1}{3}} d t

u (t) = \int (\frac{H ( t )}{t})^{\frac{1}{3}} d t

4 x^{2} + 2 x^{2} d x - 4 y \cdot d y = 8 x \cdot d x

(ln (y))^{5} \frac{d y}{d x} = x^{5} y, y (1) = e^{2}

y^{^{'}} = (1 - 11 x) y^{2}, y (0) = - \frac{1}{9}

x y (y + 2) d x - (y + 1) d y = 0

y^{^{'}} csc (t) = \frac{- y ^{17}}{16}

What is the solution for u^'=((\H(t))/t)^{1/3} ?
The solution for u^'=((\H(t))/t)^{1/3} is u(t)=integral of ((\H(t))/t)^{1/3}