What is the solution for h^'=-h^{1/3} ?

The solution for h^'=-h^{1/3} is

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h^'=-h^{1/3}

h^{^{'}} = - h^{\frac{1}{3}}

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

Solution steps

h^{'} (t) = - h (t)^{\frac{1}{3}}

Solve separable ODE:

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

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

\frac{d y}{d t} + e^{t} y = 23 e^{t}

\int \frac{ln ( x ^{25} )}{x} d x

\frac{\partial}{\partial y} (x^{5} + 2^{y} + x^{y})

\frac{d y}{d x} = 2 x (1 + y^{2})

x \to 2 - lim (x + 2)