What is the solution for y^'=-0.5y^{1.5} ?

The solution for y^'=-0.5y^{1.5} is y= 4/(0.25t^2-c_{1)t+c_{1}^2}

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y^'=-0.5y^{1.5}

y^{^{'}} = - 0.5 y^{1.5}

y = \frac{4}{0.25 t ^{2} - c _{1} t + c _{1}^{2}}

Solution steps

y^{'} (t) = - 0.5 y^{1.5}

Solve separable ODE:

y = \frac{4}{0.25 t ^{2} - c _{1} t + c _{1}^{2}}

y = \frac{4}{0.25 t ^{2} - c _{1} t + c _{1}^{2}}

\frac{\partial}{\partial y} (x arcsin (\frac{y}{x}))

f (x) = (x + 5) (9 x + 5)

\frac{d}{d x} ((cos (x) + 2)^{3} (2 x)^{(- 3 x)})

\int (e^{- x^{2}}) d x

\int t e^{- t} d t

What is the solution for y^'=-0.5y^{1.5} ?
The solution for y^'=-0.5y^{1.5} is y= 4/(0.25t^2-c_{1)t+c_{1}^2}