What is the solution for h^'=-0.02215*h^{1/2} ?

The solution for h^'=-0.02215*h^{1/2} is h(t)=(0.0004906225t^2)/4-(0.0443c_{1}t)/4+(c_{1}^2)/4

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h^'=-0.02215*h^{1/2}

h^{^{'}} = - 0.02215 \cdot h^{\frac{1}{2}}

h (t) = \frac{0.0004906225 t ^{2}}{4} - \frac{0.0443 c _{1} t}{4} + \frac{c _{1}^{2}}{4}

Solution steps

h^{'} (t) = - 0.02215 \cdot h (t)^{\frac{1}{2}}

Solve separable ODE:

h (t) = \frac{0.0004906225 t ^{2}}{4} - \frac{0.0443 c _{1} t}{4} + \frac{c _{1}^{2}}{4}

h (t) = \frac{0.0004906225 t ^{2}}{4} - \frac{0.0443 c _{1} t}{4} + \frac{c _{1}^{2}}{4}

y^{^{'}} = \frac{( x ^{3} x ^{4} - 1 )}{y ^{3}}

y^{^{'}} + 7 \frac{y}{x} = 0

\frac{d x}{d y} = \frac{x y ^{2} + x}{10 y ^{2}}

\frac{d y}{d x} = x y^{2}, y (- 3) = - \frac{1}{5}

(- 4 x y^{2} - 4 y) d x + (- 4 x^{2} y - 4 x) d y = 0

What is the solution for h^'=-0.02215*h^{1/2} ?
The solution for h^'=-0.02215*h^{1/2} is h(t)=(0.0004906225t^2)/4-(0.0443c_{1}t)/4+(c_{1}^2)/4