What is the solution for x^'=-0.0002sqrt(x) ?

The solution for x^'=-0.0002sqrt(x) is x=(0.00000004t^2)/4-(0.0004c_{1}t)/4+(c_{1}^2)/4

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x^'=-0.0002sqrt(x)

x^{^{'}} = - 0.0002 x

x = \frac{0.00000004 t ^{2}}{4} - \frac{0.0004 c _{1} t}{4} + \frac{c _{1}^{2}}{4}

Solution steps

x^{'} (t) = - 0.0002 x

Solve separable ODE:

x = \frac{0.00000004 t ^{2}}{4} - \frac{0.0004 c _{1} t}{4} + \frac{c _{1}^{2}}{4}

x = \frac{0.00000004 t ^{2}}{4} - \frac{0.0004 c _{1} t}{4} + \frac{c _{1}^{2}}{4}

x \to 0 lim (\frac{3 x}{2 x ^{2} + x})

\int (e^{4 x} - 4)^{\frac{1}{5}} \cdot 8 e^{4 x} d x

\frac{2 x}{1 - cot ( x )}

x \to 0 lim (\frac{1}{x + \frac{1}{x}})

\frac{8 s}{3 ( s ^{2} - 16 ) ^{2}}

What is the solution for x^'=-0.0002sqrt(x) ?
The solution for x^'=-0.0002sqrt(x) is x=(0.00000004t^2)/4-(0.0004c_{1}t)/4+(c_{1}^2)/4