What is the solution for x^'=(20x-0.1x^2) ?

The solution for x^'=(20x-0.1x^2) is x=-(200e^{20t+20c_{1}})/(1-e^{20(t+c_{1))}}

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x^'=(20x-0.1x^2)

x^{^{'}} = (20 x - 0.1 x^{2})

x = - \frac{200 e ^{20 t + 20 c_{1}}}{1 - e ^{20 (t + c_{1})}}

Solution steps

x^{'} (t) = (20 x - 0.1 x^{2})

Solve separable ODE:

x = - \frac{200 e ^{20 t + 20 c_{1}}}{1 - e ^{20 (t + c_{1})}}

x = - \frac{200 e ^{20 t + 20 c_{1}}}{1 - e ^{20 (t + c_{1})}}

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

y^{^{'}} = 11 t y - 16 y, y (0) = 4

3 y^{^{'}} - 3 e^{2 x} y = 0, y (0) = 1

\frac{d x}{d t} = t (x - 2)

\frac{d y}{d x} = \frac{8 sin ( x )}{sin ( y )}, y (0) = \frac{π}{2}

What is the solution for x^'=(20x-0.1x^2) ?
The solution for x^'=(20x-0.1x^2) is x=-(200e^{20t+20c_{1}})/(1-e^{20(t+c_{1))}}