What is the solution for x^'=ax-bx^2 ?

The solution for x^'=ax-bx^2 is x=-(ae^{at+c_{1}a})/(b(1-e^{a(t+c_{1))})}

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x^'=ax-bx^2

x^{^{'}} = a x - b x^{2}

x = - \frac{a e ^{a t + c_{1} a}}{b ( 1 - e ^{a (t + c_{1})} )}

Solution steps

x^{'} (t) = a x (t) - b x (t)^{2}

Solve separable ODE:

x = - \frac{a e ^{a t + c_{1} a}}{b ( 1 - e ^{a (t + c_{1})} )}

x = - \frac{a e ^{a t + c_{1} a}}{b ( 1 - e ^{a (t + c_{1})} )}

x \to - 1 lim (\frac{f ( x )}{x + 1})

y^{^{'}} = \frac{- x y ^{2}}{2}

\int \frac{1}{N} d N

\int (\frac{1}{y ^{2}}) d y

\int e^{x} sin (t - x) d x

What is the solution for x^'=ax-bx^2 ?
The solution for x^'=ax-bx^2 is x=-(ae^{at+c_{1}a})/(b(1-e^{a(t+c_{1))})}