What is the solution for (dP)/(dt)=aP(t)-bP(t)^2 ?

The solution for (dP)/(dt)=aP(t)-bP(t)^2 is P=e^{at-(bt^3)/3+c_{1}}

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(dP)/(dt)=aP(t)-bP(t)^2

\frac{d P}{d t} = a P (t) - b P (t)^{2}

P = e^{a t - \frac{b t ^{3}}{3} + c_{1}}

Solution steps

\frac{d P}{d t} = a P - b P t^{2}

Solve separable ODE:

P = e^{a t - \frac{b t ^{3}}{3} + c_{1}}

P = e^{a t - \frac{b t ^{3}}{3} + c_{1}}

y - 2 y^{^{'}} + 10 y = 0

\frac{d v}{d t} = 10 - \frac{1}{100} v, v = 0, t = 0

\frac{d y}{d x} = \frac{x 4 - y ^{2}}{1 + x ^{2}}

1 + x \frac{d y}{d x} = 1 + y

2 x sin^{2} (y) d x - (x^{2} + 10) cos (y) d y = 0

What is the solution for (dP)/(dt)=aP(t)-bP(t)^2 ?
The solution for (dP)/(dt)=aP(t)-bP(t)^2 is P=e^{at-(bt^3)/3+c_{1}}