What is the solution for P^'=(P)(t)(1-P/(10)) ?

The solution for P^'=(P)(t)(1-P/(10)) is P(t)=-(10e^{(t^2)/2+c_{1}})/(1-e^{(t^2)/2+c_{1)}}

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P^'=(P)(t)(1-P/(10))

P^{^{'}} = (P) (t) (1 - \frac{P}{10})

P (t) = - \frac{10 e ^{\frac{t ^{2}}{2} + c_{1}}}{1 - e ^{\frac{t ^{2}}{2} + c_{1}}}

Solution steps

P^{'} (t) = (P (t)) (t) (1 - \frac{P ( t )}{10})

Solve separable ODE:

P (t) = - \frac{10 e ^{\frac{t ^{2}}{2} + c_{1}}}{1 - e ^{\frac{t ^{2}}{2} + c_{1}}}

P (t) = - \frac{10 e ^{\frac{t ^{2}}{2} + c_{1}}}{1 - e ^{\frac{t ^{2}}{2} + c_{1}}}

(1 + x^{2}) \cdot y^{^{'}} + \frac{y}{arctan ( x )} = 0

\frac{d y}{d x} = 3 x^{2} y, y > 0

y^{^{'}} = \frac{y + 9}{x + 8}

\frac{d y}{d t} + 20 y = 2, 4 = \frac{6}{5} - \frac{6}{5} e^{- 20 t}

y^{^{'}} = \frac{x y}{e ^{y}}, y (0) = 0

What is the solution for P^'=(P)(t)(1-P/(10)) ?
The solution for P^'=(P)(t)(1-P/(10)) is P(t)=-(10e^{(t^2)/2+c_{1}})/(1-e^{(t^2)/2+c_{1)}}