56.1.33 problem 34

Internal problem ID [8745]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 34
Date solved : Monday, January 27, 2025 at 04:46:12 PM
CAS classification : [_quadrature]

\begin{align*} p^{\prime }&=a p-b p^{2} \end{align*}

With initial conditions

\begin{align*} p \left (\operatorname {t0} \right )&=\operatorname {p0} \end{align*}

✓ Solution by Maple

Time used: 0.045 (sec). Leaf size: 29

dsolve([diff(p(t),t)=a*p(t)-b*p(t)^2,p(t0) = p0],p(t), singsol=all)
 

\[ p = \frac {a \operatorname {p0}}{\left (-\operatorname {p0} b +a \right ) {\mathrm e}^{-a \left (t -\operatorname {t0} \right )}+\operatorname {p0} b} \]

✓ Solution by Mathematica

Time used: 0.917 (sec). Leaf size: 39

DSolve[{D[p[t],t]==a*p[t]-b*p[t]^2,p[t0]==p0},p[t],t,IncludeSingularSolutions -> True]
 

\[ p(t)\to \frac {a \text {p0} e^{a t}}{b \text {p0} \left (e^{a t}-e^{a \text {t0}}\right )+a e^{a \text {t0}}} \]