80.2.15 problem 22

Internal problem ID [21158]
Book : A Textbook on Ordinary Differential Equations by Shair Ahmad and Antonio Ambrosetti. Second edition. ISBN 978-3-319-16407-6. Springer 2015
Section : Chapter 2. Theory of first order differential equations. Excercise 2.6 at page 37
Problem number : 22
Date solved : Thursday, October 02, 2025 at 07:13:53 PM
CAS classification : [_quadrature]

\begin{align*} x^{\prime }&=\left (x+2\right ) \left (1-x^{4}\right ) \end{align*}

With initial conditions

\begin{align*} x \left (0\right )&=0 \\ \end{align*}

✓ Maple. Time used: 0.274 (sec). Leaf size: 47

ode:=diff(x(t),t) = (x(t)+2)*(1-x(t)^4); 
ic:=[x(0) = 0]; 
dsolve([ode,op(ic)],x(t), singsol=all);
 

\[ x = \operatorname {RootOf}\left (5 i \pi -4 \ln \left (\textit {\_Z} +2\right )+15 \ln \left (\textit {\_Z} +1\right )-5 \ln \left (\textit {\_Z} -1\right )-3 \ln \left (\textit {\_Z}^{2}+1\right )+12 \arctan \left (\textit {\_Z} \right )+4 \ln \left (2\right )-60 t \right ) \]

✓ Mathematica. Time used: 0.15 (sec). Leaf size: 65

ode=D[x[t],t]==(x[t]+2)*(1-x[t]^4); 
ic={x[0]==0}; 
DSolve[{ode,ic},x[t],t,IncludeSingularSolutions->True]
 

\begin{align*} x(t)&\to \text {InverseFunction}\left [\frac {1}{20} \log \left (\text {$\#$1}^2+1\right )-\frac {\arctan (\text {$\#$1})}{5}+\frac {1}{12} \log (1-\text {$\#$1})-\frac {1}{4} \log (\text {$\#$1}+1)+\frac {1}{15} \log (\text {$\#$1}+2)\&\right ]\left [\frac {\log (2)}{15}-t\right ] \end{align*}

✓ Sympy. Time used: 3.217 (sec). Leaf size: 53

from sympy import * 
t = symbols("t") 
x = Function("x") 
ode = Eq((x(t) + 2)*(x(t)**4 - 1) + Derivative(x(t), t),0) 
ics = {x(0): 0} 
dsolve(ode,func=x(t),ics=ics)
 

\[ t + \frac {\log {\left (x{\left (t \right )} - 1 \right )}}{12} - \frac {\log {\left (x{\left (t \right )} + 1 \right )}}{4} + \frac {\log {\left (x{\left (t \right )} + 2 \right )}}{15} + \frac {\log {\left (x^{2}{\left (t \right )} + 1 \right )}}{20} - \frac {\operatorname {atan}{\left (x{\left (t \right )} \right )}}{5} = \frac {\log {\left (2 \right )}}{15} + \frac {i \pi }{12} \]