2.1.7 Problem 2 (i)

Internal problem ID [19665]
Book : Elementary Differential Equations. By R.L.E. Schwarzenberger. Chapman and Hall. London. First Edition (1969)
Section : Chapter 3. Solutions of first-order equations. Exercises at page 47
Problem number : 2 (i)
Date solved : Tuesday, March 10, 2026 at 11:54:18 AM
CAS classification : [_quadrature]

2.1.7.1 Existence and uniqueness analysis

\begin{align*} x^{\prime }&=x^{2}-3 x+2 \\ x \left (0\right ) &= 1 \\ \end{align*}

This is non linear first order ODE. In canonical form it is written as

\begin{align*} x^{\prime } &= f(t,x)\\ &= x^{2}-3 x +2 \end{align*}

The \(x\) domain of \(f(t,x)\) when \(t=0\) is

\[ \{-\infty <x <\infty \} \]

And the point \(x_0 = 1\) is inside this domain. Now we will look at the continuity of

\begin{align*} \frac {\partial f}{\partial x} &= \frac {\partial }{\partial x}\left (x^{2}-3 x +2\right ) \\ &= 2 x -3 \end{align*}

The \(x\) domain of \(\frac {\partial f}{\partial x}\) when \(t=0\) is

\[ \{-\infty <x <\infty \} \]

And the point \(x_0 = 1\) is inside this domain. Therefore solution exists and is unique.

2.1.7.2 Solved using first_order_ode_autonomous

0.195 (sec)

Entering first order ode autonomous solver

\begin{align*} x^{\prime }&=x^{2}-3 x+2 \\ x \left (0\right ) &= 1 \\ \end{align*}

Since the ode has the form \(x^{\prime }=f(x)\) and initial conditions \(\left (t_0,x_0\right ) \) are given such that they satisfy the ode itself, then we can write

\begin{align*} 0 &= \left . f(x) \right |_{x=x_0}\\ 0 &= 0 \end{align*}

And the solution is immediately written as

\begin{align*}x&=x_0\\ x&=1 \end{align*}

Singular solutions are found by solving

\begin{align*} x^{2}-3 x +2&= 0 \end{align*}

for \(x\). This is because of dividing by the above earlier. This gives the following singular solution(s), which also has to satisfy the given ODE.

\begin{align*} x = 1 \end{align*}

The following diagram is the phase line diagram. It classifies each of the above equilibrium points as stable or not stable or semi-stable.

Figure 2.1: Phase line diagram

Solution plot for \(x^{\prime } = x^{2}-3 x+2\)	Direction fields for \(x^{\prime } = x^{2}-3 x+2\)
Isoclines for \(x^{\prime } = x^{2}-3 x+2\)

Summary of solutions found

\begin{align*} x &= 1 \\ \end{align*}

2.1.7.3 ✓ Maple. Time used: 0.003 (sec). Leaf size: 5

ode:=diff(x(t),t) = x(t)^2-3*x(t)+2; 
ic:=[x(0) = 1]; 
dsolve([ode,op(ic)],x(t), singsol=all);
 

\[ x = 1 \]

Maple trace

Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
<- separable successful
 

2.1.7.4 ✓ Mathematica. Time used: 0.001 (sec). Leaf size: 6

ode=D[x[t],t]==x[t]^2-3*x[t]+2; 
ic={x[0]==1}; 
DSolve[{ode,ic},x[t],t,IncludeSingularSolutions->True]
 

\begin{align*} x(t)&\to 1 \end{align*}

2.1.7.5 ✗ Sympy

from sympy import * 
t = symbols("t") 
x = Function("x") 
ode = Eq(-x(t)**2 + 3*x(t) + Derivative(x(t), t) - 2,0) 
ics = {x(0): 1} 
dsolve(ode,func=x(t),ics=ics)
 

ValueError : Couldnt solve for initial conditions
 

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0
 

classify_ode(ode,func=x(t)) 
 
('separable', '1st_exact', '1st_rational_riccati', '1st_power_series', 'lie_group', 'separable_Integral', '1st_exact_Integral')