2.1.7 Problem 2 (i)

Internal problem ID [18423]
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 : Monday, March 31, 2025 at 05:27:57 PM
CAS classification : [_quadrature]

Existence and uniqueness analysis

Solve

$$\begin{array}{rl}{x}^{\prime }& =x^2-3x+2\end{array}$$

With initial conditions

$$\begin{array}{rl}x\left(0\right)& =1\end{array}$$

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

$$\begin{array}{rl}{x}^{\prime }& =f(t,x)\\ & =x^2-3x+2\end{array}$$

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

$$\{-\mathrm{\infty }<x<\mathrm{\infty }\}$$

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

$$\begin{array}{rl}\frac{\partial f}{\partial x}& =\frac{\partial }{\partial x}(x^2-3x+2)\\ & =2x-3\end{array}$$

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

$$\{-\mathrm{\infty }<x<\mathrm{\infty }\}$$

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

Solved using first_order_ode_autonomous

Time used: 0.103 (sec)

Solve

$$\begin{array}{rl}{x}^{\prime }& =x^2-3x+2\end{array}$$

With initial conditions

$$\begin{array}{rl}x\left(0\right)& =1\end{array}$$

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

$$\begin{array}{rl}0& ={f(x)|}_{x=x_0}\\ 0& =0\end{array}$$

And the solution is immediately written as

$$\begin{array}{rl}x& =x_0\\ x& =1\end{array}$$

Singular solutions are found by solving

$$\begin{array}{rl}{x}^2-3x+2& =0\end{array}$$

for $x$. This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

$$\begin{array}{r}x=1\end{array}$$

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	Slope field $x^{\prime }=x^2-3x+2$

Summary of solutions found

$$\begin{array}{rl}x& =1\end{array}$$

✓ Maple. Time used: 0.010 (sec). Leaf size: 5

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

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & [\frac{d}{dt}x\left(t\right)=x{\left(t\right)}^2-3x\left(t\right)+2,x\left(0\right)=1]\\ \bullet & & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{d}{dt}x\left(t\right)\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dt}x\left(t\right)=x{\left(t\right)}^2-3x\left(t\right)+2\\ \bullet & & \text{Separate variables}\\ & & \frac{\frac{d}{dt}x\left(t\right)}{x{\left(t\right)}^2-3x\left(t\right)+2}=1\\ \bullet & & \text{Integrate both sides with respect to}t\\ & & \int \frac{\frac{d}{dt}x\left(t\right)}{x{\left(t\right)}^2-3x\left(t\right)+2}dt=\int 1dt+\mathit{C}\mathit{1}\\ \bullet & & \text{Evaluate integral}\\ & & \ln (x\left(t\right)-2)-\ln (x\left(t\right)-1)=t+\mathit{C}\mathit{1}\\ \bullet & & \text{Solve for}x\left(t\right)\\ & & x\left(t\right)=\frac{-2+{\mathrm{e}}^{t+\mathit{C}\mathit{1}}}{{\mathrm{e}}^{t+\mathit{C}\mathit{1}}-1}\\ \bullet & & \text{Redefine the integration constant(s)}\\ & & x\left(t\right)=\frac{-2+\mathit{C}\mathit{1}{\mathrm{e}}^t}{\mathit{C}\mathit{1}{\mathrm{e}}^t-1}\\ \bullet & & \text{Use initial condition}x\left(0\right)=1\\ & & 1=\frac{-2+\mathit{C}\mathit{1}}{\mathit{C}\mathit{1}-1}\\ \bullet & & \text{Solve for}\mathit{\text{\_C1}}\\ & & \text{No solution}\\ \bullet & & \text{Solution does not satisfy initial condition}\end{array}$$

✓ 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]
 

$$x(t)\to 1$$

✗ 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