2.1.9 Problem 2 (iii)

Internal problem ID [19667]
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 (iii)
Date solved : Tuesday, March 10, 2026 at 11:54:27 AM
CAS classification : [_quadrature]

2.1.9.1 Existence and uniqueness analysis

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

2.1.9.2 Solved using first_order_ode_autonomous

0.133 (sec)

Entering first order ode autonomous solver

\begin{align*} x^{\prime }&=\left (x-1\right )^{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*} \left (x -1\right )^{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 2.2: Phase line diagram

Solution plot for \(x^{\prime } = \left (x-1\right )^{2}\)	Direction fields for \(x^{\prime } = \left (x-1\right )^{2}\)
Isoclines for \(x^{\prime } = \left (x-1\right )^{2}\)

Summary of solutions found

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

2.1.9.3 Solved using first_order_ode_exact

0.526 (sec)

Entering first order ode exact solver

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

To solve an ode of the form

\begin{equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A}\end{equation}

We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives

\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \]

Hence

\begin{equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B}\end{equation}

Comparing (A,B) shows that

\begin{align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end{align*}

But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that

\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\]

If the above condition is satisfied, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is to write the ODE in standard form to check for exactness, which is

\[ M(t,x) \mathop {\mathrm {d}t}+ N(t,x) \mathop {\mathrm {d}x}=0 \tag {1A} \]

Therefore

\begin{align*} \mathop {\mathrm {d}x} &= \left (\left (x -1\right )^{2}\right )\mathop {\mathrm {d}t}\\ \left (-\left (x -1\right )^{2}\right ) \mathop {\mathrm {d}t} + \mathop {\mathrm {d}x} &= 0 \tag {2A} \end{align*}

Comparing (1A) and (2A) shows that

\begin{align*} M(t,x) &= -\left (x -1\right )^{2}\\ N(t,x) &= 1 \end{align*}

The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied

\[ \frac {\partial M}{\partial x} = \frac {\partial N}{\partial t} \]

Using result found above gives

\begin{align*} \frac {\partial M}{\partial x} &= \frac {\partial }{\partial x} \left (-\left (x -1\right )^{2}\right )\\ &= 2-2 x \end{align*}

And

\begin{align*} \frac {\partial N}{\partial t} &= \frac {\partial }{\partial t} \left (1\right )\\ &= 0 \end{align*}

Since \(\frac {\partial M}{\partial x} \neq \frac {\partial N}{\partial t}\), then the ODE is not exact. Since the ODE is not exact, we will try to find an integrating factor to make it exact. Let

\begin{align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial x} - \frac {\partial N}{\partial t} \right ) \\ &=1\left ( \left ( 2-2 x\right ) - \left (0 \right ) \right ) \\ &=2-2 x \end{align*}

Since \(A\) depends on \(x\), it can not be used to obtain an integrating factor. We will now try a second method to find an integrating factor. Let

\begin{align*} B &= \frac {1}{M} \left ( \frac {\partial N}{\partial t} - \frac {\partial M}{\partial x} \right ) \\ &=-\frac {1}{\left (x -1\right )^{2}}\left ( \left ( 0\right ) - \left (2-2 x \right ) \right ) \\ &=-\frac {2}{x -1} \end{align*}

Since \(B\) does not depend on \(t\), it can be used to obtain an integrating factor. Let the integrating factor be \(\mu \). Then

\begin{align*} \mu &= e^{\int B \mathop {\mathrm {d}x}} \\ &= e^{\int -\frac {2}{x -1}\mathop {\mathrm {d}x} } \end{align*}

The result of integrating gives

\begin{align*} \mu &= e^{-2 \ln \left (x -1\right ) } \\ &= \frac {1}{\left (x -1\right )^{2}} \end{align*}

\(M\) and \(N\) are now multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\) and \(\overline {N}\) so not to confuse them with the original \(M\) and \(N\).

\begin{align*} \overline {M} &=\mu M \\ &= \frac {1}{\left (x -1\right )^{2}}\left (-\left (x -1\right )^{2}\right ) \\ &= -1 \end{align*}

And

\begin{align*} \overline {N} &=\mu N \\ &= \frac {1}{\left (x -1\right )^{2}}\left (1\right ) \\ &= \frac {1}{\left (x -1\right )^{2}} \end{align*}

So now a modified ODE is obtained from the original ODE which will be exact and can be solved using the standard method. The modified ODE is

\begin{align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}x}}{\mathop {\mathrm {d}t}} &= 0 \\ \left (-1\right ) + \left (\frac {1}{\left (x -1\right )^{2}}\right ) \frac { \mathop {\mathrm {d}x}}{\mathop {\mathrm {d}t}} &= 0 \end{align*}

The following equations are now set up to solve for the function \(\phi \left (t,x\right )\)

\begin{align*} \frac {\partial \phi }{\partial t } &= \overline {M}\tag {1} \\ \frac {\partial \phi }{\partial x } &= \overline {N}\tag {2} \end{align*}

Integrating (1) w.r.t. \(t\) gives

\begin{align*} \int \frac {\partial \phi }{\partial t} \mathop {\mathrm {d}t} &= \int \overline {M}\mathop {\mathrm {d}t} \\ \int \frac {\partial \phi }{\partial t} \mathop {\mathrm {d}t} &= \int -1\mathop {\mathrm {d}t} \\ \tag{3} \phi &= -t+ f(x) \\ \end{align*}

Where \(f(x)\) is used for the constant of integration since \(\phi \) is a function of both \(t\) and \(x\). Taking derivative of equation (3) w.r.t \(x\) gives

\begin{equation} \tag{4} \frac {\partial \phi }{\partial x} = 0+f'(x) \end{equation}

But equation (2) says that \(\frac {\partial \phi }{\partial x} = \frac {1}{\left (x -1\right )^{2}}\). Therefore equation (4) becomes

\begin{equation} \tag{5} \frac {1}{\left (x -1\right )^{2}} = 0+f'(x) \end{equation}

Solving equation (5) for \( f'(x)\) gives

\[ f'(x) = \frac {1}{\left (x -1\right )^{2}} \]

Integrating the above w.r.t \(x\) gives

\begin{align*} \int f'(x) \mathop {\mathrm {d}x} &= \int \left ( \frac {1}{\left (x -1\right )^{2}}\right ) \mathop {\mathrm {d}x} \\ f(x) &= -\frac {1}{x -1}+ c_1 \\ \end{align*}

\[ \phi = -t -\frac {1}{x -1}+ c_1 \]

But since \(\phi \) itself is a constant function, then let \(\phi =c_2\) where \(c_2\) is new constant and combining \(c_1\) and \(c_2\) constants into the constant \(c_1\) gives the solution as

\[ c_1 = -t -\frac {1}{x -1} \]

Solving for the constant of integration from initial conditions, the solution becomes

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

Solution plot for \(x^{\prime } = \left (x-1\right )^{2}\)	Direction fields for \(x^{\prime } = \left (x-1\right )^{2}\)
Isoclines for \(x^{\prime } = \left (x-1\right )^{2}\)

Summary of solutions found

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

2.1.9.4 ✓ Maple. Time used: 0.004 (sec). Leaf size: 5

ode:=diff(x(t),t) = (x(t)-1)^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
 

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

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

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

2.1.9.6 ✗ Sympy

from sympy import * 
t = symbols("t") 
x = Function("x") 
ode = Eq(-(x(t) - 1)**2 + Derivative(x(t), t),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)) 
 
('factorable', 'separable', '1st_exact', '1st_rational_riccati', '1st_power_series', 'lie_group', 'separable_Integral', '1st_exact_Integral')