2.1.28 Problem 29

Internal problem ID [10014]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 29
Date solved : Monday, December 08, 2025 at 07:03:46 PM
CAS classification : [_quadrature]

2.1.28.1 Solved using first_order_ode_quadrature

0.045 (sec)

Entering first order ode quadrature solver

\begin{align*} y^{\prime }&=0 \\ \end{align*}

Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dy} &= \int {0\, dx} + c_1 \\ y &= c_1 \end{align*}

Figure 2.72: Slope field \(y^{\prime } = 0\)

Summary of solutions found

\begin{align*} y &= c_1 \\ \end{align*}

2.1.28.2 Solved using first_order_ode_homog_type_D2

0.131 (sec)

Entering first order ode homog type D2 solver

\begin{align*} y^{\prime }&=0 \\ \end{align*}

Applying change of variables \(y = u \left (x \right ) x\), then the ode becomes

\begin{align*} u^{\prime }\left (x \right ) x +u \left (x \right ) = 0 \end{align*}

Which is now solved The ode

\begin{equation} u^{\prime }\left (x \right ) = -\frac {u \left (x \right )}{x} \end{equation}

is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= -\frac {u \left (x \right )}{x}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= -\frac {1}{x}\\ g(u) &= u \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx} \\ \int { \frac {1}{u}\,du} &= \int { -\frac {1}{x} \,dx} \\ \end{align*}

\[ \ln \left (u \left (x \right )\right )=\ln \left (\frac {1}{x}\right )+c_1 \]

Taking the exponential of both sides the solution becomes

\[ u \left (x \right ) = \frac {c_1}{x} \]

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ u=0 \]

for \(u \left (x \right )\) gives

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

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} u \left (x \right ) &= \frac {c_1}{x} \\ u \left (x \right ) &= 0 \\ \end{align*}

Converting \(u \left (x \right ) = \frac {c_1}{x}\) back to \(y\) gives

\begin{align*} y = c_1 \end{align*}

Converting \(u \left (x \right ) = 0\) back to \(y\) gives

\begin{align*} y = 0 \end{align*}

Figure 2.73: Slope field \(y^{\prime } = 0\)

Summary of solutions found

\begin{align*} y &= 0 \\ y &= c_1 \\ \end{align*}

2.1.28.3 ✓ Maple. Time used: 0.001 (sec). Leaf size: 5

ode:=diff(y(x),x) = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ y = c_1 \]

Maple trace

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

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}y \left (x \right )\right )d x =\int 0d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C1} \end {array} \]

2.1.28.4 ✓ Mathematica. Time used: 0.002 (sec). Leaf size: 7

ode=D[y[x],x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to c_1 \end{align*}

2.1.28.5 ✓ Sympy. Time used: 0.011 (sec). Leaf size: 3

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
                                                                                  
                                                                                  
 

\[ y{\left (x \right )} = C_{1} \]