2.1.12 Problem 12

Internal problem ID [10270]
Book : First order enumerated odes
Section : section 1
Problem number : 12
Date solved : Thursday, November 27, 2025 at 10:29:23 AM
CAS classification : [_quadrature]

Solved using first_order_ode_quadrature

Time used: 0.039 (sec)

Solve

\begin{align*} c 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 2.16: Slope field \(c y^{\prime } = 0\)

Summary of solutions found

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

Solved using first_order_ode_homog_type_D2

Time used: 0.076 (sec)

Solve

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

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

\begin{align*} c \left (u^{\prime }\left (x \right ) x +u \left (x \right )\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 2.17: Slope field \(c y^{\prime } = 0\)

Summary of solutions found

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

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

ode:=c*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

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

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

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

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

✓ Sympy. Time used: 0.015 (sec). Leaf size: 3

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

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