2.1.27 Problem 28

Internal problem ID [8739]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 28
Date solved : Friday, February 21, 2025 at 08:26:54 PM
CAS classification : [_quadrature]

Solve

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

Solved as first order quadrature ode

Time used: 0.016 (sec)

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 2.63: Slope field \(\frac {y^{\prime }}{x} = 0\)

Summary of solutions found

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

Solved as first order homogeneous class D2 ode

Time used: 0.125 (sec)

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

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

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

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*} \ln \left (u \left (x \right )\right ) &= \ln \left (\frac {1}{x}\right )+c_1 \\ u \left (x \right ) &= 0 \\ \end{align*}

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

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

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

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

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

\begin{align*} y = {\mathrm e}^{c_1} \end{align*}

Figure 2.64: Slope field \(\frac {y^{\prime }}{x} = 0\)

Summary of solutions found

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

Solved as first order ode of type differential

Time used: 0.008 (sec)

Writing the ode as

\begin{align*} y^{\prime }&=0\tag {1} \end{align*}

Which becomes

\begin{align*} \left (1\right ) dy &= \left (0\right ) dx\tag {2} \end{align*}

But the RHS is complete differential because

\begin{align*} \left (0\right ) dx &= d\left (0\right ) \end{align*}

Hence (2) becomes

\begin{align*} \left (1\right ) dy &= d\left (0\right ) \end{align*}

Integrating gives

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

Figure 2.65: Slope field \(\frac {y^{\prime }}{x} = 0\)

Summary of solutions found

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

Maple step by step solution

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

Maple trace

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

Maple dsolve solution

Solving time : 0.002 (sec)
Leaf size : 5

dsolve(1/x*diff(y(x),x) = 0,y(x),singsol=all)
 

\[ y = c_{1} \]

✓Mathematica DSolve solution

Solving time : 0.002 (sec)
Leaf size : 7

DSolve[{1/x*D[y[x],x]==0,{}},y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to c_1 \]

✓Sympy solution

Solving time : 0.030 (sec)
Leaf size : 3

Python version: 3.13.1 (main, Dec  4 2024, 18:05:56) [GCC 14.2.1 20240910] 
Sympy version 1.13.3
 

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

Eq(y(x), C1)
 

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