1.28 problem 29
Internal
problem
ID
[7720]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
1.0
Problem
number
:
29
Date
solved
:
Monday, October 21, 2024 at 03:59:08 PM
CAS
classification
:
[_quadrature]
Solve
\begin{align*} y^{\prime }&=0 \end{align*}
1.28.1 Solved as first order quadrature ode
Time used: 0.028 (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 63: Slope field plot
\(y^{\prime } = 0\)
1.28.2 Solved as first order homogeneous class D2 ode
Time used: 0.141 (sec)
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 \(u^{\prime }\left (x \right ) = -\frac {u \left (x \right )}{x}\) 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}\\ \ln \left (u \left (x \right )\right )&=\ln \left (\frac {1}{x}\right )+c_1 \end{align*}
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 \(u \left (x \right )\) from the above solution(s) gives (after possible removing of solutions that do not
verify)
\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 64: Slope field plot
\(y^{\prime } = 0\)
1.28.3 Solved as first order ode of type differential
Time used: 0.010 (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 65: Slope field plot
\(y^{\prime } = 0\)
1.28.4 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int 0d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\mathit {C1} \end {array} \]
1.28.5 Maple trace
Methods for first order ODEs:
1.28.6 Maple dsolve solution
Solving time : 0.000
(sec)
Leaf size : 5
dsolve ( diff ( y ( x ), x ) = 0,
y(x),singsol=all)
\[
y = c_1
\]
1.28.7 Mathematica DSolve solution
Solving time : 0.002
(sec)
Leaf size : 7
DSolve [{ D [ y [ x ], x ]==0,{}},
y[x],x,IncludeSingularSolutions-> True ]
\[
y(x)\to c_1
\]