1.31 problem 31

Internal problem ID [7995]
Book : First order enumerated odes
Section : section 1
Problem number : 31
Date solved : Monday, October 21, 2024 at 04:40:41 PM
CAS classification : [_quadrature]

Solve

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

1.31.1 Solved as first order quadrature ode

Time used: 0.026 (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 28: Slope field plot
\(x y^{\prime } = 0\)

1.31.2 Solved as first order homogeneous class D2 ode

Time used: 0.123 (sec)

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

\begin{align*} x \left (u^{\prime }\left (x \right ) x +u \left (x \right )\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 29: Slope field plot
\(x y^{\prime } = 0\)

1.31.3 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 30: Slope field plot
\(x y^{\prime } = 0\)

1.31.4 Maple step by step solution

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

1.31.5 Maple trace

Methods for first order ODEs:
 

1.31.6 Maple dsolve solution

Solving time : 0.001 (sec)
Leaf size : 5

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

\[ y = c_1 \]

1.31.7 Mathematica DSolve solution

Solving time : 0.002 (sec)
Leaf size : 7

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

\[ y(x)\to c_1 \]