1.47 problem 47

Internal problem ID [8011]
Book : First order enumerated odes
Section : section 1
Problem number : 47
Date solved : Monday, October 21, 2024 at 04:40:52 PM
CAS classification : [_quadrature]

Solve

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

1.47.1 Solved as first order quadrature ode

Time used: 0.052 (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 68: Slope field plot
\({y^{\prime }}^{n} = 0\)

1.47.2 Solved as first order homogeneous class D2 ode

Time used: 0.205 (sec)

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

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

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 ) = \frac {{\mathrm e}^{c_1}}{x} \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 69: Slope field plot
\({y^{\prime }}^{n} = 0\)

1.47.3 Maple step by step solution

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

1.47.4 Maple trace

Methods for first order ODEs:
 

1.47.5 Maple dsolve solution

Solving time : 0.003 (sec)
Leaf size : 5

dsolve(diff(y(x),x)^n = 0, 
       y(x),singsol=all)
 

\[ y = c_1 \]

1.47.6 Mathematica DSolve solution

Solving time : 0.003 (sec)
Leaf size : 15

DSolve[{(D[y[x],x])^n==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to 0^{\frac {1}{n}} x+c_1 \]