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2.6.5 Problem (e)

Solved using first_order_ode_dAlembert
✓Maple
✓Mathematica
✗Sympy

Internal problem ID [20985]
Book : Ordinary Differential Equations. By Wolfgang Walter. Graduate texts in Mathematics. Springer. NY. QA372.W224 1998
Section : Chapter 1. First order equations: Some integrable cases. Excercises VIII at page 51
Problem number : (e)
Date solved : Saturday, November 29, 2025 at 01:26:05 AM
CAS classification : [_dAlembert]

Solved using first_order_ode_dAlembert

Time used: 0.449 (sec)

Solve

\begin{align*} x&=y \left (y^{\prime }+\frac {1}{y^{\prime }}\right )+{y^{\prime }}^{5} \\ \end{align*}
Let \(p=y^{\prime }\) the ode becomes
\begin{align*} x = y \left (p +\frac {1}{p}\right )+p^{5} \end{align*}

Solving for \(y\) from the above results in

\begin{align*} \tag{1} y &= \frac {p x}{p^{2}+1}-\frac {p^{6}}{p^{2}+1} \\ \end{align*}
This has the form
\begin{align*} y=x f(p)+g(p)\tag {*} \end{align*}

Where \(f,g\) are functions of \(p=y'(x)\). The above ode is dAlembert ode which is now solved.

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(y=x f + g\) to (1A) shows that

\begin{align*} f &= \frac {p}{p^{2}+1}\\ g &= -\frac {p^{6}}{p^{2}+1} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p -\frac {p}{p^{2}+1} = \left (-\frac {2 x \,p^{2}}{\left (p^{2}+1\right )^{2}}+\frac {x}{p^{2}+1}-\frac {6 p^{5}}{p^{2}+1}+\frac {2 p^{7}}{\left (p^{2}+1\right )^{2}}\right ) p^{\prime }\left (x \right ) \end{equation}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives
\begin{align*} p -\frac {p}{p^{2}+1} = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0\\ p_{2} &=0\\ p_{3} &=0 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

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

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{equation} \tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )-\frac {p \left (x \right )}{p \left (x \right )^{2}+1}}{-\frac {2 x p \left (x \right )^{2}}{\left (p \left (x \right )^{2}+1\right )^{2}}+\frac {x}{p \left (x \right )^{2}+1}-\frac {6 p \left (x \right )^{5}}{p \left (x \right )^{2}+1}+\frac {2 p \left (x \right )^{7}}{\left (p \left (x \right )^{2}+1\right )^{2}}} \end{equation}

Inverting the above ode gives

\begin{align*} \frac {d}{d p}x \left (p \right ) = \frac {-\frac {2 x \left (p \right ) p^{2}}{\left (p^{2}+1\right )^{2}}+\frac {x \left (p \right )}{p^{2}+1}-\frac {6 p^{5}}{p^{2}+1}+\frac {2 p^{7}}{\left (p^{2}+1\right )^{2}}}{p -\frac {p}{p^{2}+1}}\tag {4} \end{align*}

This ODE is now solved for \(x \left (p \right )\). The integrating factor is

\begin{align*} \mu &= {\mathrm e}^{\int -\frac {-p^{2}+1}{p^{3} \left (p^{2}+1\right )}d p}\\ \mu &= {\mathrm e}^{-\ln \left (p^{2}+1\right )+\frac {1}{2 p^{2}}+2 \ln \left (p \right )}\\ \mu &= \frac {p^{2} {\mathrm e}^{\frac {1}{2 p^{2}}}}{p^{2}+1}\tag {5} \end{align*}

Integrating gives

\begin{align*} x \left (p \right )&=\frac {\left (p^{2}+1\right ) {\mathrm e}^{-\frac {1}{2 p^{2}}} \left (\int \frac {\left (-4 p^{4}-6 p^{2}\right ) p^{2} {\mathrm e}^{\frac {1}{2 p^{2}}}}{\left (p^{2}+1\right )^{2}}d p +c_1 \right )}{p^{2}}\tag {5} \end{align*}

Unable to use this solution. skipping

Summary of solutions found

\begin{align*} y &= 0 \\ \end{align*}
✓ Maple. Time used: 0.174 (sec). Leaf size: 110
ode:=x = y(x)*(diff(y(x),x)+1/diff(y(x),x))+diff(y(x),x)^5; 
dsolve(ode,y(x), singsol=all);
\[ \left [x \left (\textit {\_T} \right ) = \frac {{\mathrm e}^{-\frac {1}{2 \textit {\_T}^{2}}} \left (\textit {\_T}^{2}+1\right ) \left (-2 \int \frac {\textit {\_T}^{4} \left (2 \textit {\_T}^{2}+3\right ) {\mathrm e}^{\frac {1}{2 \textit {\_T}^{2}}}}{\left (\textit {\_T}^{2}+1\right )^{2}}d \textit {\_T} +c_1 \right )}{\textit {\_T}^{2}}, y \left (\textit {\_T} \right ) = \frac {{\mathrm e}^{-\frac {1}{2 \textit {\_T}^{2}}} \textit {\_T} \left (1+\frac {1}{\textit {\_T}^{2}}\right ) \left (-2 \int \frac {\textit {\_T}^{4} \left (2 \textit {\_T}^{2}+3\right ) {\mathrm e}^{\frac {1}{2 \textit {\_T}^{2}}}}{\left (\textit {\_T}^{2}+1\right )^{2}}d \textit {\_T} +c_1 \right )-\textit {\_T}^{6}}{\textit {\_T}^{2}+1}\right ] \]

Maple trace 

Methods for first order ODEs: 
-> Solving 1st order ODE of high degree, 1st attempt 
trying 1st order WeierstrassP solution for high degree ODE 
trying 1st order WeierstrassPPrime solution for high degree ODE 
trying 1st order JacobiSN solution for high degree ODE 
trying 1st order ODE linearizable_by_differentiation 
trying differential order: 1; missing variables 
trying dAlembert 
<- dAlembert successful

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x =y \left (x \right ) \left (\frac {d}{d x}y \left (x \right )+\frac {1}{\frac {d}{d x}y \left (x \right )}\right )+\left (\frac {d}{d x}y \left (x \right )\right )^{5} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\mathit {RootOf}\left (\textit {\_Z}^{6}+y \left (x \right ) \textit {\_Z}^{2}-x \textit {\_Z} +y \left (x \right )\right ) \end {array} \]
✓ Mathematica. Time used: 2.788 (sec). Leaf size: 7957
ode=x==y[x]*(D[y[x],x]+1/D[y[x],x])+D[y[x],x]^5; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]

Too large to display

✗ Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x - (Derivative(y(x), x) + 1/Derivative(y(x), x))*y(x) - Derivative(y(x), x)**5,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : The given ODE x - (Derivative(y(x), x) + 1/Derivative(y(x), x))*y(x) - Derivative(y(x), x)**5 cannot be solved by the lie group method

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