2.4.33 Problem 29

Internal problem ID [10196]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 29
Date solved : Thursday, November 27, 2025 at 10:26:06 AM
CAS classification : [[_1st_order, _with_linear_symmetries], _dAlembert]

Solved using first_order_ode_dAlembert

Time used: 0.198 (sec)

Solve

\begin{align*} \left (y-2 y^{\prime } x \right )^{2}&={y^{\prime }}^{3} \\ \end{align*}

Let \(p=y^{\prime }\) the ode becomes

\begin{align*} \left (-2 p x +y \right )^{2} = p^{3} \end{align*}

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

\begin{align*} \tag{1} y &= 2 p x +p^{{3}/{2}} \\ \tag{2} y &= 2 p x -p^{{3}/{2}} \\ \end{align*}

This has the form

\begin{align*} y=x f(p)+g(p)\tag {*} \end{align*}

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

Solving ode 1A

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 &= 2 p\\ g &= p^{{3}/{2}} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} -p = \left (2 x +\frac {3 \sqrt {p}}{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 = 0 \end{align*}

Solving the above for \(p\) results in

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

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

\begin{align*} 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 )}{2 x +\frac {3 \sqrt {p \left (x \right )}}{2}} \end{equation}

Inverting the above ode gives

\begin{align*} \frac {d}{d p}x \left (p \right ) = -\frac {2 x \left (p \right )+\frac {3 \sqrt {p}}{2}}{p}\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 {2}{p}d p}\\ \mu &= p^{2}\\ \mu &= p^{2}\tag {5} \end{align*}

Integrating gives

\begin{align*} x \left (p \right )&= \frac {1}{\mu } \left ( \int { \mu \left (-\frac {3}{2 \sqrt {p}}\right ) \,dp} + c_1\right )\\ &= \frac {1}{\mu } \left (\frac {-3 p^{{5}/{2}}+5 c_1}{5 p^{2}}+c_1\right ) \\ &= \frac {-3 p^{{5}/{2}}+5 c_1}{5 p^{2}}\tag {5} \end{align*}

Now we need to eliminate \(p\) between the above solution and (1A). The first method is to solve for \(p\) from Eq. (1A) and substitute the result into Eq. (5). The Second method is to solve for \(p\) from Eq. (5) and substitute the result into (1A).

Eliminating \(p\) from the following two equations

\begin{align*} x &= \frac {-3 p^{{5}/{2}}+5 c_1}{5 p^{2}} \\ y &= 2 p x +p^{{3}/{2}} \\ \end{align*}

results in

\begin{align*} p &= \operatorname {RootOf}\left (3 \textit {\_Z}^{5}+5 x \,\textit {\_Z}^{4}-5 c_1 \right )^{2} \\ \end{align*}

Substituting the above into Eq (1A) and simplifying gives

\begin{align*} y &= 2 \operatorname {RootOf}\left (3 \textit {\_Z}^{5}+5 x \,\textit {\_Z}^{4}-5 c_1 \right )^{2} x +{\left (\operatorname {RootOf}\left (3 \textit {\_Z}^{5}+5 x \,\textit {\_Z}^{4}-5 c_1 \right )^{2}\right )}^{{3}/{2}} \\ \end{align*}

Solving ode 2A

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 &= 2 p\\ g &= -p^{{3}/{2}} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} -p = \left (2 x -\frac {3 \sqrt {p}}{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 = 0 \end{align*}

Solving the above for \(p\) results in

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

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

\begin{align*} 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 )}{2 x -\frac {3 \sqrt {p \left (x \right )}}{2}} \end{equation}

Inverting the above ode gives

\begin{align*} \frac {d}{d p}x \left (p \right ) = -\frac {2 x \left (p \right )-\frac {3 \sqrt {p}}{2}}{p}\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 {2}{p}d p}\\ \mu &= p^{2}\\ \mu &= p^{2}\tag {5} \end{align*}

Integrating gives

\begin{align*} x \left (p \right )&= \frac {1}{\mu } \left ( \int { \mu \left (\frac {3}{2 \sqrt {p}}\right ) \,dp} + c_2\right )\\ &= \frac {1}{\mu } \left (\frac {3 p^{{5}/{2}}+5 c_2}{5 p^{2}}+c_2\right ) \\ &= \frac {3 p^{{5}/{2}}+5 c_2}{5 p^{2}}\tag {5} \end{align*}

Now we need to eliminate \(p\) between the above solution and (1A). The first method is to solve for \(p\) from Eq. (1A) and substitute the result into Eq. (5). The Second method is to solve for \(p\) from Eq. (5) and substitute the result into (1A).

Eliminating \(p\) from the following two equations

\begin{align*} x &= \frac {3 p^{{5}/{2}}+5 c_2}{5 p^{2}} \\ y &= 2 p x -p^{{3}/{2}} \\ \end{align*}

results in

\begin{align*} p &= \operatorname {RootOf}\left (3 \textit {\_Z}^{5}-5 x \,\textit {\_Z}^{4}+5 c_2 \right )^{2} \\ \end{align*}

Substituting the above into Eq (1A) and simplifying gives

\begin{align*} y &= 2 \operatorname {RootOf}\left (3 \textit {\_Z}^{5}-5 x \,\textit {\_Z}^{4}+5 c_2 \right )^{2} x -{\left (\operatorname {RootOf}\left (3 \textit {\_Z}^{5}-5 x \,\textit {\_Z}^{4}+5 c_2 \right )^{2}\right )}^{{3}/{2}} \\ \end{align*}

Simplifying the above gives

\begin{align*} y &= 0 \\ y &= \operatorname {RootOf}\left (3 \textit {\_Z}^{5}+5 x \,\textit {\_Z}^{4}-5 c_1 \right )^{2} \left (\sqrt {\operatorname {RootOf}\left (3 \textit {\_Z}^{5}+5 x \,\textit {\_Z}^{4}-5 c_1 \right )^{2}}+2 x \right ) \\ y &= 0 \\ y &= 2 \operatorname {RootOf}\left (3 \textit {\_Z}^{5}-5 x \,\textit {\_Z}^{4}+5 c_2 \right )^{2} x -{\left (\operatorname {RootOf}\left (3 \textit {\_Z}^{5}-5 x \,\textit {\_Z}^{4}+5 c_2 \right )^{2}\right )}^{{3}/{2}} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= 0 \\ y &= \operatorname {RootOf}\left (3 \textit {\_Z}^{5}+5 x \,\textit {\_Z}^{4}-5 c_1 \right )^{2} \left (\sqrt {\operatorname {RootOf}\left (3 \textit {\_Z}^{5}+5 x \,\textit {\_Z}^{4}-5 c_1 \right )^{2}}+2 x \right ) \\ y &= 2 \operatorname {RootOf}\left (3 \textit {\_Z}^{5}-5 x \,\textit {\_Z}^{4}+5 c_2 \right )^{2} x -{\left (\operatorname {RootOf}\left (3 \textit {\_Z}^{5}-5 x \,\textit {\_Z}^{4}+5 c_2 \right )^{2}\right )}^{{3}/{2}} \\ \end{align*}

✓ Maple. Time used: 0.086 (sec). Leaf size: 73

ode:=(y(x)-2*diff(y(x),x)*x)^2 = diff(y(x),x)^3; 
dsolve(ode,y(x), singsol=all);
 

\begin{align*} y &= 0 \\ \left [x \left (\textit {\_T} \right ) &= \frac {3 \textit {\_T}^{{5}/{2}}+5 c_1}{5 \textit {\_T}^{2}}, y \left (\textit {\_T} \right ) = \frac {\textit {\_T}^{{5}/{2}}+10 c_1}{5 \textit {\_T}}\right ] \\ \left [x \left (\textit {\_T} \right ) &= \frac {-3 \textit {\_T}^{{5}/{2}}+5 c_1}{5 \textit {\_T}^{2}}, y \left (\textit {\_T} \right ) = \frac {-\textit {\_T}^{{5}/{2}}+10 c_1}{5 \textit {\_T}}\right ] \\ \end{align*}

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 simple symmetries for implicit equations 
Successful isolation of dy/dx: 3 solutions were found. Trying to solve each res\ 
ulting ODE. 
   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying homogeneous types: 
   trying exact 
   Looking for potential symmetries 
   trying an equivalence to an Abel ODE 
   trying 1st order ODE linearizable_by_differentiation 
-> Solving 1st order ODE of high degree, Lie methods, 1st trial 
   -> Computing symmetries using: way = 2 
   -> Computing symmetries using: way = 2 
-> Solving 1st order ODE of high degree, 2nd attempt. Trying parametric methods 
trying dAlembert 
<- dAlembert successful 
<- dAlembert successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (y \left (x \right )-2 x \left (\frac {d}{d x}y \left (x \right )\right )\right )^{2}=\left (\frac {d}{d x}y \left (x \right )\right )^{3} \\ \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} \\ {} & {} & \left [\frac {d}{d x}y \left (x \right )=\frac {\left (-576 y \left (x \right ) x^{3}+108 y \left (x \right )^{2}+512 x^{6}+12 \sqrt {-96 y \left (x \right )^{3} x^{3}+81 y \left (x \right )^{4}}\right )^{{1}/{3}}}{6}-\frac {6 \left (\frac {4 y \left (x \right ) x}{3}-\frac {16 x^{4}}{9}\right )}{\left (-576 y \left (x \right ) x^{3}+108 y \left (x \right )^{2}+512 x^{6}+12 \sqrt {-96 y \left (x \right )^{3} x^{3}+81 y \left (x \right )^{4}}\right )^{{1}/{3}}}+\frac {4 x^{2}}{3}, \frac {d}{d x}y \left (x \right )=-\frac {\left (-576 y \left (x \right ) x^{3}+108 y \left (x \right )^{2}+512 x^{6}+12 \sqrt {-96 y \left (x \right )^{3} x^{3}+81 y \left (x \right )^{4}}\right )^{{1}/{3}}}{12}+\frac {3 \left (\frac {4 y \left (x \right ) x}{3}-\frac {16 x^{4}}{9}\right )}{\left (-576 y \left (x \right ) x^{3}+108 y \left (x \right )^{2}+512 x^{6}+12 \sqrt {-96 y \left (x \right )^{3} x^{3}+81 y \left (x \right )^{4}}\right )^{{1}/{3}}}+\frac {4 x^{2}}{3}-\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (-576 y \left (x \right ) x^{3}+108 y \left (x \right )^{2}+512 x^{6}+12 \sqrt {-96 y \left (x \right )^{3} x^{3}+81 y \left (x \right )^{4}}\right )^{{1}/{3}}}{6}+\frac {6 \left (\frac {4 y \left (x \right ) x}{3}-\frac {16 x^{4}}{9}\right )}{\left (-576 y \left (x \right ) x^{3}+108 y \left (x \right )^{2}+512 x^{6}+12 \sqrt {-96 y \left (x \right )^{3} x^{3}+81 y \left (x \right )^{4}}\right )^{{1}/{3}}}\right )}{2}, \frac {d}{d x}y \left (x \right )=-\frac {\left (-576 y \left (x \right ) x^{3}+108 y \left (x \right )^{2}+512 x^{6}+12 \sqrt {-96 y \left (x \right )^{3} x^{3}+81 y \left (x \right )^{4}}\right )^{{1}/{3}}}{12}+\frac {3 \left (\frac {4 y \left (x \right ) x}{3}-\frac {16 x^{4}}{9}\right )}{\left (-576 y \left (x \right ) x^{3}+108 y \left (x \right )^{2}+512 x^{6}+12 \sqrt {-96 y \left (x \right )^{3} x^{3}+81 y \left (x \right )^{4}}\right )^{{1}/{3}}}+\frac {4 x^{2}}{3}+\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (-576 y \left (x \right ) x^{3}+108 y \left (x \right )^{2}+512 x^{6}+12 \sqrt {-96 y \left (x \right )^{3} x^{3}+81 y \left (x \right )^{4}}\right )^{{1}/{3}}}{6}+\frac {6 \left (\frac {4 y \left (x \right ) x}{3}-\frac {16 x^{4}}{9}\right )}{\left (-576 y \left (x \right ) x^{3}+108 y \left (x \right )^{2}+512 x^{6}+12 \sqrt {-96 y \left (x \right )^{3} x^{3}+81 y \left (x \right )^{4}}\right )^{{1}/{3}}}\right )}{2}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=\frac {\left (-576 y \left (x \right ) x^{3}+108 y \left (x \right )^{2}+512 x^{6}+12 \sqrt {-96 y \left (x \right )^{3} x^{3}+81 y \left (x \right )^{4}}\right )^{{1}/{3}}}{6}-\frac {6 \left (\frac {4 y \left (x \right ) x}{3}-\frac {16 x^{4}}{9}\right )}{\left (-576 y \left (x \right ) x^{3}+108 y \left (x \right )^{2}+512 x^{6}+12 \sqrt {-96 y \left (x \right )^{3} x^{3}+81 y \left (x \right )^{4}}\right )^{{1}/{3}}}+\frac {4 x^{2}}{3} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=-\frac {\left (-576 y \left (x \right ) x^{3}+108 y \left (x \right )^{2}+512 x^{6}+12 \sqrt {-96 y \left (x \right )^{3} x^{3}+81 y \left (x \right )^{4}}\right )^{{1}/{3}}}{12}+\frac {3 \left (\frac {4 y \left (x \right ) x}{3}-\frac {16 x^{4}}{9}\right )}{\left (-576 y \left (x \right ) x^{3}+108 y \left (x \right )^{2}+512 x^{6}+12 \sqrt {-96 y \left (x \right )^{3} x^{3}+81 y \left (x \right )^{4}}\right )^{{1}/{3}}}+\frac {4 x^{2}}{3}-\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (-576 y \left (x \right ) x^{3}+108 y \left (x \right )^{2}+512 x^{6}+12 \sqrt {-96 y \left (x \right )^{3} x^{3}+81 y \left (x \right )^{4}}\right )^{{1}/{3}}}{6}+\frac {6 \left (\frac {4 y \left (x \right ) x}{3}-\frac {16 x^{4}}{9}\right )}{\left (-576 y \left (x \right ) x^{3}+108 y \left (x \right )^{2}+512 x^{6}+12 \sqrt {-96 y \left (x \right )^{3} x^{3}+81 y \left (x \right )^{4}}\right )^{{1}/{3}}}\right )}{2} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=-\frac {\left (-576 y \left (x \right ) x^{3}+108 y \left (x \right )^{2}+512 x^{6}+12 \sqrt {-96 y \left (x \right )^{3} x^{3}+81 y \left (x \right )^{4}}\right )^{{1}/{3}}}{12}+\frac {3 \left (\frac {4 y \left (x \right ) x}{3}-\frac {16 x^{4}}{9}\right )}{\left (-576 y \left (x \right ) x^{3}+108 y \left (x \right )^{2}+512 x^{6}+12 \sqrt {-96 y \left (x \right )^{3} x^{3}+81 y \left (x \right )^{4}}\right )^{{1}/{3}}}+\frac {4 x^{2}}{3}+\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (-576 y \left (x \right ) x^{3}+108 y \left (x \right )^{2}+512 x^{6}+12 \sqrt {-96 y \left (x \right )^{3} x^{3}+81 y \left (x \right )^{4}}\right )^{{1}/{3}}}{6}+\frac {6 \left (\frac {4 y \left (x \right ) x}{3}-\frac {16 x^{4}}{9}\right )}{\left (-576 y \left (x \right ) x^{3}+108 y \left (x \right )^{2}+512 x^{6}+12 \sqrt {-96 y \left (x \right )^{3} x^{3}+81 y \left (x \right )^{4}}\right )^{{1}/{3}}}\right )}{2} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]

✓ Mathematica. Time used: 89.397 (sec). Leaf size: 32354

ode=(y[x]-2*x*D[y[x],x])^2== D[y[x],x]^3; 
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((-2*x*Derivative(y(x), x) + y(x))**2 - Derivative(y(x), x)**3,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

Timed Out