2.1.46 Problem 46

Internal problem ID [10405]
Book : Second order enumerated odes
Section : section 1
Problem number : 46
Date solved : Monday, December 08, 2025 at 08:46:43 PM
CAS classification : [[_2nd_order, _missing_x]]

2.1.46.1 second order ode missing x

2.570 (sec)

\begin{align*} y {y^{\prime \prime }}^{2}+{y^{\prime }}^{3}&=0 \\ \end{align*}

Entering second order ode missing \(x\) solverThis is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using

\begin{align*} y' &= p \end{align*}

Then

\begin{align*} y'' &= \frac {dp}{dx}\\ &= \frac {dp}{dy}\frac {dy}{dx}\\ &= p \frac {dp}{dy} \end{align*}

Hence the ode becomes

\begin{align*} y p \left (y \right )^{2} \left (\frac {d}{d y}p \left (y \right )\right )^{2}+p \left (y \right )^{3} = 0 \end{align*}

Which is now solved as first order ode for \(p(y)\).

2.1.46.2 Solved by factoring the differential equation

Time used: 0.397 (sec)

\begin{align*} y p^{2} {p^{\prime }}^{2}+p^{3}&=0 \\ \end{align*}

Writing the ode as

\begin{align*} \left (p^{2}\right )\left ({p^{\prime }}^{2} y +p\right )&=0 \end{align*}

Therefore we need to solve the following equations

\begin{align*} \tag{1} p^{2} &= 0 \\ \tag{2} {p^{\prime }}^{2} y +p &= 0 \\ \end{align*}

Now each of the above equations is solved in turn.

Solving equation (1)

Entering zero order ode solverSolving for \(p\) from

\begin{align*} p^{2} = 0 \end{align*}

Solving gives

\begin{align*} p &= 0 \\ \end{align*}

Solving equation (2)

Entering first order ode dAlembert solverLet \(p=p^{\prime }\) the ode becomes

\begin{align*} p^{2} y +p = 0 \end{align*}

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

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

This has the form

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

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

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

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

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

\begin{align*} f &= -p^{2}\\ g &= 0 \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p^{2}+p = -2 y p p^{\prime }\left (y \right ) \end{equation}

The singular solution is found by setting \(\frac {dp}{dy}=0\) in the above which gives

\begin{align*} p^{2}+p = 0 \end{align*}

Solving the above for \(p\) results in

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

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

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

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

\begin{equation} \tag{3} p^{\prime }\left (y \right ) = -\frac {p \left (y \right )^{2}+p \left (y \right )}{2 y p \left (y \right )} \end{equation}

This ODE is now solved for \(p \left (y \right )\). No inversion is needed.

The ode

\begin{equation} p^{\prime }\left (y \right ) = -\frac {p \left (y \right )+1}{2 y} \end{equation}

is separable as it can be written as

\begin{align*} p^{\prime }\left (y \right )&= -\frac {p \left (y \right )+1}{2 y}\\ &= f(y) g(p) \end{align*}

Where

\begin{align*} f(y) &= \frac {1}{y}\\ g(p) &= -\frac {p}{2}-\frac {1}{2} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(y) \,dy} \\ \int { \frac {1}{-\frac {p}{2}-\frac {1}{2}}\,dp} &= \int { \frac {1}{y} \,dy} \\ \end{align*}

\[ -2 \ln \left (p \left (y \right )+1\right )=\ln \left (y \right )+c_1 \]

Taking the exponential of both sides the solution becomes

\[ \frac {1}{\left (p \left (y \right )+1\right )^{2}} = c_1 y \]

We now need to find the singular solutions, these are found by finding for what values \(g(p)\) is zero, since we had to divide by this above. Solving \(g(p)=0\) or

\[ -\frac {p}{2}-\frac {1}{2}=0 \]

for \(p \left (y \right )\) gives

\begin{align*} p \left (y \right )&=-1 \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*} \frac {1}{\left (p \left (y \right )+1\right )^{2}} &= c_1 y \\ p \left (y \right ) &= -1 \\ \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} p &= -\frac {\left (\sqrt {c_1 y}-1\right )^{2}}{c_1} \\ p &= -y \\ \end{align*}

Summary of solutions found

\begin{align*} p &= 0 \\ p &= -y \\ p &= -\frac {\left (\sqrt {c_1 y}-1\right )^{2}}{c_1} \\ \end{align*}

For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

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

Entering first order ode quadrature solverSince 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_2 \\ y &= c_2 \end{align*}

For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = -y \end{align*}

Entering first order ode autonomous solverIntegrating gives

\begin{align*} \int -\frac {1}{y}d y &= dx\\ -\ln \left (y \right )&= x +c_3 \end{align*}

Singular solutions are found by solving

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

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

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

Solving for \(y\) gives

\begin{align*} y &= 0 \\ y &= {\mathrm e}^{-x -c_3} \\ \end{align*}

For solution (3) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = -\frac {\left (\sqrt {c_1 y}-1\right )^{2}}{c_1} \end{align*}

Entering first order ode autonomous solverIntegrating gives

\begin{align*} \int -\frac {c_1}{\left (\sqrt {c_1 y}-1\right )^{2}}d y &= dx\\ \frac {2}{\sqrt {c_1 y}-1}-2 \ln \left (\sqrt {c_1 y}-1\right )&= x +c_4 \end{align*}

Singular solutions are found by solving

\begin{align*} -\frac {\left (\sqrt {c_1 y}-1\right )^{2}}{c_1}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = \frac {1}{c_1} \end{align*}

Simplifying the above gives

\begin{align*} \frac {2+\left (-2 \sqrt {c_1 y}+2\right ) \ln \left (\sqrt {c_1 y}-1\right )}{\sqrt {c_1 y}-1} &= x +c_4 \\ y &= \frac {1}{c_1} \\ \end{align*}

Solving for \(y\) gives

\begin{align*} y &= \frac {1}{c_1} \\ y &= \frac {\operatorname {LambertW}\left ({\mathrm e}^{\frac {c_4}{2}+\frac {x}{2}}\right )^{2}+2 \operatorname {LambertW}\left ({\mathrm e}^{\frac {c_4}{2}+\frac {x}{2}}\right )+1}{\operatorname {LambertW}\left ({\mathrm e}^{\frac {c_4}{2}+\frac {x}{2}}\right )^{2} c_1} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= 0 \\ y &= c_2 \\ y &= \frac {1}{c_1} \\ y &= \frac {\operatorname {LambertW}\left ({\mathrm e}^{\frac {c_4}{2}+\frac {x}{2}}\right )^{2}+2 \operatorname {LambertW}\left ({\mathrm e}^{\frac {c_4}{2}+\frac {x}{2}}\right )+1}{\operatorname {LambertW}\left ({\mathrm e}^{\frac {c_4}{2}+\frac {x}{2}}\right )^{2} c_1} \\ y &= {\mathrm e}^{-x -c_3} \\ \end{align*}

2.1.46.3 ✓ Maple. Time used: 0.167 (sec). Leaf size: 178

ode:=y(x)*diff(diff(y(x),x),x)^2+diff(y(x),x)^3 = 0; 
dsolve(ode,y(x), singsol=all);
 

\begin{align*} y &= c_1 \\ y &= 0 \\ y &= {\mathrm e}^{-\int {\mathrm e}^{2 \operatorname {RootOf}\left ({\mathrm e}^{\textit {\_Z}} \ln \left (\left ({\mathrm e}^{\textit {\_Z}}-1\right )^{2}\right )+c_1 \,{\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}} \textit {\_Z} +x \,{\mathrm e}^{\textit {\_Z}}-\ln \left (\left ({\mathrm e}^{\textit {\_Z}}-1\right )^{2}\right )-c_1 +2 \textit {\_Z} -x +2\right )}d x +2 \int {\mathrm e}^{\operatorname {RootOf}\left ({\mathrm e}^{\textit {\_Z}} \ln \left (\left ({\mathrm e}^{\textit {\_Z}}-1\right )^{2}\right )+c_1 \,{\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}} \textit {\_Z} +x \,{\mathrm e}^{\textit {\_Z}}-\ln \left (\left ({\mathrm e}^{\textit {\_Z}}-1\right )^{2}\right )-c_1 +2 \textit {\_Z} -x +2\right )}d x -x +c_2} \\ y &= \frac {c_2 {\left (\operatorname {LambertW}\left (c_1 \,{\mathrm e}^{-1+\frac {x}{2}}\right )+1\right )}^{2}}{\operatorname {LambertW}\left (c_1 \,{\mathrm e}^{-1+\frac {x}{2}}\right )^{2}} \\ y &= \frac {c_2 {\left (\operatorname {LambertW}\left (-c_1 \,{\mathrm e}^{-1+\frac {x}{2}}\right )+1\right )}^{2}}{\operatorname {LambertW}\left (-c_1 \,{\mathrm e}^{-1+\frac {x}{2}}\right )^{2}} \\ \end{align*}

Maple trace

Methods for second order ODEs: 
   *** Sublevel 2 *** 
   Methods for second order ODEs: 
   Successful isolation of d^2y/dx^2: 2 solutions were found. Trying to solve e\ 
ach resulting ODE. 
      *** Sublevel 3 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying 2nd order Liouville 
      trying 2nd order WeierstrassP 
      trying 2nd order JacobiSN 
      differential order: 2; trying a linearization to 3rd order 
      trying 2nd order ODE linearizable_by_differentiation 
      trying 2nd order, 2 integrating factors of the form mu(x,y) 
      trying differential order: 2; missing variables 
         -> Computing symmetries using: way = 3 
      Try integration with the canonical coordinates of the symmetry [0, y] 
      -> Calling odsolve with the ODE, diff(_b(_a),_a) = -(-_b(_a))^(3/2)-_b(_a 
)^2, _b(_a), explicit, HINT = [[1, 0]] 
         *** Sublevel 4 *** 
         symmetry methods on request 
         1st order, trying reduction of order with given symmetries: 
[1, 0] 
         1st order, trying the canonical coordinates of the invariance group 
         <- 1st order, canonical coordinates successful 
      <- differential order: 2; canonical coordinates successful 
      <- differential order 2; missing variables successful 
   ------------------- 
   * Tackling next ODE. 
      *** Sublevel 3 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying 2nd order Liouville 
      trying 2nd order WeierstrassP 
      trying 2nd order JacobiSN 
      differential order: 2; trying a linearization to 3rd order 
      trying 2nd order ODE linearizable_by_differentiation 
      trying 2nd order, 2 integrating factors of the form mu(x,y) 
      trying differential order: 2; missing variables 
         -> Computing symmetries using: way = 3 
      Try integration with the canonical coordinates of the symmetry [0, y] 
      -> Calling odsolve with the ODE, diff(_b(_a),_a) = (-_b(_a))^(3/2)-_b(_a) 
^2, _b(_a), explicit, HINT = [[1, 0]] 
         *** Sublevel 4 *** 
         symmetry methods on request 
         1st order, trying reduction of order with given symmetries: 
[1, 0] 
         1st order, trying the canonical coordinates of the invariance group 
         <- 1st order, canonical coordinates successful 
      <- differential order: 2; canonical coordinates successful 
      <- differential order 2; missing variables successful 
-> Calling odsolve with the ODE, diff(y(x),x) = 0, y(x), singsol = none 
   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   <- 1st order linear successful
 

2.1.46.4 ✓ Mathematica. Time used: 1.187 (sec). Leaf size: 361

ode=y[x]*D[y[x],{x,2}]^2+D[y[x],x]^3==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to \text {InverseFunction}\left [-4 \left (\frac {1}{2} \log \left (2 \sqrt {\text {$\#$1}}-i c_1\right )-\frac {i c_1}{2 \left (2 \sqrt {\text {$\#$1}}-i c_1\right )}\right )\&\right ][x+c_2]\\ y(x)&\to \text {InverseFunction}\left [-4 \left (\frac {i c_1}{2 \left (2 \sqrt {\text {$\#$1}}+i c_1\right )}+\frac {1}{2} \log \left (2 \sqrt {\text {$\#$1}}+i c_1\right )\right )\&\right ][x+c_2]\\ y(x)&\to \text {InverseFunction}\left [-4 \left (\frac {1}{2} \log \left (2 \sqrt {\text {$\#$1}}-i (-c_1)\right )-\frac {i (-c_1)}{2 \left (2 \sqrt {\text {$\#$1}}-i (-c_1)\right )}\right )\&\right ][x+c_2]\\ y(x)&\to \text {InverseFunction}\left [-4 \left (\frac {i (-c_1)}{2 \left (2 \sqrt {\text {$\#$1}}+i (-1) c_1\right )}+\frac {1}{2} \log \left (2 \sqrt {\text {$\#$1}}+i (-1) c_1\right )\right )\&\right ][x+c_2]\\ y(x)&\to \text {InverseFunction}\left [-4 \left (\frac {1}{2} \log \left (2 \sqrt {\text {$\#$1}}-i c_1\right )-\frac {i c_1}{2 \left (2 \sqrt {\text {$\#$1}}-i c_1\right )}\right )\&\right ][x+c_2]\\ y(x)&\to \text {InverseFunction}\left [-4 \left (\frac {i c_1}{2 \left (2 \sqrt {\text {$\#$1}}+i c_1\right )}+\frac {1}{2} \log \left (2 \sqrt {\text {$\#$1}}+i c_1\right )\right )\&\right ][x+c_2] \end{align*}

2.1.46.5 ✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(y(x)*Derivative(y(x), (x, 2))**2 + Derivative(y(x), x)**3,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : The given ODE (-y(x)*Derivative(y(x), (x, 2))**2)**(1/3)/2 - sqrt(3)*I*(-y(x)*Derivative(y(x), (x, 2))**2)**(1/3)/2 + Derivative(y(x), x) cannot be solved by the factorable group method