2.1.47 Problem 47

Internal problem ID [10406]
Book : Second order enumerated odes
Section : section 1
Problem number : 47
Date solved : Monday, December 08, 2025 at 08:46:53 PM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]

2.1.47.1 second order ode missing x

2.474 (sec)

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

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

2.1.47.2 Solved by factoring the differential equation

Time used: 0.332 (sec)

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

Writing the ode as

\begin{align*} \left (p\right )\left ({p^{\prime }}^{2} p y^{2}+1\right )&=0 \end{align*}

Therefore we need to solve the following equations

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

Solving gives

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

Solving equation (2)

Entering first order ode homog type G solverMultiplying the right side of the ode, which is \(-\frac {1}{\sqrt {-p}\, y}\) by \(\frac {y}{p}\) gives

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

Since \(F \left (y , p\right )\) has \(p\), then let

\begin{align*} f_y&= y \left (\frac {\partial }{\partial y}F \left (y , p\right )\right )\\ &= 0\\ f_p&= p \left (\frac {\partial }{\partial p}F \left (y , p\right )\right )\\ &= -\frac {3}{2 \left (-p \right )^{{3}/{2}}}\\ \alpha &= \frac {f_y}{f_p} \\ &=0 \end{align*}

Since \(\alpha \) is independent of \(y,p\) then this is Homogeneous type G.

Let

\begin{align*} p&=\frac {z}{y^ \alpha }\\ &=\frac {z}{1} \end{align*}

Substituting the above back into \(F(y,p)\) gives

\begin{align*} F \left (z \right ) &=\frac {1}{\left (-z \right )^{{3}/{2}}} \end{align*}

We see that \(F \left (z \right )\) does not depend on \(y\) nor on \(p\). If this was not the case, then this method will not work.

Therefore, the implicit solution is given by

\begin{align*} \ln \left (y \right )- c_1 - \int ^{p y^\alpha } \frac {1}{z \left (\alpha + F(z)\right ) } \,dz & = 0 \end{align*}

Which gives

\[ \ln \left (y \right )-c_1 +\int _{}^{p}-\frac {\left (-z \right )^{{3}/{2}}}{z}d z = 0 \]

Multiplying the right side of the ode, which is \(\frac {1}{\sqrt {-p}\, y}\) by \(\frac {y}{p}\) gives

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

Since \(F \left (y , p\right )\) has \(p\), then let

\begin{align*} f_y&= y \left (\frac {\partial }{\partial y}F \left (y , p\right )\right )\\ &= 0\\ f_p&= p \left (\frac {\partial }{\partial p}F \left (y , p\right )\right )\\ &= \frac {3}{2 \left (-p \right )^{{3}/{2}}}\\ \alpha &= \frac {f_y}{f_p} \\ &=0 \end{align*}

Since \(\alpha \) is independent of \(y,p\) then this is Homogeneous type G.

Let

\begin{align*} p&=\frac {z}{y^ \alpha }\\ &=\frac {z}{1} \end{align*}

Substituting the above back into \(F(y,p)\) gives

\begin{align*} F \left (z \right ) &=-\frac {1}{\left (-z \right )^{{3}/{2}}} \end{align*}

We see that \(F \left (z \right )\) does not depend on \(y\) nor on \(p\). If this was not the case, then this method will not work.

Therefore, the implicit solution is given by

\begin{align*} \ln \left (y \right )- c_1 - \int ^{p y^\alpha } \frac {1}{z \left (\alpha + F(z)\right ) } \,dz & = 0 \end{align*}

Which gives

\[ \ln \left (y \right )-c_2 +\int _{}^{p}\frac {\left (-z \right )^{{3}/{2}}}{z}d z = 0 \]

Summary of solutions found

\begin{align*} \ln \left (y \right )-c_1 +\int _{}^{p}-\frac {\left (-z \right )^{{3}/{2}}}{z}d z &= 0 \\ \ln \left (y \right )-c_2 +\int _{}^{p}\frac {\left (-z \right )^{{3}/{2}}}{z}d z &= 0 \\ p &= 0 \\ \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*} \ln \left (y\right )-c_1 +\int _{}^{y^{\prime }}-\frac {\left (-z \right )^{{3}/{2}}}{z}d z = 0 \end{align*}

Entering first order ode autonomous solverUnable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{y}\frac {1}{\operatorname {RootOf}\left (-\ln \left (\tau \right )+c_1 -\int _{}^{\textit {\_Z}}-\frac {\left (-z \right )^{{3}/{2}}}{z}d z \right )}d \tau = x +c_3 \]

Solving for \(y\) gives

\begin{align*} y &= \operatorname {RootOf}\left (-\int _{}^{\textit {\_Z}}\frac {1}{\operatorname {RootOf}\left (-\ln \left (\tau \right )+c_1 -\int _{}^{\textit {\_Z}}-\frac {\left (-z \right )^{{3}/{2}}}{z}d z \right )}d \tau +x +c_3 \right ) \\ \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*} \ln \left (y\right )-c_2 +\int _{}^{y^{\prime }}\frac {\left (-z \right )^{{3}/{2}}}{z}d z = 0 \end{align*}

Entering first order ode autonomous solverUnable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{y}\frac {1}{\operatorname {RootOf}\left (-\ln \left (\tau \right )+c_2 -\int _{}^{\textit {\_Z}}\frac {\left (-z \right )^{{3}/{2}}}{z}d z \right )}d \tau = x +c_4 \]

Solving for \(y\) gives

\begin{align*} y &= \operatorname {RootOf}\left (-\int _{}^{\textit {\_Z}}\frac {1}{\operatorname {RootOf}\left (-\ln \left (\tau \right )+c_2 -\int _{}^{\textit {\_Z}}\frac {\left (-z \right )^{{3}/{2}}}{z}d z \right )}d \tau +x +c_4 \right ) \\ \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 } = 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_5 \\ y &= c_5 \end{align*}

Summary of solutions found

\begin{align*} y &= c_5 \\ y &= \operatorname {RootOf}\left (-\int _{}^{\textit {\_Z}}\frac {1}{\operatorname {RootOf}\left (-\ln \left (\tau \right )+c_2 -\int _{}^{\textit {\_Z}}\frac {\left (-z \right )^{{3}/{2}}}{z}d z \right )}d \tau +x +c_4 \right ) \\ y &= \operatorname {RootOf}\left (-\int _{}^{\textit {\_Z}}\frac {1}{\operatorname {RootOf}\left (-\ln \left (\tau \right )+c_1 -\int _{}^{\textit {\_Z}}-\frac {\left (-z \right )^{{3}/{2}}}{z}d z \right )}d \tau +x +c_3 \right ) \\ \end{align*}

2.1.47.3 ✓ Maple. Time used: 0.036 (sec). Leaf size: 233

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

\begin{align*} y &= c_1 \\ y &= 0 \\ -4 \int _{}^{y}\frac {1}{\left (-12 \ln \left (\textit {\_a} \right )+8 c_1 \right )^{{2}/{3}}}d \textit {\_a} -x -c_2 &= 0 \\ -4 \int _{}^{y}\frac {1}{\left (12 \ln \left (\textit {\_a} \right )-8 c_1 \right )^{{2}/{3}}}d \textit {\_a} -x -c_2 &= 0 \\ \frac {-16 \int _{}^{y}\frac {1}{\left (-12 \ln \left (\textit {\_a} \right )+8 c_1 \right )^{{2}/{3}}}d \textit {\_a} -2 i \left (c_2 +x \right ) \sqrt {3}+2 x +2 c_2}{\left (-1-i \sqrt {3}\right )^{2}} &= 0 \\ \frac {-16 \int _{}^{y}\frac {1}{\left (-12 \ln \left (\textit {\_a} \right )+8 c_1 \right )^{{2}/{3}}}d \textit {\_a} +2 i \left (c_2 +x \right ) \sqrt {3}+2 x +2 c_2}{\left (1-i \sqrt {3}\right )^{2}} &= 0 \\ \frac {-16 \int _{}^{y}\frac {1}{\left (12 \ln \left (\textit {\_a} \right )-8 c_1 \right )^{{2}/{3}}}d \textit {\_a} -2 i \left (c_2 +x \right ) \sqrt {3}+2 x +2 c_2}{\left (-1-i \sqrt {3}\right )^{2}} &= 0 \\ \frac {-16 \int _{}^{y}\frac {1}{\left (12 \ln \left (\textit {\_a} \right )-8 c_1 \right )^{{2}/{3}}}d \textit {\_a} +2 i \left (c_2 +x \right ) \sqrt {3}+2 x +2 c_2}{\left (1-i \sqrt {3}\right )^{2}} &= 0 \\ \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 
      -> Calling odsolve with the ODE, diff(_b(_a),_a)*_b(_a)-(-_b(_a))^(1/2)/ 
_a = 0, _b(_a), HINT = [[_a, 0]] 
         *** Sublevel 4 *** 
         symmetry methods on request 
         1st order, trying reduction of order with given symmetries: 
[_a, 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 
      -> Calling odsolve with the ODE, diff(_b(_a),_a)*_b(_a)+(-_b(_a))^(1/2)/ 
_a = 0, _b(_a), HINT = [[_a, 0]] 
         *** Sublevel 4 *** 
         symmetry methods on request 
         1st order, trying reduction of order with given symmetries: 
[_a, 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
 

2.1.47.4 ✓ Mathematica. Time used: 1.385 (sec). Leaf size: 449

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

\begin{align*} y(x)&\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{-i c_1} (-\log (\text {$\#$1})-i c_1){}^{2/3} \Gamma \left (\frac {1}{3},-i c_1-\log (\text {$\#$1})\right )}{(c_1-i \log (\text {$\#$1})){}^{2/3}}\&\right ][x+c_2]\\ y(x)&\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{i c_1} (-\log (\text {$\#$1})+i c_1){}^{2/3} \Gamma \left (\frac {1}{3},i c_1-\log (\text {$\#$1})\right )}{(i \log (\text {$\#$1})+c_1){}^{2/3}}\&\right ][x+c_2]\\ y(x)&\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{-i (-c_1)} (-\log (\text {$\#$1})-i (-1) c_1){}^{2/3} \Gamma \left (\frac {1}{3},-i (-1) c_1-\log (\text {$\#$1})\right )}{(-i \log (\text {$\#$1})-c_1){}^{2/3}}\&\right ][x+c_2]\\ y(x)&\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{-i c_1} (-\log (\text {$\#$1})-i c_1){}^{2/3} \Gamma \left (\frac {1}{3},-i c_1-\log (\text {$\#$1})\right )}{(c_1-i \log (\text {$\#$1})){}^{2/3}}\&\right ][x+c_2]\\ y(x)&\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{i (-c_1)} (-\log (\text {$\#$1})+i (-c_1)){}^{2/3} \Gamma \left (\frac {1}{3},i (-c_1)-\log (\text {$\#$1})\right )}{(i \log (\text {$\#$1})-c_1){}^{2/3}}\&\right ][x+c_2]\\ y(x)&\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{i c_1} (-\log (\text {$\#$1})+i c_1){}^{2/3} \Gamma \left (\frac {1}{3},i c_1-\log (\text {$\#$1})\right )}{(i \log (\text {$\#$1})+c_1){}^{2/3}}\&\right ][x+c_2] \end{align*}

2.1.47.5 ✗ Sympy

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

NotImplementedError : solve: Cannot solve y(x)**2*Derivative(y(x), (x, 2))**2 + Derivative(y(x), x)