2.1.49 Problem 49

Internal problem ID [9120]
Book : Second order enumerated odes
Section : section 1
Problem number : 49
Date solved : Friday, February 21, 2025 at 09:16:09 PM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]

Solve

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

Factoring the ode gives these factors

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

Now each of the above equations is solved in turn.

Solving equation (1)

Solving for \(y\) from

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

Solving gives \(y = 0\)

Solving equation (2)

Solved as second order missing x ode

Time used: 5.033 (sec)

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

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

Factoring the ode gives these factors

\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)

Solving for \(p\) from

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

Solving gives \(p = 0\)

Solving equation (2)

Solving for the derivative gives these ODE’s to solve

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

Now each of the above is solved separately.

Solving Eq. (1)

The ode

\begin{equation} p^{\prime } = -\frac {1}{\sqrt {-p}\, y} \end{equation}

is separable as it can be written as

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

Where

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

Integrating gives

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

\[ -\frac {2 \left (-p\right )^{{3}/{2}}}{3}=\ln \left (\frac {1}{y}\right )+c_1 \]

Solving for \(p\) gives

\begin{align*} p &= -\frac {\left (-12 \ln \left (\frac {1}{y}\right )-12 c_1 \right )^{{2}/{3}}}{4} \\ \end{align*}

Solving Eq. (2)

The ode

\begin{equation} p^{\prime } = \frac {1}{\sqrt {-p}\, y} \end{equation}

is separable as it can be written as

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

Where

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

Integrating gives

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

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

Solving for \(p\) gives

\begin{align*} p &= -\frac {\left (-12 \ln \left (y \right )-12 c_2 \right )^{{2}/{3}}}{4} \\ \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*}

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_3 \\ y &= c_3 \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 } = -\frac {\left (-12 \ln \left (y\right )-12 c_2 \right )^{{2}/{3}}}{4} \end{align*}

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{y}-\frac {4}{\left (-12 \ln \left (\tau \right )-12 c_2 \right )^{{2}/{3}}}d \tau = x +c_4 \]

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 (-12 \ln \left (\frac {1}{y}\right )-12 c_1 \right )^{{2}/{3}}}{4} \end{align*}

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{y}-\frac {4}{\left (-12 \ln \left (\frac {1}{\tau }\right )-12 c_1 \right )^{{2}/{3}}}d \tau = x +c_5 \]

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} \int _{}^{y}-\frac {4}{\left (-12 \ln \left (\tau \right )-12 c_2 \right )^{{2}/{3}}}d \tau &= x +c_4 \\ \int _{}^{y}-\frac {4}{\left (-12 \ln \left (\frac {1}{\tau }\right )-12 c_1 \right )^{{2}/{3}}}d \tau &= x +c_5 \\ y &= c_3 \\ \end{align*}

Maple step by step solution

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 each 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 
      <- 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 
      <- differential order: 2; canonical coordinates successful 
      <- differential order 2; missing variables successful`
 

Maple dsolve solution

Solving time : 0.102 (sec)
Leaf size : 233

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

\begin{align*} y &= c_{1} \\ y &= 0 \\ -4 \left (\int _{}^{y}\frac {1}{\left (-12 \ln \left (\textit {\_a} \right )+8 c_{1} \right )^{{2}/{3}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ -4 \left (\int _{}^{y}\frac {1}{\left (12 \ln \left (\textit {\_a} \right )-8 c_{1} \right )^{{2}/{3}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ \frac {-16 \left (\int _{}^{y}\frac {1}{\left (-12 \ln \left (\textit {\_a} \right )+8 c_{1} \right )^{{2}/{3}}}d \textit {\_a} \right )+2 i \left (-x -c_{2} \right ) \sqrt {3}+2 x +2 c_{2}}{\left (-i \sqrt {3}-1\right )^{2}} &= 0 \\ \frac {-16 \left (\int _{}^{y}\frac {1}{\left (-12 \ln \left (\textit {\_a} \right )+8 c_{1} \right )^{{2}/{3}}}d \textit {\_a} \right )+2 i \left (x +c_{2} \right ) \sqrt {3}+2 x +2 c_{2}}{\left (1-i \sqrt {3}\right )^{2}} &= 0 \\ \frac {-16 \left (\int _{}^{y}\frac {1}{\left (12 \ln \left (\textit {\_a} \right )-8 c_{1} \right )^{{2}/{3}}}d \textit {\_a} \right )+2 i \left (-x -c_{2} \right ) \sqrt {3}+2 x +2 c_{2}}{\left (-i \sqrt {3}-1\right )^{2}} &= 0 \\ \frac {-16 \left (\int _{}^{y}\frac {1}{\left (12 \ln \left (\textit {\_a} \right )-8 c_{1} \right )^{{2}/{3}}}d \textit {\_a} \right )+2 i \left (x +c_{2} \right ) \sqrt {3}+2 x +2 c_{2}}{\left (1-i \sqrt {3}\right )^{2}} &= 0 \\ \end{align*}

✓Mathematica DSolve solution

Solving time : 2.307 (sec)
Leaf size : 459

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

\begin{align*} y(x)\to 0 \\ y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{-i c_1} (-\log (\text {$\#$1})-i c_1){}^{2/3} \Gamma \left (\frac {1}{3},-i c_1-\log (\text {$\#$1})\right )}{(c_1-i \log (\text {$\#$1})){}^{2/3}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{i c_1} (-\log (\text {$\#$1})+i c_1){}^{2/3} \Gamma \left (\frac {1}{3},i c_1-\log (\text {$\#$1})\right )}{(i \log (\text {$\#$1})+c_1){}^{2/3}}\&\right ][x+c_2] \\ y(x)\to 0 \\ y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{-i (-c_1)} (-\log (\text {$\#$1})-i (-1) c_1){}^{2/3} \Gamma \left (\frac {1}{3},-i (-1) c_1-\log (\text {$\#$1})\right )}{(-i \log (\text {$\#$1})-c_1){}^{2/3}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{-i c_1} (-\log (\text {$\#$1})-i c_1){}^{2/3} \Gamma \left (\frac {1}{3},-i c_1-\log (\text {$\#$1})\right )}{(c_1-i \log (\text {$\#$1})){}^{2/3}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{i (-c_1)} (-\log (\text {$\#$1})+i (-c_1)){}^{2/3} \Gamma \left (\frac {1}{3},i (-c_1)-\log (\text {$\#$1})\right )}{(i \log (\text {$\#$1})-c_1){}^{2/3}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{i c_1} (-\log (\text {$\#$1})+i c_1){}^{2/3} \Gamma \left (\frac {1}{3},i c_1-\log (\text {$\#$1})\right )}{(i \log (\text {$\#$1})+c_1){}^{2/3}}\&\right ][x+c_2] \\ \end{align*}

✓Sympy solution

Solving time : 0.228 (sec)
Leaf size : 3

Python version: 3.13.1 (main, Dec  4 2024, 18:05:56) [GCC 14.2.1 20240910] 
Sympy version 1.13.3
 

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

Eq(y(x), 0)
 

\[ y{\left (x \right )} = 0 \]