2.1.49 Problem 49
Internal
problem
ID
[9120]
Book
:
Second
order
enumerated
odes
Section
:
section
1
Problem
number
:
49
Date
solved
:
Sunday, March 30, 2025 at 02:08:02 PM
CAS
classification
:
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]
Solved as second order missing x ode
Time used: 1.872 (sec)
Solve
\begin{align*} y^{3} {y^{\prime \prime }}^{2}+y y^{\prime }&=0 \end{align*}
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 \(p(y)\).
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 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*}
\[
-\frac {2 \left (-p\right )^{{3}/{2}}}{3}=\ln \left (y \right )+c_2
\]
For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} -\frac {2 \left (-y^{\prime }\right )^{{3}/{2}}}{3} = \ln \left (\frac {1}{y}\right )+c_1 \end{align*}
Solving for the derivative gives these ODE’s to solve
\begin{align*}
\tag{1} y^{\prime }&=-\frac {\left (-12 \ln \left (\frac {1}{y}\right )-12 c_1 \right )^{{2}/{3}}}{4} \\
\tag{2} y^{\prime }&=-\left (-\frac {\left (-12 \ln \left (\frac {1}{y}\right )-12 c_1 \right )^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, \left (-12 \ln \left (\frac {1}{y}\right )-12 c_1 \right )^{{1}/{3}}}{4}\right )^{2} \\
\tag{3} y^{\prime }&=-\left (-\frac {\left (-12 \ln \left (\frac {1}{y}\right )-12 c_1 \right )^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, \left (-12 \ln \left (\frac {1}{y}\right )-12 c_1 \right )^{{1}/{3}}}{4}\right )^{2} \\
\end{align*}
Now each of the above is solved separately.
Solving Eq. (1)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}-\frac {4}{\left (-12 \ln \left (\frac {1}{\tau }\right )-12 c_1 \right )^{{2}/{3}}}d \tau = x +c_3 \]
Singular solutions are found by solving
\begin{align*} -\frac {\left (-12 \ln \left (\frac {1}{y}\right )-12 c_1 \right )^{{2}/{3}}}{4}&= 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 = {\mathrm e}^{c_1} \end{align*}
Solving Eq. (2)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}-\frac {16}{\left (-12 \ln \left (\frac {1}{\tau }\right )-12 c_1 \right )^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau = x +c_4 \]
Singular solutions are found by solving
\begin{align*} -\frac {\left (-12 \ln \left (\frac {1}{y}\right )-12 c_1 \right )^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}{16}&= 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 = {\mathrm e}^{c_1} \end{align*}
Solving Eq. (3)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}-\frac {16}{\left (-12 \ln \left (\frac {1}{\tau }\right )-12 c_1 \right )^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau = x +c_5 \]
Singular solutions are found by solving
\begin{align*} -\frac {\left (-12 \ln \left (\frac {1}{y}\right )-12 c_1 \right )^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}{16}&= 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 = {\mathrm e}^{c_1} \end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
\int _{}^{y}-\frac {4}{\left (-12 \ln \left (\frac {1}{\tau }\right )-12 c_1 \right )^{{2}/{3}}}d \tau &= x +c_3 \\
y &= {\mathrm e}^{c_1} \\
\end{align*}
✓ Maple. Time used: 0.088 (sec). Leaf size: 225
ode:=y(x)^3*diff(diff(y(x),x),x)^2+y(x)*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
✓ Mathematica. Time used: 2.307 (sec). Leaf size: 459
ode=y[x]^3*D[y[x],{x,2}]^2+y[x]*D[y[x],x]==0;
ic={};
DSolve[{ode,ic},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. Time used: 0.228 (sec). Leaf size: 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)
\[
y{\left (x \right )} = 0
\]