2.1.49 Problem 49
Internal
problem
ID
[9120]
Book
:
Second
order
enumerated
odes
Section
:
section
1
Problem
number
:
49
Date
solved
:
Wednesday, March 05, 2025 at 07:26:13 AM
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: 2.879 (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 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 (y\right )+c_2 \end{align*}
Solving for the derivative gives these ODE’s to solve
\begin{align*}
\tag{1} y^{\prime }&=-\frac {\left (-12 \ln \left (y\right )-12 c_2 \right )^{{2}/{3}}}{4} \\
\tag{2} y^{\prime }&=-\left (-\frac {\left (-12 \ln \left (y\right )-12 c_2 \right )^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, \left (-12 \ln \left (y\right )-12 c_2 \right )^{{1}/{3}}}{4}\right )^{2} \\
\tag{3} y^{\prime }&=-\left (-\frac {\left (-12 \ln \left (y\right )-12 c_2 \right )^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, \left (-12 \ln \left (y\right )-12 c_2 \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
\[ \int _{}^{y}-\frac {4}{\left (-12 \ln \left (\tau \right )-12 c_2 \right )^{{2}/{3}}}d \tau = x +c_3 \]
Solving Eq. (2)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}-\frac {16}{\left (-12 \ln \left (\tau \right )-12 c_2 \right )^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau = x +c_4 \]
Solving Eq. (3)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}-\frac {16}{\left (-12 \ln \left (\tau \right )-12 c_2 \right )^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau = x +c_5 \]
For solution (2) 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
\[ \int _{}^{y}-\frac {4}{\left (-12 \ln \left (\frac {1}{\tau }\right )-12 c_1 \right )^{{2}/{3}}}d \tau = x +c_6 \]
Solving Eq. (2)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}-\frac {16}{\left (-12 \ln \left (\frac {1}{\tau }\right )-12 c_1 \right )^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau = x +c_7 \]
Solving Eq. (3)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}-\frac {16}{\left (-12 \ln \left (\frac {1}{\tau }\right )-12 c_1 \right )^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau = x +c_8 \]
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*}
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_9 \\ y &= c_9 \end{align*}
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_3 \\
\int _{}^{y}-\frac {4}{\left (-12 \ln \left (\frac {1}{\tau }\right )-12 c_1 \right )^{{2}/{3}}}d \tau &= x +c_6 \\
\int _{}^{y}-\frac {16}{\left (-12 \ln \left (\tau \right )-12 c_2 \right )^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau &= x +c_5 \\
\int _{}^{y}-\frac {16}{\left (-12 \ln \left (\tau \right )-12 c_2 \right )^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau &= x +c_4 \\
\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_8 \\
\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_7 \\
y &= c_9 \\
\end{align*}
✓ Maple. Time used: 0.102 (sec). Leaf size: 233
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 \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*}
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`
✓ 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)
\[
y{\left (x \right )} = 0
\]