2.2.6 Problem 6
Internal
problem
ID
[9129]
Book
:
Second
order
enumerated
odes
Section
:
section
2
Problem
number
:
6
Date
solved
:
Friday, February 21, 2025 at 09:17:50 PM
CAS
classification
:
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]
Solve
\begin{align*} y^{\prime \prime } y^{\prime }+y^{n}&=0 \end{align*}
Solved as second order missing x ode
Time used: 1.208 (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*} p \left (y \right )^{2} \left (\frac {d}{d y}p \left (y \right )\right )+y^{n} = 0 \end{align*}
Which is now solved as first order ode for \(p(y)\).
The ode
\begin{equation}
p^{\prime } = -\frac {y^{n}}{p^{2}}
\end{equation}
is separable as it can be written as
\begin{align*} p^{\prime }&= -\frac {y^{n}}{p^{2}}\\ &= f(y) g(p) \end{align*}
Where
\begin{align*} f(y) &= -y^{n}\\ g(p) &= \frac {1}{p^{2}} \end{align*}
Integrating gives
\begin{align*}
\int { \frac {1}{g(p)} \,dp} &= \int { f(y) \,dy} \\
\int { p^{2}\,dp} &= \int { -y^{n} \,dy} \\
\end{align*}
\[
\frac {p^{3}}{3}=-\frac {y^{n +1}}{n +1}+c_1
\]
Solving for \(p\) gives
\begin{align*}
p &= \frac {{\left (\left (3 c_1 n -3 y^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{n +1} \\
p &= -\frac {{\left (\left (3 c_1 n -3 y^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{2 \left (n +1\right )}-\frac {i \sqrt {3}\, {\left (\left (3 c_1 n -3 y^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{2 \left (n +1\right )} \\
p &= -\frac {{\left (\left (3 c_1 n -3 y^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{2 \left (n +1\right )}+\frac {i \sqrt {3}\, {\left (\left (3 c_1 n -3 y^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{2 n +2} \\
\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 } = \frac {{\left (\left (3 c_1 n -3 y^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{n +1} \end{align*}
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {n +1}{{\left (\left (3 c_1 n -3 \tau ^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}d \tau = x +c_2 \]
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 (\left (3 c_1 n -3 y^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{2 \left (n +1\right )}-\frac {i \sqrt {3}\, {\left (\left (3 c_1 n -3 y^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{2 \left (n +1\right )} \end{align*}
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}-\frac {2 \left (n +1\right )}{{\left (\left (3 c_1 n -3 \tau ^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}d \tau = x +c_3 \]
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 (\left (3 c_1 n -3 y^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{2 \left (n +1\right )}+\frac {i \sqrt {3}\, {\left (\left (3 c_1 n -3 y^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{2 n +2} \end{align*}
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {2 n +2}{{\left (\left (3 c_1 n -3 \tau ^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}} \left (i \sqrt {3}-1\right )}d \tau = x +c_4 \]
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
\int _{}^{y}\frac {n +1}{{\left (\left (3 c_1 n -3 \tau ^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}d \tau &= x +c_2 \\
\int _{}^{y}\frac {2 n +2}{{\left (\left (3 c_1 n -3 \tau ^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}} \left (i \sqrt {3}-1\right )}d \tau &= x +c_4 \\
\int _{}^{y}-\frac {2 \left (n +1\right )}{{\left (\left (3 c_1 n -3 \tau ^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}d \tau &= x +c_3 \\
\end{align*}
Solved as second order can be made integrable
Time used: 1.372 (sec)
Multiplying the ode by \(y^{\prime }\) gives
\[ {y^{\prime }}^{2} y^{\prime \prime }+y^{n -1} y^{\prime } y = 0 \]
Integrating the above w.r.t \(x\) gives
\begin{align*} \int \left ({y^{\prime }}^{2} y^{\prime \prime }+y^{n -1} y^{\prime } y\right )d x &= 0 \\ \frac {{y^{\prime }}^{3}}{3}+\frac {y^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (y\right )}}{n +1} &= c_1 \end{align*}
Which is now solved for \(y\). Solving for the derivative gives these ODE’s to solve
\begin{align*}
\tag{1} y^{\prime }&=\frac {{\left (\left (-3 y^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (y\right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{n +1} \\
\tag{2} y^{\prime }&=-\frac {{\left (\left (-3 y^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (y\right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{2 \left (n +1\right )}-\frac {i \sqrt {3}\, {\left (\left (-3 y^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (y\right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{2 \left (n +1\right )} \\
\tag{3} y^{\prime }&=-\frac {{\left (\left (-3 y^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (y\right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{2 \left (n +1\right )}+\frac {i \sqrt {3}\, {\left (\left (-3 y^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (y\right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{2 n +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
\[ \int _{}^{y}\frac {n +1}{{\left (\left (-3 \tau ^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (\tau \right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}d \tau = x +c_2 \]
Solving Eq. (2)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}-\frac {2 \left (n +1\right )}{{\left (\left (-3 \tau ^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (\tau \right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}d \tau = x +c_3 \]
Solving Eq. (3)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{y}\frac {2 n +2}{{\left (\left (-3 \tau ^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (\tau \right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}} \left (i \sqrt {3}-1\right )}d \tau = x +c_4 \]
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
\int _{}^{y}\frac {n +1}{{\left (\left (-3 \tau ^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (\tau \right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}d \tau &= x +c_2 \\
\int _{}^{y}-\frac {2 \left (n +1\right )}{{\left (\left (-3 \tau ^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (\tau \right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}d \tau &= x +c_3 \\
\int _{}^{y}\frac {2 n +2}{{\left (\left (-3 \tau ^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (\tau \right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}} \left (i \sqrt {3}-1\right )}d \tau &= x +c_4 \\
\end{align*}
Maple step by step solution
Maple trace
`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)+_a^n/_b(_a) = 0, _b(_a), HINT = [[-3*_a/(n-2), -_b*(n+1)/(n-2)]]` ***
symmetry methods on request
`, `1st order, trying reduction of order with given symmetries:`[-3/(n-2)*_a, -_b*(n+1)/(n-2)]
Maple dsolve solution
Solving time : 0.028 (sec)
Leaf size : 169
dsolve(diff(y(x),x)*diff(diff(y(x),x),x)+y(x)^n = 0,y(x),singsol=all)
\begin{align*}
\frac {\left (-2 n -2\right ) \left (\int _{}^{y}\frac {1}{{\left (-\left (3 \textit {\_a}^{n +1}-c_{1} \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}d \textit {\_a} \right )-\left (1+i \sqrt {3}\right ) \left (x +c_{2} \right )}{1+i \sqrt {3}} &= 0 \\
-\frac {2 i \left (n +1\right ) \left (\int _{}^{y}\frac {1}{{\left (-\left (3 \textit {\_a}^{n +1}-c_{1} \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}d \textit {\_a} \right )+\left (x +c_{2} \right ) \left (\sqrt {3}+i\right )}{\sqrt {3}+i} &= 0 \\
\left (\int _{}^{y}\frac {1}{{\left (-\left (3 \textit {\_a}^{n +1}-c_{1} \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}d \textit {\_a} \right ) n +\int _{}^{y}\frac {1}{{\left (-\left (3 \textit {\_a}^{n +1}-c_{1} \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}d \textit {\_a} -c_{2} -x &= 0 \\
\end{align*}
✓Mathematica DSolve solution
Solving time : 2.216 (sec)
Leaf size : 910
DSolve[{D[y[x],{x,2}]*D[y[x],x]+y[x]^n==0,{}},y[x],x,IncludeSingularSolutions->True]
\begin{align*}
y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \sqrt [3]{n+1} \sqrt [3]{1-\frac {\text {$\#$1}^{n+1}}{c_1 (n+1)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n+1},1+\frac {1}{n+1},\frac {\text {$\#$1}^{n+1}}{(n+1) c_1}\right )}{\sqrt [3]{-3 \text {$\#$1}^{n+1}+3 c_1 (n+1)}}\&\right ][x+c_2] \\
y(x)\to \text {InverseFunction}\left [\frac {(-1)^{2/3} \text {$\#$1} \sqrt [3]{n+1} \sqrt [3]{1-\frac {\text {$\#$1}^{n+1}}{c_1 (n+1)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n+1},1+\frac {1}{n+1},\frac {\text {$\#$1}^{n+1}}{(n+1) c_1}\right )}{\sqrt [3]{-3 \text {$\#$1}^{n+1}+3 c_1 (n+1)}}\&\right ][x+c_2] \\
y(x)\to \text {InverseFunction}\left [-\frac {\sqrt [3]{-\frac {1}{3}} \text {$\#$1} \sqrt [3]{n+1} \sqrt [3]{1-\frac {\text {$\#$1}^{n+1}}{c_1 (n+1)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n+1},1+\frac {1}{n+1},\frac {\text {$\#$1}^{n+1}}{(n+1) c_1}\right )}{\sqrt [3]{-\text {$\#$1}^{n+1}+c_1 (n+1)}}\&\right ][x+c_2] \\
y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \sqrt [3]{n+1} \sqrt [3]{1-\frac {\text {$\#$1}^{n+1}}{(-c_1) (n+1)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n+1},1+\frac {1}{n+1},\frac {\text {$\#$1}^{n+1}}{(n+1) (-c_1)}\right )}{\sqrt [3]{-3 \text {$\#$1}^{n+1}+3 (-c_1) (n+1)}}\&\right ][x+c_2] \\
y(x)\to \text {InverseFunction}\left [\frac {(-1)^{2/3} \text {$\#$1} \sqrt [3]{n+1} \sqrt [3]{1-\frac {\text {$\#$1}^{n+1}}{(-c_1) (n+1)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n+1},1+\frac {1}{n+1},\frac {\text {$\#$1}^{n+1}}{(n+1) (-c_1)}\right )}{\sqrt [3]{-3 \text {$\#$1}^{n+1}+3 (-c_1) (n+1)}}\&\right ][x+c_2] \\
y(x)\to \text {InverseFunction}\left [-\frac {\sqrt [3]{-\frac {1}{3}} \text {$\#$1} \sqrt [3]{n+1} \sqrt [3]{1-\frac {\text {$\#$1}^{n+1}}{(-c_1) (n+1)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n+1},1+\frac {1}{n+1},\frac {\text {$\#$1}^{n+1}}{(n+1) (-c_1)}\right )}{\sqrt [3]{-\text {$\#$1}^{n+1}+(-c_1) (n+1)}}\&\right ][x+c_2] \\
y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \sqrt [3]{n+1} \sqrt [3]{1-\frac {\text {$\#$1}^{n+1}}{c_1 (n+1)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n+1},1+\frac {1}{n+1},\frac {\text {$\#$1}^{n+1}}{(n+1) c_1}\right )}{\sqrt [3]{-3 \text {$\#$1}^{n+1}+3 c_1 (n+1)}}\&\right ][x+c_2] \\
y(x)\to \text {InverseFunction}\left [\frac {(-1)^{2/3} \text {$\#$1} \sqrt [3]{n+1} \sqrt [3]{1-\frac {\text {$\#$1}^{n+1}}{c_1 (n+1)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n+1},1+\frac {1}{n+1},\frac {\text {$\#$1}^{n+1}}{(n+1) c_1}\right )}{\sqrt [3]{-3 \text {$\#$1}^{n+1}+3 c_1 (n+1)}}\&\right ][x+c_2] \\
y(x)\to \text {InverseFunction}\left [-\frac {\sqrt [3]{-\frac {1}{3}} \text {$\#$1} \sqrt [3]{n+1} \sqrt [3]{1-\frac {\text {$\#$1}^{n+1}}{c_1 (n+1)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n+1},1+\frac {1}{n+1},\frac {\text {$\#$1}^{n+1}}{(n+1) c_1}\right )}{\sqrt [3]{-\text {$\#$1}^{n+1}+c_1 (n+1)}}\&\right ][x+c_2] \\
\end{align*}
✗Sympy solution
Solving time : 0.000 (sec)
Leaf size : 0
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")
n = symbols("n")
y = Function("y")
ode = Eq(y(x)**n + Derivative(y(x), x)*Derivative(y(x), (x, 2)),0)
ics = {}
dsolve(ode,func=y(x),ics=ics)
NotImplementedError : The given ODE y(x)**n/Derivative(y(x), (x, 2)) + Derivative(y(x), x) cannot be solved by the factorable group method