2.2.6 Problem 6

Internal problem ID [9129]
Book : Second order enumerated odes
Section : section 2
Problem number : 6
Date solved : Wednesday, March 05, 2025 at 07:30:48 AM
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: 1.835 (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*}

\[ \frac {p^{3}}{3}=-\frac {y^{n +1}}{n +1}+c_1 \]

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 {{y^{\prime }}^{3}}{3} = -\frac {y^{n +1}}{n +1}+c_1 \end{align*}

Let \(p=y^{\prime }\) the ode becomes

\begin{align*} \frac {p^{3}}{3} = -\frac {y^{n +1}}{n +1}+c_1 \end{align*}

Solving for \(y\) from the above results in

\begin{align*} \tag{1} y &= {\mathrm e}^{\frac {\ln \left (-\frac {1}{3} p^{3} n -\frac {1}{3} p^{3}+c_1 n +c_1 \right )}{n +1}} \\ \end{align*}

This has the form

\begin{align*} y=xf(p)+g(p)\tag {*} \end{align*}

Where \(f,g\) are functions of \(p=y'(x)\). The above ode is dAlembert ode which is now solved.

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(y=x f + g\) to (1A) shows that

\begin{align*} f &= 0\\ g &= 3^{-\frac {1}{n +1}} {\left (-\left (n +1\right ) \left (p^{3}-3 c_1 \right )\right )}^{\frac {1}{n +1}} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p = \frac {3 \,3^{-\frac {1}{n +1}} \left (-p^{3} n -p^{3}+3 c_1 n +3 c_1 \right )^{\frac {1}{n +1}} p^{2} p^{\prime }\left (x \right )}{\left (n +1\right ) \left (p^{3}-3 c_1 \right )} \end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

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

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} y = \left (c_1 \left (n +1\right )\right )^{\frac {1}{n +1}} \end{align*}

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{equation} \tag{3} p^{\prime }\left (x \right ) = \frac {3^{\frac {1}{n +1}} \left (-p \left (x \right )^{3} n -p \left (x \right )^{3}+3 c_1 n +3 c_1 \right )^{-\frac {1}{n +1}} \left (n +1\right ) \left (p \left (x \right )^{3}-3 c_1 \right )}{3 p \left (x \right )} \end{equation}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed.

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

\[ \int _{}^{p \left (x \right )}\frac {3^{\frac {n}{n +1}} \tau {\left (\left (n +1\right ) \left (-\tau ^{3}+3 c_1 \right )\right )}^{\frac {1}{n +1}}}{\left (n +1\right ) \left (\tau ^{3}-3 c_1 \right )}d \tau = x +c_2 \]

Singular solutions are found by solving

\begin{align*} \frac {3^{-\frac {n}{n +1}} \left (n +1\right ) \left (p^{3}-3 c_1 \right ) {\left (\left (n +1\right ) \left (-p^{3}+3 c_1 \right )\right )}^{-\frac {1}{n +1}}}{p}&= 0 \end{align*}

for \(p \left (x \right )\). 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*} p \left (x \right ) = 3^{{1}/{3}} c_1^{{1}/{3}}\\ p \left (x \right ) = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}-\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{2}\\ p \left (x \right ) = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}+\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{2} \end{align*}

Substituing the above solution for \(p\) in (2A) gives

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

Will add steps showing solving for IC soon.

The solution

\[ y = \left (c_1 \left (n +1\right )\right )^{\frac {1}{n +1}} \]

was found not to satisfy the ode or the IC. Hence it is removed. The solution

\[ y = 3^{-\frac {1}{n +1}} {\left (-\left (n +1\right ) \left (\operatorname {RootOf}\left (-\int _{}^{\textit {\_Z}}-\frac {\tau 3^{\frac {n}{n +1}} {\left (\left (n +1\right ) \left (-\tau ^{3}+3 c_1 \right )\right )}^{\frac {1}{n +1}}}{\left (n +1\right ) \left (-\tau ^{3}+3 c_1 \right )}d \tau +x +c_2 \right )^{3}-3 c_1 \right )\right )}^{\frac {1}{n +1}} \]

was found not to satisfy the ode or the IC. Hence it is removed.

Summary of solutions found

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

Solved as second order can be made integrable

Time used: 2.317 (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 \(y\). Let \(p=y^{\prime }\) the ode becomes

\begin{align*} \frac {p^{3}}{3}+\frac {y^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (y \right )}}{n +1} = c_1 \end{align*}

Solving for \(y\) from the above results in

\begin{align*} \tag{1} y &= {\mathrm e}^{\frac {\ln \left (-\frac {1}{3} p^{3} n -\frac {1}{3} p^{3}+c_1 n +c_1 \right )}{n +1}} \\ \end{align*}

This has the form

\begin{align*} y=xf(p)+g(p)\tag {*} \end{align*}

Where \(f,g\) are functions of \(p=y'(x)\). The above ode is dAlembert ode which is now solved.

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(y=x f + g\) to (1A) shows that

\begin{align*} f &= 0\\ g &= 3^{-\frac {1}{n +1}} {\left (-\left (n +1\right ) \left (p^{3}-3 c_1 \right )\right )}^{\frac {1}{n +1}} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p = \frac {3 \,3^{-\frac {1}{n +1}} \left (-p^{3} n -p^{3}+3 c_1 n +3 c_1 \right )^{\frac {1}{n +1}} p^{2} p^{\prime }\left (x \right )}{\left (n +1\right ) \left (p^{3}-3 c_1 \right )} \end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

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

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} y = \left (c_1 \left (n +1\right )\right )^{\frac {1}{n +1}} \end{align*}

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{equation} \tag{3} p^{\prime }\left (x \right ) = \frac {3^{\frac {1}{n +1}} \left (-p \left (x \right )^{3} n -p \left (x \right )^{3}+3 c_1 n +3 c_1 \right )^{-\frac {1}{n +1}} \left (n +1\right ) \left (p \left (x \right )^{3}-3 c_1 \right )}{3 p \left (x \right )} \end{equation}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed.

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

\[ \int _{}^{p \left (x \right )}\frac {3^{\frac {n}{n +1}} \tau {\left (\left (n +1\right ) \left (-\tau ^{3}+3 c_1 \right )\right )}^{\frac {1}{n +1}}}{\left (n +1\right ) \left (\tau ^{3}-3 c_1 \right )}d \tau = x +c_2 \]

Singular solutions are found by solving

\begin{align*} \frac {3^{-\frac {n}{n +1}} \left (n +1\right ) \left (p^{3}-3 c_1 \right ) {\left (\left (n +1\right ) \left (-p^{3}+3 c_1 \right )\right )}^{-\frac {1}{n +1}}}{p}&= 0 \end{align*}

for \(p \left (x \right )\). 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*} p \left (x \right ) = 3^{{1}/{3}} c_1^{{1}/{3}}\\ p \left (x \right ) = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}-\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{2}\\ p \left (x \right ) = -\frac {3^{{1}/{3}} c_1^{{1}/{3}}}{2}+\frac {i 3^{{5}/{6}} c_1^{{1}/{3}}}{2} \end{align*}

Substituing the above solution for \(p\) in (2A) gives

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

Will add steps showing solving for IC soon.

The solution

\[ y = \left (c_1 \left (n +1\right )\right )^{\frac {1}{n +1}} \]

was found not to satisfy the ode or the IC. Hence it is removed. The solution

\[ y = 3^{-\frac {1}{n +1}} {\left (-\left (n +1\right ) \left (\operatorname {RootOf}\left (-\int _{}^{\textit {\_Z}}-\frac {\tau 3^{\frac {n}{n +1}} {\left (\left (n +1\right ) \left (-\tau ^{3}+3 c_1 \right )\right )}^{\frac {1}{n +1}}}{\left (n +1\right ) \left (-\tau ^{3}+3 c_1 \right )}d \tau +x +c_2 \right )^{3}-3 c_1 \right )\right )}^{\frac {1}{n +1}} \]

was found not to satisfy the ode or the IC. Hence it is removed.

Summary of solutions found

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

✓ Maple. Time used: 0.028 (sec). Leaf size: 169

ode:=diff(diff(y(x),x),x)*diff(y(x),x)+y(x)^n = 0; 
dsolve(ode,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*}

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)]
 

✓ Mathematica. Time used: 2.216 (sec). Leaf size: 910

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

\begin{align*} \text {Solution too large to show}\end{align*}

✗ Sympy

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