2.2.6 Problem 6

Internal problem ID [9129]
Book : Second order enumerated odes
Section : section 2
Problem number : 6
Date solved : Sunday, March 30, 2025 at 02:17:13 PM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]

Solved as second order missing x ode

Time used: 1.831 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }{y}^{\prime }+y^n& =0\end{array}$$

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

$$\begin{array}{rl}{y}^{\prime }& =p\end{array}$$

Then

$$\begin{array}{rl}{y}^{″}& =\frac{dp}{dx}\\ & =\frac{dp}{dy}\frac{dy}{dx}\\ & =p\frac{dp}{dy}\end{array}$$

Hence the ode becomes

$$\begin{array}{r}p{\left(y\right)}^2\left(\frac{d}{dy}p\left(y\right)\right)+y^n=0\end{array}$$

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

The ode

$$\begin{array}{}\text{(1)}& p^{\prime }=-\frac{y^n}{p^2}\end{array}$$

is separable as it can be written as

$$\begin{array}{rl}{p}^{\prime }& =-\frac{y^n}{p^2}\\ & =f(y)g(p)\end{array}$$

Where

$$\begin{array}{rl}f(y)& =-y^n\\ g(p)& =\frac{1}{p^2}\end{array}$$

Integrating gives

$$\begin{array}{rl}\int \frac{1}{g(p)}dp& =\int f(y)dy\\ \int p^{2}dp& =\int -y^{n}dy\end{array}$$

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

For solution (1) found earlier, since $p=y^{\prime }$ then we now have a new first order ode to solve which is

$$\begin{array}{r}\frac{{y^{\prime }}^3}{3}=-\frac{y^{n+1}}{n+1}+c_1\end{array}$$

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

$$\begin{array}{r}\frac{p^3}{3}=-\frac{y^{n+1}}{n+1}+c_1\end{array}$$

Solving for $y$ from the above results in

$$\begin{array}{}\text{(1)}& y& ={\mathrm{e}}^{\frac{\ln (-\frac{1}{3}{p}^{3}n-\frac{1}{3}{p}^3+c_{1}n+c_1)}{n+1}}\end{array}$$

This has the form

$$\begin{array}{}\text{(*)}& y=xf(p)+g(p)\end{array}$$

Where $f,g$ are functions of $p=y^{\prime }(x)$. The above ode is dAlembert ode which is now solved.

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

$$\begin{array}{rl}p& =f+(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\\ \text{(2)}& p-f& =(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\end{array}$$

Comparing the form $y=xf+g$ to (1A) shows that

$$\begin{array}{rl}f& =0\\ g& =3^{-\frac{1}{n+1}}{(-(n+1)(p^3-3c_1))}^{\frac{1}{n+1}}\end{array}$$

Hence (2) becomes

$$\begin{array}{}\text{(2A)}& p=\frac{33^{-\frac{1}{n+1}}{(-p^{3}n-p^3+3c_{1}n+3c_1)}^{\frac{1}{n+1}}{p}^2{p}^{\prime }\left(x\right)}{(n+1)(p^3-3c_1)}\end{array}$$

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

$$\begin{array}{r}p=0\end{array}$$

Solving the above for $p$ results in

$$\begin{array}{rl}{p}_1& =0\end{array}$$

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

$$\begin{array}{r}y={\left(c_1(n+1)\right)}^{\frac{1}{n+1}}\end{array}$$

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

$$\begin{array}{}\text{(3)}& p^{\prime }\left(x\right)=\frac{3^{\frac{1}{n+1}}{(-p{\left(x\right)}^{3}n-p{\left(x\right)}^3+3c_{1}n+3c_1)}^{-\frac{1}{n+1}}(n+1)(p{\left(x\right)}^3-3c_1)}{3p\left(x\right)}\end{array}$$

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

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

Singular solutions are found by solving

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

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.

$$\begin{array}{r}p\left(x\right)=3^{1/3}{c}_1^{1/3}\\ p\left(x\right)=-\frac{3^{1/3}{c}_1^{1/3}}{2}-\frac{i{3}^{5/6}{c}_1^{1/3}}{2}\\ p\left(x\right)=-\frac{3^{1/3}{c}_1^{1/3}}{2}+\frac{i{3}^{5/6}{c}_1^{1/3}}{2}\end{array}$$

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

$$\begin{array}{rl}y& =3^{-\frac{1}{n+1}}{(-(n+1)(\mathrm{RootOf}{(-{\int }_{}^{\mathit{\text{\_Z}}}-\frac{\tau 3^{\frac{n}{n+1}}{\left((n+1)(-{\tau }^3+3c_1)\right)}^{\frac{1}{n+1}}}{(n+1)(-{\tau }^3+3c_1)}d\tau +x+c_2)}^3-3c_1))}^{\frac{1}{n+1}}\\ y& =0\\ y& =0\\ y& =0\end{array}$$

Will add steps showing solving for IC soon.

The solution

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

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

Summary of solutions found

$$\begin{array}{rl}y& =0\\ y& =3^{-\frac{1}{n+1}}{(-(n+1)(\mathrm{RootOf}{(-{\int }_{}^{\mathit{\text{\_Z}}}-\frac{\tau 3^{\frac{n}{n+1}}{\left((n+1)(-{\tau }^3+3c_1)\right)}^{\frac{1}{n+1}}}{(n+1)(-{\tau }^3+3c_1)}d\tau +x+c_2)}^3-3c_1))}^{\frac{1}{n+1}}\end{array}$$

Solved as second order can be made integrable

Time used: 2.149 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }{y}^{\prime }+y^n& =0\end{array}$$

Multiplying the ode by $y^{\prime }$ gives

$${y^{\prime }}^2{y}^{\prime \prime }+y^{\prime }{y}^{n-1}y=0$$

Integrating the above w.r.t $x$ gives

$$\begin{array}{rl}\int ({y^{\prime }}^2{y}^{\prime \prime }+y^{\prime }{y}^{n-1}y)dx& =0\\ \frac{{y^{\prime }}^3}{3}+\frac{y^2{\mathrm{e}}^{(n-1)\ln \left(y\right)}}{n+1}& =c_1\end{array}$$

Which is now solved for $y$. Let $p=y^{\prime }$ the ode becomes

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

Solving for $y$ from the above results in

$$\begin{array}{}\text{(1)}& y& ={\mathrm{e}}^{\frac{\ln (-\frac{1}{3}{p}^{3}n-\frac{1}{3}{p}^3+c_{1}n+c_1)}{n+1}}\end{array}$$

This has the form

$$\begin{array}{}\text{(*)}& y=xf(p)+g(p)\end{array}$$

Where $f,g$ are functions of $p=y^{\prime }(x)$. The above ode is dAlembert ode which is now solved.

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

$$\begin{array}{rl}p& =f+(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\\ \text{(2)}& p-f& =(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\end{array}$$

Comparing the form $y=xf+g$ to (1A) shows that

$$\begin{array}{rl}f& =0\\ g& =3^{-\frac{1}{n+1}}{(-(n+1)(p^3-3c_1))}^{\frac{1}{n+1}}\end{array}$$

Hence (2) becomes

$$\begin{array}{}\text{(2A)}& p=\frac{33^{-\frac{1}{n+1}}{(-p^{3}n-p^3+3c_{1}n+3c_1)}^{\frac{1}{n+1}}{p}^2{p}^{\prime }\left(x\right)}{(n+1)(p^3-3c_1)}\end{array}$$

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

$$\begin{array}{r}p=0\end{array}$$

Solving the above for $p$ results in

$$\begin{array}{rl}{p}_1& =0\end{array}$$

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

$$\begin{array}{r}y={\left(c_1(n+1)\right)}^{\frac{1}{n+1}}\end{array}$$

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

$$\begin{array}{}\text{(3)}& p^{\prime }\left(x\right)=\frac{3^{\frac{1}{n+1}}{(-p{\left(x\right)}^{3}n-p{\left(x\right)}^3+3c_{1}n+3c_1)}^{-\frac{1}{n+1}}(n+1)(p{\left(x\right)}^3-3c_1)}{3p\left(x\right)}\end{array}$$

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

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

Singular solutions are found by solving

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

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.

$$\begin{array}{r}p\left(x\right)=3^{1/3}{c}_1^{1/3}\\ p\left(x\right)=-\frac{3^{1/3}{c}_1^{1/3}}{2}-\frac{i{3}^{5/6}{c}_1^{1/3}}{2}\\ p\left(x\right)=-\frac{3^{1/3}{c}_1^{1/3}}{2}+\frac{i{3}^{5/6}{c}_1^{1/3}}{2}\end{array}$$

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

$$\begin{array}{rl}y& =3^{-\frac{1}{n+1}}{(-(n+1)(\mathrm{RootOf}{(-{\int }_{}^{\mathit{\text{\_Z}}}-\frac{\tau 3^{\frac{n}{n+1}}{\left((n+1)(-{\tau }^3+3c_1)\right)}^{\frac{1}{n+1}}}{(n+1)(-{\tau }^3+3c_1)}d\tau +x+c_2)}^3-3c_1))}^{\frac{1}{n+1}}\\ y& =0\\ y& =0\\ y& =0\end{array}$$

Will add steps showing solving for IC soon.

The solution

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

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

Summary of solutions found

$$\begin{array}{rl}y& =0\\ y& =3^{-\frac{1}{n+1}}{(-(n+1)(\mathrm{RootOf}{(-{\int }_{}^{\mathit{\text{\_Z}}}-\frac{\tau 3^{\frac{n}{n+1}}{\left((n+1)(-{\tau }^3+3c_1)\right)}^{\frac{1}{n+1}}}{(n+1)(-{\tau }^3+3c_1)}d\tau +x+c_2)}^3-3c_1))}^{\frac{1}{n+1}}\end{array}$$

✓ Maple. Time used: 0.019 (sec). Leaf size: 169

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

$$\begin{array}{rl}\frac{(-2-2n){\int }_{}^y\frac{1}{{(-(3{\mathit{\text{\_a}}}^{1+n}-c_1){(1+n)}^2)}^{1/3}}d\mathit{\text{\_a}}-(1+i\sqrt{3})(x+c_2)}{1+i\sqrt{3}}& =0\\ -\frac{2i(1+n){\int }_{}^y\frac{1}{{(-(3{\mathit{\text{\_a}}}^{1+n}-c_1){(1+n)}^2)}^{1/3}}d\mathit{\text{\_a}}+(x+c_2)(\sqrt{3}+i)}{\sqrt{3}+i}& =0\\ {\int }_{}^y\frac{1}{{(-(3{\mathit{\text{\_a}}}^{1+n}-c_1){(1+n)}^2)}^{1/3}}d\mathit{\text{\_a}}n+{\int }_{}^y\frac{1}{{(-(3{\mathit{\text{\_a}}}^{1+n}-c_1){(1+n)}^2)}^{1/3}}d\mathit{\text{\_a}}-c_2-x& =0\end{array}$$

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/(n-2)*_a, -_b*(1+n)/(n-2)]] 
   *** Sublevel 2 *** 
   symmetry methods on request 
   1st order, trying reduction of order with given symmetries: 
[-3/(n-2)*_a, -_b*(1+n)/(n-2)] 
   1st order, trying the canonical coordinates of the invariance group 
      -> Calling odsolve with the ODE, diff(y(x),x) = 1/3*y(x)*(1+n)/x, y(x) 
         *** Sublevel 3 *** 
         Methods for first order ODEs: 
         --- Trying classification methods --- 
         trying a quadrature 
         trying 1st order linear 
         <- 1st order linear successful 
   <- 1st order, canonical coordinates successful 
<- differential order: 2; canonical coordinates successful 
<- differential order 2; missing variables successful
 

✓ 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]
 

$$\begin{array}{r}\text{Solution too large to show}\end{array}$$

✗ 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