2.1.50 Problem 50

Internal problem ID [9121]
Book : Second order enumerated odes
Section : section 1
Problem number : 50
Date solved : Sunday, March 30, 2025 at 02:08:05 PM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_y_y1]]

Solved as second order missing x ode

Time used: 0.475 (sec)

Solve

$$\begin{array}{rl}y{y}^{\prime \prime }+{y^{\prime }}^3& =0\end{array}$$

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

$$\begin{array}{r}yp\left(y\right)\left(\frac{d}{dy}p\left(y\right)\right)+p{\left(y\right)}^3=0\end{array}$$

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

The ode

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

is separable as it can be written as

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

Where

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

Integrating gives

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

$$-\frac{1}{p}=\ln \left(\frac{1}{y}\right)+c_1$$

We now need to find the singular solutions, these are found by finding for what values $g(p)$ is zero, since we had to divide by this above. Solving $g(p)=0$ or

$$p^2=0$$

for $p$ gives

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

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

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

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

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

Integrating gives

$$\begin{array}{rl}\int (-\ln \left(\frac{1}{y}\right)-c_1)dy& =dx\\ -y(\ln \left(\frac{1}{y}\right)+c_1+1)& =x+c_2\end{array}$$

For solution (2) found earlier, since $p=y^{\prime }$ then we now have a new first order ode to solve which is

$$\begin{array}{r}{y}^{\prime }=0\end{array}$$

Since the ode has the form $y^{\prime }=f(x)$, then we only need to integrate $f(x)$.

$$\begin{array}{rl}\int dy& =\int 0dx+c_3\\ y& =c_3\end{array}$$

Will add steps showing solving for IC soon.

Solving for $y$ from the above solution(s) gives (after possible removing of solutions that do not verify)

$$\begin{array}{rl}y& =c_3\\ y& =\frac{x+c_2}{\mathrm{LambertW}\left((x+c_2){\mathrm{e}}^{-1-c_1}\right)}\end{array}$$

Summary of solutions found

$$\begin{array}{rl}y& =c_3\\ y& =\frac{x+c_2}{\mathrm{LambertW}\left((x+c_2){\mathrm{e}}^{-1-c_1}\right)}\end{array}$$

✓ Maple. Time used: 0.032 (sec). Leaf size: 27

ode:=y(x)*diff(diff(y(x),x),x)+diff(y(x),x)^3 = 0; 
dsolve(ode,y(x), singsol=all);
 

$$\begin{array}{rl}y& =0\\ y& =c_1\\ y& =\frac{c_2+x}{\mathrm{LambertW}\left((c_2+x){\mathrm{e}}^{c_1-1}\right)}\end{array}$$

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 
-> Calling odsolve with the ODE, diff(_b(_a),_a)*_b(_a)+_b(_a)^3/_a = 0, _b(_a) 
   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   trying Bernoulli 
   <- Bernoulli successful 
<- differential order: 2; canonical coordinates successful 
<- differential order 2; missing variables successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & y\left(x\right)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+{\left(\frac{d}{dx}y\left(x\right)\right)}^3=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Define new dependent variable}u\\ & & u\left(x\right)=\frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Compute}\frac{d}{dx}\frac{d}{dx}y\left(x\right)\\ & & \frac{d}{dx}u\left(x\right)=\frac{d^2}{d{x}^2}y\left(x\right)\\ \bullet & & \text{Use chain rule on the lhs}\\ & & \left(\frac{d}{dx}y\left(x\right)\right)\left(\frac{d}{dy}u\left(y\right)\right)=\frac{d^2}{d{x}^2}y\left(x\right)\\ \bullet & & \text{Substitute in the definition of}u\\ & & u\left(y\right)\left(\frac{d}{dy}u\left(y\right)\right)=\frac{d^2}{d{x}^2}y\left(x\right)\\ \bullet & & \text{Make substitutions}\frac{d}{dx}y\left(x\right)=u\left(y\right),\frac{d}{dx}\frac{d}{dx}y\left(x\right)=u\left(y\right)\left(\frac{d}{dy}u\left(y\right)\right)\text{to reduce order of ODE}\\ & & yu\left(y\right)\left(\frac{d}{dy}u\left(y\right)\right)+u{\left(y\right)}^3=0\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dy}u\left(y\right)=-\frac{u{\left(y\right)}^2}{y}\\ \bullet & & \text{Separate variables}\\ & & \frac{\frac{d}{dy}u\left(y\right)}{u{\left(y\right)}^2}=-\frac{1}{y}\\ \bullet & & \text{Integrate both sides with respect to}y\\ & & \int \frac{\frac{d}{dy}u\left(y\right)}{u{\left(y\right)}^2}dy=\int -\frac{1}{y}dy+\mathit{C}\mathit{1}\\ \bullet & & \text{Evaluate integral}\\ & & -\frac{1}{u\left(y\right)}=-\ln \left(y\right)+\mathit{C}\mathit{1}\\ \bullet & & \text{Solve for}u\left(y\right)\\ & & u\left(y\right)=\frac{1}{\ln \left(y\right)-\mathit{C}\mathit{1}}\\ \bullet & & \text{Redefine the integration constant(s)}\\ & & u\left(y\right)=\frac{1}{\ln \left(y\right)+\mathit{C}\mathit{1}}\\ \bullet & & \text{Solve 1st ODE for}u\left(y\right)\\ & & u\left(y\right)=\frac{1}{\ln \left(y\right)+\mathit{C}\mathit{1}}\\ \bullet & & \text{Revert to original variables with substitution}u\left(y\right)=\frac{d}{dx}y\left(x\right),y=y\left(x\right)\\ & & \frac{d}{dx}y\left(x\right)=\frac{1}{\ln \left(y\left(x\right)\right)+\mathit{C}\mathit{1}}\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dx}y\left(x\right)=\frac{1}{\ln \left(y\left(x\right)\right)+\mathit{C}\mathit{1}}\\ \bullet & & \text{Separate variables}\\ & & \left(\frac{d}{dx}y\left(x\right)\right)(\ln \left(y\left(x\right)\right)+\mathit{C}\mathit{1})=1\\ \bullet & & \text{Integrate both sides with respect to}x\\ & & \int \left(\frac{d}{dx}y\left(x\right)\right)(\ln \left(y\left(x\right)\right)+\mathit{C}\mathit{1})dx=\int 1dx+\mathit{C}\mathit{2}\\ \bullet & & \text{Evaluate integral}\\ & & y\left(x\right)\mathit{C}\mathit{1}+y\left(x\right)\ln \left(y\left(x\right)\right)-y\left(x\right)=x+\mathit{C}\mathit{2}\\ \bullet & & \text{Solve for}y\left(x\right)\\ & & y\left(x\right)={\mathrm{e}}^{\mathit{LambertW}\left((x+\mathit{C}\mathit{2}){\mathrm{e}}^{\mathit{C}\mathit{1}-1}\right)-\mathit{C}\mathit{1}+1}\\ \bullet & & \text{Simplify}\\ & & y\left(x\right)=\frac{x+\mathit{C}\mathit{2}}{\mathit{LambertW}\left((x+\mathit{C}\mathit{2}){\mathrm{e}}^{\mathit{C}\mathit{1}-1}\right)}\\ \bullet & & \text{Redefine the integration constant(s)}\\ & & y\left(x\right)=\frac{x+\mathit{C}\mathit{2}}{\mathit{LambertW}\left((x+\mathit{C}\mathit{2})\mathit{C}\mathit{1}\right)}\end{array}$$

✓ Mathematica. Time used: 60.104 (sec). Leaf size: 26

ode=y[x]*D[y[x],{x,2}]+D[y[x],x]^3==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \frac{x+c_2}{W(e^{-1-c_1}(x+c_2))}$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(y(x)*Derivative(y(x), (x, 2)) + Derivative(y(x), x)**3,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : The given ODE (-y(x)*Derivative(y(x), (x, 2)))**(1/3)/2 - sqrt(3)*I*(-y(x)*Derivative(y(x), (x, 2)))**(1/3)/2 + Derivative(y(x), x) cannot be solved by the factorable group method