2.1.54 Problem 54

Internal problem ID [9038]
Book : First order enumerated odes
Section : section 1
Problem number : 54
Date solved : Sunday, March 30, 2025 at 01:59:55 PM
CAS classification : [[_homogeneous, `class G`], _rational]

Solved using first_order_nonlinear_p_but_separable

Time used: 0.758 (sec)

Solve

$$\begin{array}{rl}{y^{\prime }}^3& =\frac{y^2}{x}\end{array}$$

The ode has the form

$$\begin{array}{}\text{(1)}& (y^{\prime }{)}^{\frac{n}{m}}& =f(x)g(y)\end{array}$$

Where $n=3,m=1,f=\frac{1}{x},g=y^2$. Hence the ode is

$$\begin{array}{rl}(y^{\prime }{)}^3& =\frac{y^2}{x}\end{array}$$

Solving for $y^{\prime }$ from (1) gives

$$\begin{array}{rl}{y}^{\prime }& ={\left(fg\right)}^{1/3}\\ y^{\prime }& =-\frac{{\left(fg\right)}^{1/3}}{2}+\frac{i\sqrt{3}{\left(fg\right)}^{1/3}}{2}\\ y^{\prime }& =-\frac{{\left(fg\right)}^{1/3}}{2}-\frac{i\sqrt{3}{\left(fg\right)}^{1/3}}{2}\end{array}$$

To be able to solve as separable ode, we have to now assume that $f>0,g>0$.

$$\begin{array}{rl}\frac{1}{x}& >0\\ y^2& >0\end{array}$$

Under the above assumption the differential equations become separable and can be written as

$$\begin{array}{rl}{y}^{\prime }& =f^{1/3}{g}^{1/3}\\ y^{\prime }& =\frac{f^{1/3}{g}^{1/3}(-1+i\sqrt{3})}{2}\\ y^{\prime }& =-\frac{f^{1/3}{g}^{1/3}(1+i\sqrt{3})}{2}\end{array}$$

Therefore

$$\begin{array}{rl}\frac{1}{g^{1/3}}dy& =\left(f^{1/3}\right)dx\\ \frac{2}{g^{1/3}(-1+i\sqrt{3})}dy& =\left(f^{1/3}\right)dx\\ -\frac{2}{g^{1/3}(1+i\sqrt{3})}dy& =\left(f^{1/3}\right)dx\end{array}$$

Replacing $f(x),g(y)$ by their values gives

$$\begin{array}{rl}\frac{1}{{\left(y^2\right)}^{1/3}}dy& =\left({\left(\frac{1}{x}\right)}^{1/3}\right)dx\\ \frac{2}{{\left(y^2\right)}^{1/3}(-1+i\sqrt{3})}dy& =\left({\left(\frac{1}{x}\right)}^{1/3}\right)dx\\ -\frac{2}{{\left(y^2\right)}^{1/3}(1+i\sqrt{3})}dy& =\left({\left(\frac{1}{x}\right)}^{1/3}\right)dx\end{array}$$

Integrating now gives the following solutions

$$\begin{array}{rl}\int \frac{1}{{\left(y^2\right)}^{1/3}}dy& =\int {\left(\frac{1}{x}\right)}^{1/3}dx+c_1\\ \frac{3{\left(y^2\right)}^{2/3}}{y}& =\frac{3x{\left(\frac{1}{x}\right)}^{1/3}}{2}\\ \int \frac{2}{{\left(y^2\right)}^{1/3}(-1+i\sqrt{3})}dy& =\int {\left(\frac{1}{x}\right)}^{1/3}dx+c_1\\ -\frac{3{\left(y^2\right)}^{2/3}(1+i\sqrt{3})}{2y}& =\frac{3x{\left(\frac{1}{x}\right)}^{1/3}}{2}\\ \int -\frac{2}{{\left(y^2\right)}^{1/3}(1+i\sqrt{3})}dy& =\int {\left(\frac{1}{x}\right)}^{1/3}dx+c_1\\ \frac{3{\left(y^2\right)}^{2/3}(-1+i\sqrt{3})}{2y}& =\frac{3x{\left(\frac{1}{x}\right)}^{1/3}}{2}\end{array}$$

Therefore

$$\begin{array}{rl}\frac{3{\left(y^2\right)}^{2/3}}{y}& =\frac{3x{\left(\frac{1}{x}\right)}^{1/3}}{2}+c_1\\ y& =\frac{x^2}{8}+\frac{{\left(\frac{1}{x}\right)}^{2/3}{c}_1{x}^2}{4}+\frac{{\left(\frac{1}{x}\right)}^{1/3}{c}_1^{2}x}{6}+\frac{c_1^3}{27}\\ y& =\frac{x^2}{8}+\frac{{\left(\frac{1}{x}\right)}^{2/3}{c}_1{x}^2}{4}+\frac{{\left(\frac{1}{x}\right)}^{1/3}{c}_1^{2}x}{6}+\frac{c_1^3}{27}\end{array}$$

Summary of solutions found

$$\begin{array}{rl}\frac{3{\left(y^2\right)}^{2/3}}{y}& =\frac{3x{\left(\frac{1}{x}\right)}^{1/3}}{2}+c_1\\ y& =\frac{x^2}{8}+\frac{{\left(\frac{1}{x}\right)}^{2/3}{c}_1{x}^2}{4}+\frac{{\left(\frac{1}{x}\right)}^{1/3}{c}_1^{2}x}{6}+\frac{c_1^3}{27}\end{array}$$

✓ Maple. Time used: 0.925 (sec). Leaf size: 341

ode:=diff(y(x),x)^3 = y(x)^2/x; 
dsolve(ode,y(x), singsol=all);
 

$$\begin{array}{rl}y& =0\\ y& =-\frac{3x^{4/3}{c}_1}{8}+\frac{3x^{2/3}{c}_1^2}{8}-\frac{c_1^3}{8}+\frac{x^2}{8}\\ y& =\frac{3(-1-i\sqrt{3})c_1^2{x}^{2/3}}{16}+\frac{3c_1(1-i\sqrt{3})x^{4/3}}{16}-\frac{c_1^3}{8}+\frac{x^2}{8}\\ y& =\frac{3(i\sqrt{3}-1)c_1^2{x}^{2/3}}{16}+\frac{3c_1(1+i\sqrt{3})x^{4/3}}{16}-\frac{c_1^3}{8}+\frac{x^2}{8}\\ y& =\frac{3x^{4/3}{c}_1}{16}+\frac{3x^{2/3}{c}_1^2}{32}+\frac{c_1^3}{64}+\frac{x^2}{8}\\ y& =\frac{3(-1-i\sqrt{3})c_1^2{x}^{2/3}}{64}+\frac{3(i\sqrt{3}-1)c_1{x}^{4/3}}{32}+\frac{c_1^3}{64}+\frac{x^2}{8}\\ y& =\frac{3(i\sqrt{3}-1)c_1^2{x}^{2/3}}{64}-\frac{3c_1(1+i\sqrt{3})x^{4/3}}{32}+\frac{c_1^3}{64}+\frac{x^2}{8}\\ y& =-\frac{3x^{4/3}{c}_1}{16}+\frac{3x^{2/3}{c}_1^2}{32}-\frac{c_1^3}{64}+\frac{x^2}{8}\\ y& =\frac{3(-1-i\sqrt{3})c_1^2{x}^{2/3}}{64}+\frac{3c_1(1-i\sqrt{3})x^{4/3}}{32}-\frac{c_1^3}{64}+\frac{x^2}{8}\\ y& =\frac{3(i\sqrt{3}-1)c_1^2{x}^{2/3}}{64}+\frac{3c_1(1+i\sqrt{3})x^{4/3}}{32}-\frac{c_1^3}{64}+\frac{x^2}{8}\end{array}$$

Maple trace

Methods for first order ODEs: 
   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   -> Solving 1st order ODE of high degree, 1st attempt 
   trying 1st order WeierstrassP solution for high degree ODE 
   trying 1st order WeierstrassPPrime solution for high degree ODE 
   trying 1st order JacobiSN solution for high degree ODE 
   trying 1st order ODE linearizable_by_differentiation 
   trying differential order: 1; missing variables 
   trying simple symmetries for implicit equations 
   Successful isolation of dy/dx: 3 solutions were found. Trying to solve each \ 
resulting ODE. 
      *** Sublevel 3 *** 
      Methods for first order ODEs: 
      --- Trying classification methods --- 
      trying homogeneous types: 
      trying homogeneous G 
      trying an integrating factor from the invariance group 
      <- integrating factor successful 
      <- homogeneous successful 
   ------------------- 
   * Tackling next ODE. 
      *** Sublevel 3 *** 
      Methods for first order ODEs: 
      --- Trying classification methods --- 
      trying homogeneous types: 
      trying homogeneous G 
      trying an integrating factor from the invariance group 
      <- integrating factor successful 
      <- homogeneous successful 
   ------------------- 
   * Tackling next ODE. 
      *** Sublevel 3 *** 
      Methods for first order ODEs: 
      --- Trying classification methods --- 
      trying homogeneous types: 
      trying homogeneous G 
      trying an integrating factor from the invariance group 
      <- integrating factor successful 
      <- homogeneous successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & {\left(\frac{d}{dx}y\left(x\right)\right)}^3=\frac{y{\left(x\right)}^2}{x}\\ \bullet & & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Solve for the highest derivative}\\ & & [\frac{d}{dx}y\left(x\right)=\frac{{\left(y{\left(x\right)}^2{x}^2\right)}^{1/3}}{x},\frac{d}{dx}y\left(x\right)=-\frac{{\left(y{\left(x\right)}^2{x}^2\right)}^{1/3}}{2x}-\frac{\mathrm{I}\sqrt{3}{\left(y{\left(x\right)}^2{x}^2\right)}^{1/3}}{2x},\frac{d}{dx}y\left(x\right)=-\frac{{\left(y{\left(x\right)}^2{x}^2\right)}^{1/3}}{2x}+\frac{\mathrm{I}\sqrt{3}{\left(y{\left(x\right)}^2{x}^2\right)}^{1/3}}{2x}]\\ \bullet & & \text{Solve the equation}\frac{d}{dx}y\left(x\right)=\frac{{\left(y{\left(x\right)}^2{x}^2\right)}^{1/3}}{x}\\ \bullet & & \text{Solve the equation}\frac{d}{dx}y\left(x\right)=-\frac{{\left(y{\left(x\right)}^2{x}^2\right)}^{1/3}}{2x}-\frac{\mathrm{I}\sqrt{3}{\left(y{\left(x\right)}^2{x}^2\right)}^{1/3}}{2x}\\ \bullet & & \text{Solve the equation}\frac{d}{dx}y\left(x\right)=-\frac{{\left(y{\left(x\right)}^2{x}^2\right)}^{1/3}}{2x}+\frac{\mathrm{I}\sqrt{3}{\left(y{\left(x\right)}^2{x}^2\right)}^{1/3}}{2x}\\ \bullet & & \text{Set of solutions}\\ & & \{\mathit{workingODE},\mathit{workingODE},\mathit{workingODE}\}\end{array}$$

✓ Mathematica. Time used: 0.088 (sec). Leaf size: 152

ode=(D[y[x],x])^3==y[x]^2/x; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$\begin{array}{r}y(x)\to \frac{1}{216}(3x^{2/3}+2c_1){}^3\\ y(x)\to \frac{1}{216}(18i(\sqrt{3}+i)c_1{}^2{x}^{2/3}-27i(\sqrt{3}-i)c_1{x}^{4/3}+27x^2+8c_1{}^3)\\ y(x)\to \frac{1}{216}(-18i(\sqrt{3}-i)c_1{}^2{x}^{2/3}+27i(\sqrt{3}+i)c_1{x}^{4/3}+27x^2+8c_1{}^3)\\ y(x)\to 0\end{array}$$

✓ Sympy. Time used: 2.293 (sec). Leaf size: 109

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

$$[-\frac{3x\sqrt[3]{\frac{y^2\left(x\right)}{x}}}{2y^{\frac{2}{3}}\left(x\right)}+3\sqrt[3]{y\left(x\right)}=C_1,\frac{3x\sqrt[3]{\frac{y^2\left(x\right)}{x}}(1+\sqrt{3}i)}{4y^{\frac{2}{3}}\left(x\right)}+3\sqrt[3]{y\left(x\right)}=C_1,\frac{3x\sqrt[3]{\frac{y^2\left(x\right)}{x}}(1-\sqrt{3}i)}{4y^{\frac{2}{3}}\left(x\right)}+3\sqrt[3]{y\left(x\right)}=C_1]$$