2.1.53 Problem 53

Internal problem ID [10311]
Book : First order enumerated odes
Section : section 1
Problem number : 53
Date solved : Monday, January 26, 2026 at 09:37:09 PM
CAS classification : [[_homogeneous, `class G`]]

2.1.53.1 Solved using first_order_ode_homog_type_G

1.675 (sec)

Entering first order ode homog type G solver

\begin{align*} {y^{\prime }}^{2}&=\frac {y^{3}}{x} \\ \end{align*}

Multiplying the right side of the ode, which is \(\frac {\sqrt {x y}\, y}{x}\) by \(\frac {x}{y}\) gives

\begin{align*} y^{\prime } &= \left (\frac {x}{y}\right ) \frac {\sqrt {x y}\, y}{x}\\ &= \sqrt {x y}\\ &= F(x,y) \end{align*}

Since \(F \left (x , y\right )\) has \(y\), then let

\begin{align*} f_x&= x \left (\frac {\partial }{\partial x}F \left (x , y\right )\right )\\ &= \frac {x y}{2 \sqrt {x y}}\\ f_y&= y \left (\frac {\partial }{\partial y}F \left (x , y\right )\right )\\ &= \frac {x y}{2 \sqrt {x y}}\\ \alpha &= \frac {f_x}{f_y} \\ &=1 \end{align*}

Since \(\alpha \) is independent of \(x,y\) then this is Homogeneous type G.

Let

\begin{align*} y&=\frac {z}{x^ \alpha }\\ &=\frac {z}{x} \end{align*}

Substituting the above back into \(F(x,y)\) gives

\begin{align*} F \left (z \right ) &=\sqrt {z} \end{align*}

We see that \(F \left (z \right )\) does not depend on \(x\) nor on \(y\). If this was not the case, then this method will not work.

Therefore, the implicit solution is given by

\begin{align*} \ln \left (x \right )- c_1 - \int ^{y x^\alpha } \frac {1}{z \left (\alpha + F(z)\right ) } \,dz & = 0 \end{align*}

Which gives

\[ \ln \left (x \right )-c_1 +\int _{}^{y x}\frac {1}{z \left (-1-\sqrt {z}\right )}d z = 0 \]

The value of the above is

\[ \ln \left (x \right )-c_1 +2 \ln \left (1+\sqrt {y x}\right )-\ln \left (y x \right ) = 0 \]

Multiplying the right side of the ode, which is \(-\frac {\sqrt {x y}\, y}{x}\) by \(\frac {x}{y}\) gives

\begin{align*} y^{\prime } &= \left (\frac {x}{y}\right ) -\frac {\sqrt {x y}\, y}{x}\\ &= -\sqrt {x y}\\ &= F(x,y) \end{align*}

Since \(F \left (x , y\right )\) has \(y\), then let

\begin{align*} f_x&= x \left (\frac {\partial }{\partial x}F \left (x , y\right )\right )\\ &= -\frac {x y}{2 \sqrt {x y}}\\ f_y&= y \left (\frac {\partial }{\partial y}F \left (x , y\right )\right )\\ &= -\frac {x y}{2 \sqrt {x y}}\\ \alpha &= \frac {f_x}{f_y} \\ &=1 \end{align*}

Since \(\alpha \) is independent of \(x,y\) then this is Homogeneous type G.

Let

\begin{align*} y&=\frac {z}{x^ \alpha }\\ &=\frac {z}{x} \end{align*}

Substituting the above back into \(F(x,y)\) gives

\begin{align*} F \left (z \right ) &=-\sqrt {z} \end{align*}

We see that \(F \left (z \right )\) does not depend on \(x\) nor on \(y\). If this was not the case, then this method will not work.

Therefore, the implicit solution is given by

\begin{align*} \ln \left (x \right )- c_1 - \int ^{y x^\alpha } \frac {1}{z \left (\alpha + F(z)\right ) } \,dz & = 0 \end{align*}

Which gives

\[ \ln \left (x \right )-c_2 +\int _{}^{y x}\frac {1}{z \left (-1+\sqrt {z}\right )}d z = 0 \]

The value of the above is

\[ \ln \left (x \right )-c_2 +2 \ln \left (-1+\sqrt {y x}\right )-\ln \left (y x \right ) = 0 \]

Solving for \(y\) gives

\begin{align*} y &= \frac {\frac {-2 x +2 \sqrt {{\mathrm e}^{c_2} x}}{-x +{\mathrm e}^{c_2}}-1}{x -{\mathrm e}^{c_2}} \\ y &= \frac {-\frac {2 \left (x +\sqrt {{\mathrm e}^{c_2} x}\right )}{-x +{\mathrm e}^{c_2}}-1}{x -{\mathrm e}^{c_2}} \\ y &= -\frac {-\frac {2 \left (-x +\sqrt {x \,{\mathrm e}^{c_1}}\right )}{{\mathrm e}^{c_1}-x}+1}{-{\mathrm e}^{c_1}+x} \\ y &= -\frac {\frac {2 x +2 \sqrt {x \,{\mathrm e}^{c_1}}}{{\mathrm e}^{c_1}-x}+1}{-{\mathrm e}^{c_1}+x} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {\frac {-2 x +2 \sqrt {{\mathrm e}^{c_2} x}}{-x +{\mathrm e}^{c_2}}-1}{x -{\mathrm e}^{c_2}} \\ y &= \frac {-\frac {2 \left (x +\sqrt {{\mathrm e}^{c_2} x}\right )}{-x +{\mathrm e}^{c_2}}-1}{x -{\mathrm e}^{c_2}} \\ y &= -\frac {-\frac {2 \left (-x +\sqrt {x \,{\mathrm e}^{c_1}}\right )}{{\mathrm e}^{c_1}-x}+1}{-{\mathrm e}^{c_1}+x} \\ y &= -\frac {\frac {2 x +2 \sqrt {x \,{\mathrm e}^{c_1}}}{{\mathrm e}^{c_1}-x}+1}{-{\mathrm e}^{c_1}+x} \\ \end{align*}

2.1.53.2 Solved using first_order_nonlinear_p_but_separable

2.517 (sec)

Entering first order nonlinear \(p\) but separable solver

\begin{align*} {y^{\prime }}^{2}&=\frac {y^{3}}{x} \\ \end{align*}

The ode has the form

\begin{align*} (y')^{\frac {n}{m}} &= f(x) g(y)\tag {1} \end{align*}

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

\begin{align*} (y')^{2} &= \frac {y^{3}}{x} \end{align*}

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

\begin{align*} y^{\prime } &=\sqrt {f g}\\ y^{\prime } &=-\sqrt {f g} \end{align*}

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

\begin{align*} \frac {1}{x} &> 0\\ y^{3} &> 0 \end{align*}

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

\begin{align*} y^{\prime } &=\sqrt {f}\, \sqrt {g}\\ y^{\prime } &=-\sqrt {f}\, \sqrt {g} \end{align*}

Therefore

\begin{align*} \frac {1}{\sqrt {g}} \, dy &= \left (\sqrt {f}\right )\,dx\\ -\frac {1}{\sqrt {g}} \, dy &= \left (\sqrt {f}\right )\,dx \end{align*}

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

\begin{align*} \frac {1}{\sqrt {y^{3}}} \, dy &= \left (\sqrt {\frac {1}{x}}\right )\,dx\\ -\frac {1}{\sqrt {y^{3}}} \, dy &= \left (\sqrt {\frac {1}{x}}\right )\,dx \end{align*}

Integrating now gives the following solutions

\begin{align*} \int \frac {1}{\sqrt {y^{3}}}d y &= \int \sqrt {\frac {1}{x}}d x +c_1\\ -\frac {2 y}{\sqrt {y^{3}}} &= 2 x \sqrt {\frac {1}{x}}\\ \int -\frac {1}{\sqrt {y^{3}}}d y &= \int \sqrt {\frac {1}{x}}d x +c_1\\ \frac {2 y}{\sqrt {y^{3}}} &= 2 x \sqrt {\frac {1}{x}} \end{align*}

Therefore

\begin{align*} y &= \frac {8 x \sqrt {\frac {1}{x}}+4 c_1}{8 x^{3} \left (\frac {1}{x}\right )^{{3}/{2}}+6 x \sqrt {\frac {1}{x}}\, c_1^{2}+c_1^{3}+12 c_1 x} \\ y &= \frac {8 x \sqrt {\frac {1}{x}}+4 c_1}{8 x^{3} \left (\frac {1}{x}\right )^{{3}/{2}}+6 x \sqrt {\frac {1}{x}}\, c_1^{2}+c_1^{3}+12 c_1 x} \\ \end{align*}

Simplifying the above gives

\begin{align*} y &= \frac {8 x \sqrt {\frac {1}{x}}+4 c_1}{6 x \sqrt {\frac {1}{x}}\, c_1^{2}+c_1^{3}+8 x^{2} \sqrt {\frac {1}{x}}+12 c_1 x} \\ y &= \frac {8 x \sqrt {\frac {1}{x}}+4 c_1}{6 x \sqrt {\frac {1}{x}}\, c_1^{2}+c_1^{3}+8 x^{2} \sqrt {\frac {1}{x}}+12 c_1 x} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {8 x \sqrt {\frac {1}{x}}+4 c_1}{6 x \sqrt {\frac {1}{x}}\, c_1^{2}+c_1^{3}+8 x^{2} \sqrt {\frac {1}{x}}+12 c_1 x} \\ \end{align*}

2.1.53.3 Solved using first_order_ode_parametric method

1.878 (sec)

Entering first order ode parametric solver

\begin{align*} {y^{\prime }}^{2}&=\frac {y^{3}}{x} \\ \end{align*}

Let \(y^{\prime }\) be a parameter \(\lambda \). The ode becomes

\begin{align*} \lambda ^{2}-\frac {y^{3}}{x} = 0 \end{align*}

Isolating \(x\) gives

\begin{align*} x = \frac {y^{3}}{\lambda ^{2}}\\ x = F \left (y , \lambda \right ) \end{align*}

Now we generate an ode in \(y \left (\lambda \right )\) using

\begin{align*} \frac {d}{d \lambda }y \left (\lambda \right ) &= \frac { \lambda \frac {\partial F}{\partial \lambda }} { 1- \frac {\partial F}{\partial y} } \\ &= -\frac {2 y^{3}}{\lambda ^{2} \left (1-\frac {3 y^{2}}{\lambda }\right )}\\ &= -\frac {2 y \left (\lambda \right )^{3}}{\lambda \left (-3 y \left (\lambda \right )^{2}+\lambda \right )} \end{align*}

Which is now solved for \(y\).

Entering first order ode isobaric solverSolving for \(y'\) gives

\[ y' = \frac {2 y \left (\lambda \right )^{3}}{\lambda \left (3 y \left (\lambda \right )^{2}-\lambda \right )} \]

An ode \(\frac {d}{d \lambda }y \left (\lambda \right )=f(\lambda ,y)\) is isobaric if

\[ f(t \lambda , t^m y) = t^{m-1} f(\lambda ,y)\tag {1} \]

Where here

\[ f(\lambda ,y) = \frac {2 y \left (\lambda \right )^{3}}{\lambda \left (3 y \left (\lambda \right )^{2}-\lambda \right )}\tag {2} \]

\(m\) is the order of isobaric. Substituting (2) into (1) and solving for \(m\) gives

\[ m = {\frac {1}{2}} \]

Since the ode is isobaric of order \(m={\frac {1}{2}}\), then the substitution

\begin{align*} y&=u \lambda ^m \\ &=u \sqrt {\lambda } \end{align*}

Converts the ODE to a separable in \(u \left (\lambda \right )\). Performing this substitution gives

\[ \frac {u \left (\lambda \right )}{2 \sqrt {\lambda }}+\sqrt {\lambda }\, \left (\frac {d}{d \lambda }u \left (\lambda \right )\right ) = \frac {2 \sqrt {\lambda }\, u \left (\lambda \right )^{3}}{3 \lambda u \left (\lambda \right )^{2}-\lambda } \]

Entering first order ode separable solverThe ode

\begin{equation} \frac {d}{d \lambda }u \left (\lambda \right ) = \frac {u \left (\lambda \right ) \left (u \left (\lambda \right )^{2}+1\right )}{2 \lambda \left (3 u \left (\lambda \right )^{2}-1\right )} \end{equation}

is separable as it can be written as

\begin{align*} \frac {d}{d \lambda }u \left (\lambda \right )&= \frac {u \left (\lambda \right ) \left (u \left (\lambda \right )^{2}+1\right )}{2 \lambda \left (3 u \left (\lambda \right )^{2}-1\right )}\\ &= f(\lambda ) g(u) \end{align*}

Where

\begin{align*} f(\lambda ) &= \frac {1}{2 \lambda }\\ g(u) &= \frac {u \left (u^{2}+1\right )}{3 u^{2}-1} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(\lambda ) \,d\lambda } \\ \int { \frac {3 u^{2}-1}{u \left (u^{2}+1\right )}\,du} &= \int { \frac {1}{2 \lambda } \,d\lambda } \\ \end{align*}

\[ -\ln \left (\frac {u \left (\lambda \right )}{\left (u \left (\lambda \right )^{2}+1\right )^{2}}\right )=\ln \left (\sqrt {\lambda }\right )+c_1 \]

Taking the exponential of both sides the solution becomes

\[ \frac {\left (u \left (\lambda \right )^{2}+1\right )^{2}}{u \left (\lambda \right )} = c_1 \sqrt {\lambda } \]

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

\[ \frac {u \left (u^{2}+1\right )}{3 u^{2}-1}=0 \]

for \(u \left (\lambda \right )\) gives

\begin{align*} u \left (\lambda \right )&=0\\ u \left (\lambda \right )&=-i \end{align*}

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{align*} \frac {\left (u \left (\lambda \right )^{2}+1\right )^{2}}{u \left (\lambda \right )} &= c_1 \sqrt {\lambda } \\ u \left (\lambda \right ) &= 0 \\ u \left (\lambda \right ) &= -i \\ \end{align*}

Converting \(\frac {\left (u \left (\lambda \right )^{2}+1\right )^{2}}{u \left (\lambda \right )} = c_1 \sqrt {\lambda }\) back to \(y \left (\lambda \right )\) gives

\begin{align*} \frac {\sqrt {\lambda }\, \left (\frac {y \left (\lambda \right )^{2}}{\lambda }+1\right )^{2}}{y \left (\lambda \right )} = c_1 \sqrt {\lambda } \end{align*}

Converting \(u \left (\lambda \right ) = 0\) back to \(y \left (\lambda \right )\) gives

\begin{align*} \frac {y \left (\lambda \right )}{\sqrt {\lambda }} = 0 \end{align*}

Converting \(u \left (\lambda \right ) = -i\) back to \(y \left (\lambda \right )\) gives

\begin{align*} \frac {y \left (\lambda \right )}{\sqrt {\lambda }} = -i \end{align*}

Simplifying the above gives

\begin{align*} \frac {\left (y \left (\lambda \right )^{2}+\lambda \right )^{2}}{\lambda ^{{3}/{2}} y \left (\lambda \right )} &= c_1 \sqrt {\lambda } \\ \frac {y \left (\lambda \right )}{\sqrt {\lambda }} &= 0 \\ \frac {y \left (\lambda \right )}{\sqrt {\lambda }} &= -i \\ \end{align*}

Solving for \(y \left (\lambda \right )\) gives

\begin{align*} \frac {\left (y \left (\lambda \right )^{2}+\lambda \right )^{2}}{\lambda ^{{3}/{2}} y \left (\lambda \right )} &= c_1 \sqrt {\lambda } \\ y \left (\lambda \right ) &= 0 \\ y \left (\lambda \right ) &= -i \sqrt {\lambda } \\ \end{align*}

Now that we have found solution \(y\), we have two equations with parameter \(\lambda \). They are

\begin{align*} \frac {\left (y^{2}+\lambda \right )^{2}}{\lambda ^{{3}/{2}} y} &= c_1 \sqrt {\lambda } \\ x &= \frac {y^{3}}{\lambda ^{2}} \\ \end{align*}

Eliminating \(\lambda \) gives the solution for \(y\). Solving for \(y\) gives

\begin{align*} y &= \frac {c_1 +x +2 \sqrt {c_1 x}}{c_1^{2}-2 c_1 x +x^{2}} \\ y &= -\frac {-x -c_1 +2 \sqrt {c_1 x}}{c_1^{2}-2 c_1 x +x^{2}} \\ \end{align*}

Simplifying the above gives

\begin{align*} y &= \frac {c_1 +x +2 \sqrt {c_1 x}}{\left (x -c_1 \right )^{2}} \\ y &= \frac {c_1 +x -2 \sqrt {c_1 x}}{\left (x -c_1 \right )^{2}} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {c_1 +x -2 \sqrt {c_1 x}}{\left (x -c_1 \right )^{2}} \\ y &= \frac {c_1 +x +2 \sqrt {c_1 x}}{\left (x -c_1 \right )^{2}} \\ \end{align*}

2.1.53.4 ✓ Maple. Time used: 0.017 (sec). Leaf size: 27

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

\begin{align*} y &= 0 \\ y &= \frac {\operatorname {WeierstrassP}\left (1, 0, 0\right ) 2^{{2}/{3}}}{\left (\sqrt {x}\, 2^{{1}/{3}}+c_1 \right )^{2}} \\ \end{align*}

Maple trace

Methods for first order ODEs: 
-> Solving 1st order ODE of high degree, 1st attempt 
trying 1st order WeierstrassP solution for high degree ODE 
<- 1st_order WeierstrassP successful
                                                                                   
                                                                                   
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y \left (x \right )\right )^{2}=\frac {y \left (x \right )^{3}}{x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}y \left (x \right )=\frac {\sqrt {x y \left (x \right )}\, y \left (x \right )}{x}, \frac {d}{d x}y \left (x \right )=-\frac {\sqrt {x y \left (x \right )}\, y \left (x \right )}{x}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=\frac {\sqrt {x y \left (x \right )}\, y \left (x \right )}{x} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=-\frac {\sqrt {x y \left (x \right )}\, y \left (x \right )}{x} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]

2.1.53.5 ✓ Mathematica. Time used: 0.044 (sec). Leaf size: 42

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

\begin{align*} y(x)&\to \frac {4}{\left (-2 \sqrt {x}+c_1\right ){}^2}\\ y(x)&\to \frac {4}{\left (2 \sqrt {x}+c_1\right ){}^2}\\ y(x)&\to 0 \end{align*}

2.1.53.6 ✓ Sympy. Time used: 7.110 (sec). Leaf size: 80

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

\[ y{\left (x \right )} = \frac {C_{1}^{6} - 2 C_{1}^{5} \sqrt {x} - C_{1}^{4} x + 4 C_{1}^{3} x^{\frac {3}{2}} - C_{1}^{2} x^{2} - 2 C_{1} x^{\frac {5}{2}} + x^{3}}{C_{1}^{8} - 4 C_{1}^{6} x + 6 C_{1}^{4} x^{2} - 4 C_{1}^{2} x^{3} + x^{4}} \]

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0
 

classify_ode(ode,func=y(x)) 
 
('factorable', 'lie_group')