2.1.59 Problem 59

Internal problem ID [10317]
Book : First order enumerated odes
Section : section 1
Problem number : 59
Date solved : Monday, January 26, 2026 at 09:39:01 PM
CAS classification : [_separable]

2.1.59.1 Solved using first_order_ode_homog_type_G

8.351 (sec)

Entering first order ode homog type G solver

\begin{align*} {y^{\prime }}^{2}&=\frac {1}{y^{4} x^{3}} \\ \end{align*}

Multiplying the right side of the ode, which is \(\frac {1}{x^{{3}/{2}} y^{2}}\) by \(\frac {x}{y}\) gives

\begin{align*} y^{\prime } &= \left (\frac {x}{y}\right ) \frac {1}{x^{{3}/{2}} y^{2}}\\ &= \frac {1}{\sqrt {x}\, y^{3}}\\ &= 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 {1}{2 \sqrt {x}\, y^{3}}\\ f_y&= y \left (\frac {\partial }{\partial y}F \left (x , y\right )\right )\\ &= -\frac {3}{\sqrt {x}\, y^{3}}\\ \alpha &= \frac {f_x}{f_y} \\ &={\frac {1}{6}} \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^{{1}/{6}}} \end{align*}

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

\begin{align*} F \left (z \right ) &=\frac {1}{z^{3}} \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

\[ \ln \left (x \right )-c_1 +\int _{}^{y x^{{1}/{6}}}\frac {1}{z \left (-\frac {1}{6}-\frac {1}{z^{3}}\right )}d z = 0 \]

The value of the above is

\[ \ln \left (x \right )-c_1 -2 \ln \left (y^{3} \sqrt {x}+6\right ) = 0 \]

Multiplying the right side of the ode, which is \(-\frac {1}{x^{{3}/{2}} y^{2}}\) by \(\frac {x}{y}\) gives

\begin{align*} y^{\prime } &= \left (\frac {x}{y}\right ) -\frac {1}{x^{{3}/{2}} y^{2}}\\ &= -\frac {1}{\sqrt {x}\, y^{3}}\\ &= 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 {1}{2 \sqrt {x}\, y^{3}}\\ f_y&= y \left (\frac {\partial }{\partial y}F \left (x , y\right )\right )\\ &= \frac {3}{\sqrt {x}\, y^{3}}\\ \alpha &= \frac {f_x}{f_y} \\ &={\frac {1}{6}} \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^{{1}/{6}}} \end{align*}

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

\begin{align*} F \left (z \right ) &=-\frac {1}{z^{3}} \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

\[ \ln \left (x \right )-c_2 +\int _{}^{y x^{{1}/{6}}}\frac {1}{z \left (-\frac {1}{6}+\frac {1}{z^{3}}\right )}d z = 0 \]

The value of the above is

\[ \ln \left (x \right )-c_2 -2 \ln \left (y^{3} \sqrt {x}-6\right ) = 0 \]

Summary of solutions found

\begin{align*} \ln \left (x \right )-c_1 -2 \ln \left (y^{3} \sqrt {x}+6\right ) &= 0 \\ \ln \left (x \right )-c_2 -2 \ln \left (y^{3} \sqrt {x}-6\right ) &= 0 \\ \end{align*}

2.1.59.2 Solved using first_order_nonlinear_p_but_separable

1.523 (sec)

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

\begin{align*} {y^{\prime }}^{2}&=\frac {1}{y^{4} x^{3}} \\ \end{align*}

The ode has the form

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

Where \(n=2, m=1, f=\frac {1}{x^{3}} , g=\frac {1}{y^{4}}\). Hence the ode is

\begin{align*} (y')^{2} &= \frac {1}{y^{4} x^{3}} \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^{3}} &> 0\\ \frac {1}{y^{4}} &> 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 {\frac {1}{y^{4}}}} \, dy &= \left (\sqrt {\frac {1}{x^{3}}}\right )\,dx\\ -\frac {1}{\sqrt {\frac {1}{y^{4}}}} \, dy &= \left (\sqrt {\frac {1}{x^{3}}}\right )\,dx \end{align*}

Integrating now gives the following solutions

\begin{align*} \int \frac {1}{\sqrt {\frac {1}{y^{4}}}}d y &= \int \sqrt {\frac {1}{x^{3}}}d x +c_1\\ \frac {y}{3 \sqrt {\frac {1}{y^{4}}}} &= -2 x \sqrt {\frac {1}{x^{3}}}\\ \int -\frac {1}{\sqrt {\frac {1}{y^{4}}}}d y &= \int \sqrt {\frac {1}{x^{3}}}d x +c_1\\ -\frac {y}{3 \sqrt {\frac {1}{y^{4}}}} &= -2 x \sqrt {\frac {1}{x^{3}}} \end{align*}

Therefore

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

Summary of solutions found

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

2.1.59.3 ✓ Maple. Time used: 0.023 (sec). Leaf size: 137

ode:=diff(y(x),x)^2 = 1/x^3/y(x)^4; 
dsolve(ode,y(x), singsol=all);
 

\begin{align*} y &= \left (\frac {c_1 \sqrt {x}-6}{\sqrt {x}}\right )^{{1}/{3}} \\ y &= -\frac {\left (\frac {c_1 \sqrt {x}-6}{\sqrt {x}}\right )^{{1}/{3}} \left (1+i \sqrt {3}\right )}{2} \\ y &= \frac {\left (\frac {c_1 \sqrt {x}-6}{\sqrt {x}}\right )^{{1}/{3}} \left (-1+i \sqrt {3}\right )}{2} \\ y &= \left (\frac {c_1 \sqrt {x}+6}{\sqrt {x}}\right )^{{1}/{3}} \\ y &= -\frac {\left (\frac {c_1 \sqrt {x}+6}{\sqrt {x}}\right )^{{1}/{3}} \left (1+i \sqrt {3}\right )}{2} \\ y &= \frac {\left (\frac {c_1 \sqrt {x}+6}{\sqrt {x}}\right )^{{1}/{3}} \left (-1+i \sqrt {3}\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 
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: 2 solutions were found. Trying to solve each res\ 
ulting ODE. 
   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   trying Bernoulli 
   <- Bernoulli successful 
------------------- 
* Tackling next ODE. 
   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   trying Bernoulli 
   <- Bernoulli 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 {1}{x^{3} y \left (x \right )^{4}} \\ \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 {1}{x^{{3}/{2}} y \left (x \right )^{2}}, \frac {d}{d x}y \left (x \right )=-\frac {1}{x^{{3}/{2}} y \left (x \right )^{2}}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=\frac {1}{x^{{3}/{2}} y \left (x \right )^{2}} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y \left (x \right )\right ) y \left (x \right )^{2}=\frac {1}{x^{{3}/{2}}} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}y \left (x \right )\right ) y \left (x \right )^{2}d x =\int \frac {1}{x^{{3}/{2}}}d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {y \left (x \right )^{3}}{3}=-\frac {2}{\sqrt {x}}+\textit {\_C1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & \left \{y \left (x \right )=\left (\frac {3 \textit {\_C1} \sqrt {x}-6}{\sqrt {x}}\right )^{{1}/{3}}, y \left (x \right )=-\frac {\left (\frac {3 \textit {\_C1} \sqrt {x}-6}{\sqrt {x}}\right )^{{1}/{3}}}{2}-\frac {\mathrm {I} \sqrt {3}\, \left (\frac {3 \textit {\_C1} \sqrt {x}-6}{\sqrt {x}}\right )^{{1}/{3}}}{2}, y \left (x \right )=-\frac {\left (\frac {3 \textit {\_C1} \sqrt {x}-6}{\sqrt {x}}\right )^{{1}/{3}}}{2}+\frac {\mathrm {I} \sqrt {3}\, \left (\frac {3 \textit {\_C1} \sqrt {x}-6}{\sqrt {x}}\right )^{{1}/{3}}}{2}\right \} \\ {} & \circ & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \left \{y \left (x \right )=\left (\frac {3 \textit {\_C1} \sqrt {x}-6}{\sqrt {x}}\right )^{{1}/{3}}, y \left (x \right )=\frac {\left (\frac {\textit {\_C1} \sqrt {x}-2}{\sqrt {x}}\right )^{{1}/{3}} \left (-3^{{1}/{3}}+\mathrm {I} \,3^{{5}/{6}}\right )}{2}, y \left (x \right )=-\frac {\left (\frac {\textit {\_C1} \sqrt {x}-2}{\sqrt {x}}\right )^{{1}/{3}} \left (3^{{1}/{3}}+\mathrm {I} \,3^{{5}/{6}}\right )}{2}\right \} \\ {} & \circ & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & \left \{y \left (x \right )=\left (\frac {\textit {\_C1} \sqrt {x}-6}{\sqrt {x}}\right )^{{1}/{3}}, y \left (x \right )=\frac {\left (\frac {\textit {\_C1} \sqrt {x}-2}{\sqrt {x}}\right )^{{1}/{3}} \left (-3^{{1}/{3}}+\mathrm {I} \,3^{{5}/{6}}\right )}{2}, y \left (x \right )=-\frac {\left (\frac {\textit {\_C1} \sqrt {x}-2}{\sqrt {x}}\right )^{{1}/{3}} \left (3^{{1}/{3}}+\mathrm {I} \,3^{{5}/{6}}\right )}{2}\right \} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=-\frac {1}{x^{{3}/{2}} y \left (x \right )^{2}} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y \left (x \right )\right ) y \left (x \right )^{2}=-\frac {1}{x^{{3}/{2}}} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}y \left (x \right )\right ) y \left (x \right )^{2}d x =\int -\frac {1}{x^{{3}/{2}}}d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {y \left (x \right )^{3}}{3}=\frac {2}{\sqrt {x}}+\textit {\_C1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & \left \{y \left (x \right )=\left (\frac {3 \textit {\_C1} \sqrt {x}+6}{\sqrt {x}}\right )^{{1}/{3}}, y \left (x \right )=-\frac {\left (\frac {3 \textit {\_C1} \sqrt {x}+6}{\sqrt {x}}\right )^{{1}/{3}}}{2}-\frac {\mathrm {I} \sqrt {3}\, \left (\frac {3 \textit {\_C1} \sqrt {x}+6}{\sqrt {x}}\right )^{{1}/{3}}}{2}, y \left (x \right )=-\frac {\left (\frac {3 \textit {\_C1} \sqrt {x}+6}{\sqrt {x}}\right )^{{1}/{3}}}{2}+\frac {\mathrm {I} \sqrt {3}\, \left (\frac {3 \textit {\_C1} \sqrt {x}+6}{\sqrt {x}}\right )^{{1}/{3}}}{2}\right \} \\ {} & \circ & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \left \{y \left (x \right )=\left (\frac {3 \textit {\_C1} \sqrt {x}+6}{\sqrt {x}}\right )^{{1}/{3}}, y \left (x \right )=\frac {\left (\frac {\textit {\_C1} \sqrt {x}+2}{\sqrt {x}}\right )^{{1}/{3}} \left (-3^{{1}/{3}}+\mathrm {I} \,3^{{5}/{6}}\right )}{2}, y \left (x \right )=-\frac {\left (\frac {\textit {\_C1} \sqrt {x}+2}{\sqrt {x}}\right )^{{1}/{3}} \left (3^{{1}/{3}}+\mathrm {I} \,3^{{5}/{6}}\right )}{2}\right \} \\ {} & \circ & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & \left \{y \left (x \right )=\left (\frac {\textit {\_C1} \sqrt {x}+6}{\sqrt {x}}\right )^{{1}/{3}}, y \left (x \right )=\frac {\left (\frac {\textit {\_C1} \sqrt {x}+2}{\sqrt {x}}\right )^{{1}/{3}} \left (-3^{{1}/{3}}+\mathrm {I} \,3^{{5}/{6}}\right )}{2}, y \left (x \right )=-\frac {\left (\frac {\textit {\_C1} \sqrt {x}+2}{\sqrt {x}}\right )^{{1}/{3}} \left (3^{{1}/{3}}+\mathrm {I} \,3^{{5}/{6}}\right )}{2}\right \} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y \left (x \right )=\left (\frac {\mathit {C1} \sqrt {x}-6}{\sqrt {x}}\right )^{{1}/{3}}, y \left (x \right )=\left (\frac {\mathit {C1} \sqrt {x}+6}{\sqrt {x}}\right )^{{1}/{3}}, y \left (x \right )=\frac {\left (\frac {\mathit {C1} \sqrt {x}-2}{\sqrt {x}}\right )^{{1}/{3}} \left (-3^{{1}/{3}}+\mathrm {I} \,3^{{5}/{6}}\right )}{2}, y \left (x \right )=-\frac {\left (\frac {\mathit {C1} \sqrt {x}-2}{\sqrt {x}}\right )^{{1}/{3}} \left (3^{{1}/{3}}+\mathrm {I} \,3^{{5}/{6}}\right )}{2}, y \left (x \right )=\frac {\left (\frac {\mathit {C1} \sqrt {x}+2}{\sqrt {x}}\right )^{{1}/{3}} \left (-3^{{1}/{3}}+\mathrm {I} \,3^{{5}/{6}}\right )}{2}, y \left (x \right )=-\frac {\left (\frac {\mathit {C1} \sqrt {x}+2}{\sqrt {x}}\right )^{{1}/{3}} \left (3^{{1}/{3}}+\mathrm {I} \,3^{{5}/{6}}\right )}{2}\right \} \end {array} \]

2.1.59.4 ✓ Mathematica. Time used: 3.343 (sec). Leaf size: 157

ode=(D[y[x],x])^2==1/(x^3*y[x]^4); 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to -\sqrt [3]{-3} \sqrt [3]{-\frac {2}{\sqrt {x}}+c_1}\\ y(x)&\to \sqrt [3]{3} \sqrt [3]{-\frac {2}{\sqrt {x}}+c_1}\\ y(x)&\to (-1)^{2/3} \sqrt [3]{3} \sqrt [3]{-\frac {2}{\sqrt {x}}+c_1}\\ y(x)&\to -\sqrt [3]{-3} \sqrt [3]{\frac {2}{\sqrt {x}}+c_1}\\ y(x)&\to \sqrt [3]{3} \sqrt [3]{\frac {2}{\sqrt {x}}+c_1}\\ y(x)&\to (-1)^{2/3} \sqrt [3]{3} \sqrt [3]{\frac {2}{\sqrt {x}}+c_1} \end{align*}

2.1.59.5 ✓ Sympy. Time used: 2.808 (sec). Leaf size: 170

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(Derivative(y(x), x)**2 - 1/(x**3*y(x)**4),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

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

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')