2.1.59 Problem 59
Internal
problem
ID
[10317]
Book
:
First
order
enumerated
odes
Section
:
section
1
Problem
number
:
59
Date
solved
:
Thursday, November 27, 2025 at 10:33:26 AM
CAS
classification
:
[_separable]
Solved using first_order_ode_homog_type_G
Time used: 0.129 (sec)
Solve
\begin{align*}
{y^{\prime }}^{2}&=\frac {1}{x^{3} y^{4}} \\
\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
\[
\ln \left (x \right )-c_1 +\int _{}^{y x^{{1}/{6}}}\frac {1}{z \left (-\frac {1}{6}-\frac {1}{z^{3}}\right )}d z = 0
\]
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
\[
\ln \left (x \right )-c_2 +\int _{}^{y x^{{1}/{6}}}\frac {1}{z \left (-\frac {1}{6}+\frac {1}{z^{3}}\right )}d z = 0
\]
Summary of solutions found
\begin{align*}
\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 \\
\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 \\
\end{align*}
Solved using first_order_nonlinear_p_but_separable
Time used: 0.352 (sec)
Solve
\begin{align*}
{y^{\prime }}^{2}&=\frac {1}{x^{3} y^{4}} \\
\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}{x^{3} y^{4}} \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*}
✓ 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} \]
✓ 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*}
✓ Sympy. Time used: 2.906 (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)
\[
\left [ y{\left (x \right )} = \sqrt [3]{C_{1} + 6 x \sqrt {\frac {1}{x^{3}}}}, \ y{\left (x \right )} = \frac {\left (- \sqrt [3]{3} - 3^{\frac {5}{6}} i\right ) \sqrt [3]{C_{1} + 2 x \sqrt {\frac {1}{x^{3}}}}}{2}, \ y{\left (x \right )} = \frac {\left (- \sqrt [3]{3} + 3^{\frac {5}{6}} i\right ) \sqrt [3]{C_{1} + 2 x \sqrt {\frac {1}{x^{3}}}}}{2}, \ y{\left (x \right )} = \sqrt [3]{C_{1} - 6 x \sqrt {\frac {1}{x^{3}}}}, \ y{\left (x \right )} = \frac {\left (- \sqrt [3]{3} - 3^{\frac {5}{6}} i\right ) \sqrt [3]{C_{1} - 2 x \sqrt {\frac {1}{x^{3}}}}}{2}, \ y{\left (x \right )} = \frac {\left (- \sqrt [3]{3} + 3^{\frac {5}{6}} i\right ) \sqrt [3]{C_{1} - 2 x \sqrt {\frac {1}{x^{3}}}}}{2}\right ]
\]