1.54 problem 54

Internal problem ID [8018]
Book : First order enumerated odes
Section : section 1
Problem number : 54
Date solved : Monday, October 21, 2024 at 04:41:00 PM
CAS classification : [[_homogeneous, `class G`], _rational]

Solve

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

1.54.1 Solved as first order ode of type nonlinear p but separable

Time used: 1.761 (sec)

The ode has the form

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

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

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

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

\begin{align*} y^{\prime } &=\left (f g \right )^{{1}/{3}}\\ y^{\prime } &=-\frac {\left (f g \right )^{{1}/{3}}}{2}+\frac {i \sqrt {3}\, \left (f g \right )^{{1}/{3}}}{2}\\ y^{\prime } &=-\frac {\left (f g \right )^{{1}/{3}}}{2}-\frac {i \sqrt {3}\, \left (f g \right )^{{1}/{3}}}{2} \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^{2} &> 0 \end{align*}

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

\begin{align*} y^{\prime } &=f^{{1}/{3}} g^{{1}/{3}}\\ y^{\prime } &=\frac {f^{{1}/{3}} g^{{1}/{3}} \left (-1+i \sqrt {3}\right )}{2}\\ y^{\prime } &=-\frac {f^{{1}/{3}} g^{{1}/{3}} \left (1+i \sqrt {3}\right )}{2} \end{align*}

Therefore

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

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

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

Integrating now gives the following solutions

\begin{align*} \int \frac {1}{\left (y^{2}\right )^{{1}/{3}}}d y &= \int \left (\frac {1}{x}\right )^{{1}/{3}}d x +c_1\\ \frac {3 \left (y^{2}\right )^{{2}/{3}}}{y} &= \frac {3 x \left (\frac {1}{x}\right )^{{1}/{3}}}{2}\\ \int \frac {2}{\left (y^{2}\right )^{{1}/{3}} \left (-1+i \sqrt {3}\right )}d y &= \int \left (\frac {1}{x}\right )^{{1}/{3}}d x +c_1\\ -\frac {3 \left (y^{2}\right )^{{2}/{3}} \left (1+i \sqrt {3}\right )}{2 y} &= \frac {3 x \left (\frac {1}{x}\right )^{{1}/{3}}}{2}\\ \int -\frac {2}{\left (y^{2}\right )^{{1}/{3}} \left (1+i \sqrt {3}\right )}d y &= \int \left (\frac {1}{x}\right )^{{1}/{3}}d x +c_1\\ \frac {3 \left (y^{2}\right )^{{2}/{3}} \left (-1+i \sqrt {3}\right )}{2 y} &= \frac {3 x \left (\frac {1}{x}\right )^{{1}/{3}}}{2} \end{align*}

Therefore

\begin{align*} \frac {3 \left (y^{2}\right )^{{2}/{3}}}{y} &= \frac {3 x \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{align*}

1.54.2 Maple step by step solution

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

1.54.3 Maple trace

Methods for first order ODEs:
 

1.54.4 Maple dsolve solution

Solving time : 0.283 (sec)
Leaf size : 341

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

\begin{align*} y &= 0 \\ y &= -\frac {3 x^{{4}/{3}} c_1}{8}+\frac {3 x^{{2}/{3}} c_1^{2}}{8}-\frac {c_1^{3}}{8}+\frac {x^{2}}{8} \\ y &= \frac {3 \left (-1-i \sqrt {3}\right ) c_1^{2} x^{{2}/{3}}}{16}+\frac {3 c_1 \left (1-i \sqrt {3}\right ) x^{{4}/{3}}}{16}-\frac {c_1^{3}}{8}+\frac {x^{2}}{8} \\ y &= \frac {3 \left (-1+i \sqrt {3}\right ) c_1^{2} x^{{2}/{3}}}{16}+\frac {3 \left (1+i \sqrt {3}\right ) c_1 \,x^{{4}/{3}}}{16}-\frac {c_1^{3}}{8}+\frac {x^{2}}{8} \\ y &= \frac {3 x^{{4}/{3}} c_1}{16}+\frac {3 x^{{2}/{3}} c_1^{2}}{32}+\frac {c_1^{3}}{64}+\frac {x^{2}}{8} \\ y &= \frac {3 \left (-1-i \sqrt {3}\right ) c_1^{2} x^{{2}/{3}}}{64}+\frac {3 \left (-1+i \sqrt {3}\right ) c_1 \,x^{{4}/{3}}}{32}+\frac {c_1^{3}}{64}+\frac {x^{2}}{8} \\ y &= \frac {3 \left (-1+i \sqrt {3}\right ) c_1^{2} x^{{2}/{3}}}{64}+\frac {3 c_1 \left (-1-i \sqrt {3}\right ) x^{{4}/{3}}}{32}+\frac {c_1^{3}}{64}+\frac {x^{2}}{8} \\ y &= -\frac {3 x^{{4}/{3}} c_1}{16}+\frac {3 x^{{2}/{3}} c_1^{2}}{32}-\frac {c_1^{3}}{64}+\frac {x^{2}}{8} \\ y &= \frac {3 \left (-1-i \sqrt {3}\right ) c_1^{2} x^{{2}/{3}}}{64}+\frac {3 c_1 \left (1-i \sqrt {3}\right ) x^{{4}/{3}}}{32}-\frac {c_1^{3}}{64}+\frac {x^{2}}{8} \\ y &= \frac {3 \left (-1+i \sqrt {3}\right ) c_1^{2} x^{{2}/{3}}}{64}+\frac {3 \left (1+i \sqrt {3}\right ) c_1 \,x^{{4}/{3}}}{32}-\frac {c_1^{3}}{64}+\frac {x^{2}}{8} \\ \end{align*}

1.54.5 Mathematica DSolve solution

Solving time : 0.084 (sec)
Leaf size : 152

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

\begin{align*} y(x)\to \frac {1}{216} \left (3 x^{2/3}+2 c_1\right ){}^3 \\ y(x)\to \frac {1}{216} \left (18 i \left (\sqrt {3}+i\right ) c_1{}^2 x^{2/3}-27 i \left (\sqrt {3}-i\right ) c_1 x^{4/3}+27 x^2+8 c_1{}^3\right ) \\ y(x)\to \frac {1}{216} \left (-18 i \left (\sqrt {3}-i\right ) c_1{}^2 x^{2/3}+27 i \left (\sqrt {3}+i\right ) c_1 x^{4/3}+27 x^2+8 c_1{}^3\right ) \\ y(x)\to 0 \\ \end{align*}