1.53 problem 53

Internal problem ID [8017]
Book : First order enumerated odes
Section : section 1
Problem number : 53
Date solved : Monday, October 21, 2024 at 04:40:59 PM
CAS classification : [[_homogeneous, `class G`]]

Solve

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

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

Time used: 0.434 (sec)

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 \sqrt {y^{3}}}{y^{2}} &= 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 \sqrt {y^{3}}}{y^{2}} &= 2 x \sqrt {\frac {1}{x}} \end{align*}

Therefore

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

1.53.2 Maple step by step solution

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

1.53.3 Maple trace

Methods for first order ODEs:
 

1.53.4 Maple dsolve solution

Solving time : 0.023 (sec)
Leaf size : 27

dsolve(diff(y(x),x)^2 = y(x)^3/x, 
       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*}

1.53.5 Mathematica DSolve solution

Solving time : 0.071 (sec)
Leaf size : 42

DSolve[{(D[y[x],x])^2==y[x]^3/x,{}}, 
       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*}