2.5.1 Problem 1

Internal problem ID [10236]
Book : Own collection of miscellaneous problems
Section : section 5.0
Problem number : 1
Date solved : Thursday, November 27, 2025 at 10:27:24 AM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]

Solved as second_order_ode_missing_x algorithm

Time used: 1.262 (sec)

Solve

\begin{align*} y^{\prime \prime }&=A y^{{2}/{3}} \\ \end{align*}

This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using

\begin{align*} y' &= p \end{align*}

Then

\begin{align*} y'' &= \frac {dp}{dx}\\ &= \frac {dp}{dy}\frac {dy}{dx}\\ &= p \frac {dp}{dy} \end{align*}

Hence the ode becomes

\begin{align*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) = A \,y^{{2}/{3}} \end{align*}

Which is now solved as first order ode for \(p(y)\).

Solve The ode

\begin{equation} p^{\prime } = \frac {A \,y^{{2}/{3}}}{p} \end{equation}

is separable as it can be written as

\begin{align*} p^{\prime }&= \frac {A \,y^{{2}/{3}}}{p}\\ &= f(y) g(p) \end{align*}

Where

\begin{align*} f(y) &= A \,y^{{2}/{3}}\\ g(p) &= \frac {1}{p} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(y) \,dy} \\ \int { p\,dp} &= \int { A \,y^{{2}/{3}} \,dy} \\ \end{align*}

\[ \frac {p^{2}}{2}=\frac {3 y^{{5}/{3}} A}{5}+c_1 \]

Solving for \(p\) gives

\begin{align*} p &= -\frac {\sqrt {30 y^{{5}/{3}} A +50 c_1}}{5} \\ p &= \frac {\sqrt {30 y^{{5}/{3}} A +50 c_1}}{5} \\ \end{align*}

For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = -\frac {\sqrt {30 y^{{5}/{3}} A +50 c_1}}{5} \end{align*}

Solve Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{y}-\frac {5}{\sqrt {30 \tau ^{{5}/{3}} A +50 c_1}}d \tau = x +c_2 \]

For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = \frac {\sqrt {30 y^{{5}/{3}} A +50 c_1}}{5} \end{align*}

Solve Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{y}\frac {5}{\sqrt {30 \tau ^{{5}/{3}} A +50 c_1}}d \tau = x +c_3 \]

✓ Maple. Time used: 0.026 (sec). Leaf size: 61

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

\begin{align*} y &= 0 \\ -5 \int _{}^{y}\frac {1}{\sqrt {30 \textit {\_a}^{{5}/{3}} A -5 c_1}}d \textit {\_a} -x -c_2 &= 0 \\ 5 \int _{}^{y}\frac {1}{\sqrt {30 \textit {\_a}^{{5}/{3}} A -5 c_1}}d \textit {\_a} -x -c_2 &= 0 \\ \end{align*}

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying 2nd order Liouville 
trying 2nd order WeierstrassP 
trying 2nd order JacobiSN 
differential order: 2; trying a linearization to 3rd order 
trying 2nd order ODE linearizable_by_differentiation 
trying 2nd order, 2 integrating factors of the form mu(x,y) 
trying differential order: 2; missing variables 
   -> Computing symmetries using: way = 3 
-> Calling odsolve with the ODE, diff(_b(_a),_a)*_b(_a)-A*_a^(2/3) = 0, _b(_a), 
HINT = [[_a, 5/6*_b]] 
   *** Sublevel 2 *** 
   symmetry methods on request 
   1st order, trying reduction of order with given symmetries: 
[_a, 5/6*_b] 
   1st order, trying the canonical coordinates of the invariance group 
      -> Calling odsolve with the ODE, diff(y(x),x) = 5/6*y(x)/x, y(x) 
         *** Sublevel 3 *** 
         Methods for first order ODEs: 
         --- Trying classification methods --- 
         trying a quadrature 
         trying 1st order linear 
         <- 1st order linear successful 
   <- 1st order, canonical coordinates successful 
<- differential order: 2; canonical coordinates successful 
<- differential order 2; missing variables successful
 

✓ Mathematica. Time used: 0.056 (sec). Leaf size: 75

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

\[ \text {Solve}\left [\frac {y(x)^2 \left (1+\frac {6 A y(x)^{5/3}}{5 c_1}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{5},\frac {8}{5},-\frac {6 A y(x)^{5/3}}{5 c_1}\right ){}^2}{\frac {6}{5} A y(x)^{5/3}+c_1}=(x+c_2){}^2,y(x)\right ] \]

✓ Sympy. Time used: 8.429 (sec). Leaf size: 99

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

\[ \left [ \frac {3 y{\left (x \right )} \Gamma \left (\frac {3}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{5} \\ \frac {8}{5} \end {matrix}\middle | {\frac {6 A e^{i \pi } y^{\frac {5}{3}}{\left (x \right )}}{5 C_{1}}} \right )}}{5 \sqrt {C_{1}} \Gamma \left (\frac {8}{5}\right )} = C_{2} + x, \ \frac {3 y{\left (x \right )} \Gamma \left (\frac {3}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{5} \\ \frac {8}{5} \end {matrix}\middle | {\frac {6 A e^{i \pi } y^{\frac {5}{3}}{\left (x \right )}}{5 C_{1}}} \right )}}{5 \sqrt {C_{1}} \Gamma \left (\frac {8}{5}\right )} = C_{2} - x\right ] \]