2.1.67 Problem 67

Internal problem ID [10053]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 67
Date solved : Sunday, March 22, 2026 at 09:14:03 PM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]

2.1.67.1 second order ode missing x

20.399 (sec)

\begin{align*} 3 y y^{\prime \prime }+y&=5 \\ \end{align*}

Entering second order ode missing \(x\) solverThis 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*} 3 y p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+y = 5 \end{align*}

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

Entering first order ode separable solverThe ode

\begin{equation} p^{\prime } = -\frac {y -5}{3 y p} \end{equation}

is separable as it can be written as

\begin{align*} p^{\prime }&= -\frac {y -5}{3 y p}\\ &= f(y) g(p) \end{align*}

Where

\begin{align*} f(y) &= -\frac {y -5}{3 y}\\ 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 { -\frac {y -5}{3 y} \,dy} \\ \end{align*}

\[ \frac {p^{2}}{2}=-\frac {y}{3}+\ln \left (y^{{5}/{3}}\right )+c_1 \]

Solving for \(p\) gives

\begin{align*} p &= -\frac {\sqrt {18 \ln \left (y^{{5}/{3}}\right )+18 c_1 -6 y}}{3} \\ p &= \frac {\sqrt {18 \ln \left (y^{{5}/{3}}\right )+18 c_1 -6 y}}{3} \\ \end{align*}

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

\begin{align*} y^{\prime } = -\frac {\sqrt {18 \ln \left (y^{{5}/{3}}\right )+18 c_1 -6 y}}{3} \end{align*}

Entering first order ode autonomous solverUnable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{y}-\frac {3}{\sqrt {18 \ln \left (\tau ^{{5}/{3}}\right )+18 c_1 -6 \tau }}d \tau = x +c_2 \]

Singular solutions are found by solving

\begin{align*} -\frac {\sqrt {18 \ln \left (y^{{5}/{3}}\right )+18 c_1 -6 y}}{3}&= 0 \end{align*}

for \(y\). This is because of dividing by the above earlier. This gives the following singular solution(s), which also has to satisfy the given ODE.

\begin{align*} y = \left ({\mathrm e}^{-\frac {5 \operatorname {LambertW}\left (-\frac {243^{{4}/{5}} {\mathrm e}^{-\frac {3 c_1}{5}}}{405}\right )}{3}-c_1}\right )^{{3}/{5}} \end{align*}

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

\begin{align*} y^{\prime } = \frac {\sqrt {18 \ln \left (y^{{5}/{3}}\right )+18 c_1 -6 y}}{3} \end{align*}

Entering first order ode autonomous solverUnable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{y}\frac {3}{\sqrt {18 \ln \left (\tau ^{{5}/{3}}\right )+18 c_1 -6 \tau }}d \tau = x +c_3 \]

Singular solutions are found by solving

\begin{align*} \frac {\sqrt {18 \ln \left (y^{{5}/{3}}\right )+18 c_1 -6 y}}{3}&= 0 \end{align*}

for \(y\). This is because of dividing by the above earlier. This gives the following singular solution(s), which also has to satisfy the given ODE.

\begin{align*} y = \left ({\mathrm e}^{-\frac {5 \operatorname {LambertW}\left (-\frac {243^{{4}/{5}} {\mathrm e}^{-\frac {3 c_1}{5}}}{405}\right )}{3}-c_1}\right )^{{3}/{5}} \end{align*}

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

Summary of solutions found

\begin{align*} \int _{}^{y}-\frac {3}{\sqrt {18 \ln \left (\tau ^{{5}/{3}}\right )+18 c_1 -6 \tau }}d \tau &= x +c_2 \\ \int _{}^{y}\frac {3}{\sqrt {18 \ln \left (\tau ^{{5}/{3}}\right )+18 c_1 -6 \tau }}d \tau &= x +c_3 \\ \end{align*}

2.1.67.2 ✓ Maple. Time used: 0.013 (sec). Leaf size: 59

ode:=3*y(x)*diff(diff(y(x),x),x)+y(x) = 5; 
dsolve(ode,y(x), singsol=all);
 

\begin{align*} -3 \int _{}^{y}\frac {1}{\sqrt {30 \ln \left (\textit {\_a} \right )+9 c_1 -6 \textit {\_a}}}d \textit {\_a} -x -c_2 &= 0 \\ 3 \int _{}^{y}\frac {1}{\sqrt {30 \ln \left (\textit {\_a} \right )+9 c_1 -6 \textit {\_a}}}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 
   -> Computing symmetries using: way = exp_sym 
-> Calling odsolve with the ODE, diff(_b(_a),_a)*_b(_a)+1/3*(_a-5)/_a = 0, _b( 
_a) 
   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   trying Bernoulli 
   <- Bernoulli successful 
<- differential order: 2; canonical coordinates successful 
<- differential order 2; missing variables successful
 

2.1.67.3 ✓ Mathematica. Time used: 0.201 (sec). Leaf size: 41

ode=3*y[x]*D[y[x],{x,2}]+y[x]==5; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\[ \text {Solve}\left [\int _1^{y(x)}\frac {1}{\sqrt {c_1+\frac {2}{3} (5 \log (K[1])-K[1])}}dK[1]{}^2=(x+c_2){}^2,y(x)\right ] \]

2.1.67.4 ✓ Sympy. Time used: 166.040 (sec). Leaf size: 51

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(3*y(x)*Derivative(y(x), (x, 2)) + y(x) - 5,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

\[ \left [ \sqrt {3} \int \limits ^{y{\left (x \right )}} \frac {1}{\sqrt {- 2 u + C_{1} + 10 \log {\left (u \right )}}}\, du = C_{2} + x, \ \sqrt {3} \int \limits ^{y{\left (x \right )}} \frac {1}{\sqrt {- 2 u + C_{1} + 10 \log {\left (u \right )}}}\, du = C_{2} - x\right ] \]

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)) 
 
('2nd_nonlinear_autonomous_conserved', '2nd_nonlinear_autonomous_conserved_Integral')