2.1.19 Problem 19

Internal problem ID [10378]
Book : Second order enumerated odes
Section : section 1
Problem number : 19
Date solved : Wednesday, February 11, 2026 at 06:46:03 AM
CAS classification : [[_2nd_order, _missing_y]]

2.1.19.1 second order ode missing y

1.176 (sec)

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

Entering second order ode missing \(y\) solverThis is second order ode with missing dependent variable \(y\). Let

\begin{align*} u(x) &= y^{\prime } \end{align*}

Then

\begin{align*} u'(x) &= y^{\prime \prime } \end{align*}

Hence the ode becomes

\begin{align*} {u^{\prime }\left (x \right )}^{2}+u \left (x \right )-x = 0 \end{align*}

Which is now solved for \(u(x)\) as first order ode.

Entering first order ode dAlembert solverLet \(p=u^{\prime }\left (x \right )\) the ode becomes

\begin{align*} p^{2}+u -x = 0 \end{align*}

Solving for \(u \left (x \right )\) from the above results in

\begin{align*} \tag{1} u \left (x \right ) &= -p^{2}+x \\ \end{align*}

This has the form

\begin{align*} u=x f(p)+g(p)\tag {*} \end{align*}

Where \(f,g\) are functions of \(p=u'(x)\). The above ode is dAlembert ode which is now solved.

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(u \left (x \right )=x f + g\) to (1A) shows that

\begin{align*} f &= 1\\ g &= -p^{2} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p -1 = -2 p p^{\prime }\left (x \right ) \end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p -1 = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=1 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} u \left (x \right ) = x -1 \end{align*}

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{equation} \tag{3} p^{\prime }\left (x \right ) = -\frac {p \left (x \right )-1}{2 p \left (x \right )} \end{equation}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed.

Integrating gives

\begin{align*} \int \frac {2 p}{1-p}d p &= dx\\ -2 p -2 \ln \left (p -1\right )&= x +c_1 \end{align*}

Singular solutions are found by solving

\begin{align*} \frac {1-p}{2 p}&= 0 \end{align*}

for \(p \left (x \right )\). 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*} p \left (x \right ) = 1 \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} u \left (x \right ) &= -\left ({\mathrm e}^{-\operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {c_1}{2}-\frac {x}{2}}\right )-1-\frac {x}{2}-\frac {c_1}{2}}+1\right )^{2}+x \\ u \left (x \right ) &= x -1 \\ \end{align*}

Simplifying the above gives

\begin{align*} u \left (x \right ) &= x -1 \\ u \left (x \right ) &= -{\left (\operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {c_1}{2}-\frac {x}{2}}\right )+1\right )}^{2}+x \\ u \left (x \right ) &= x -1 \\ \end{align*}

In summary, these are the solution found for \(y\)

\begin{align*} u \left (x \right ) &= x -1 \\ u \left (x \right ) &= -{\left (\operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {c_1}{2}-\frac {x}{2}}\right )+1\right )}^{2}+x \\ \end{align*}

For solution \(u \left (x \right ) = x -1\), since \(u=y^{\prime }\) then the new first order ode to solve is

\begin{align*} y^{\prime } = x -1 \end{align*}

Entering first order ode quadrature solverSince the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dy} &= \int {x -1\, dx}\\ y &= \frac {1}{2} x^{2}-x + c_2 \end{align*}

\begin{align*} y&= \frac {1}{2} x^{2}-x +c_2 \end{align*}

In summary, these are the solution found for \((y)\)

\begin{align*} y &= \frac {1}{2} x^{2}-x +c_2 \\ \end{align*}

For solution \(u \left (x \right ) = -{\left (\operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {c_1}{2}-\frac {x}{2}}\right )+1\right )}^{2}+x\), since \(u=y^{\prime }\) then the new first order ode to solve is

\begin{align*} y^{\prime } = -{\left (\operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {c_1}{2}-\frac {x}{2}}\right )+1\right )}^{2}+x \end{align*}

Entering first order ode quadrature solverSince the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dy} &= \int {-{\left (\operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {c_1}{2}-\frac {x}{2}}\right )+1\right )}^{2}+x\, dx}\\ y &= \frac {2 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {c_1}{2}-\frac {x}{2}}\right )^{3}}{3}+3 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {c_1}{2}-\frac {x}{2}}\right )^{2}-x +4 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {c_1}{2}-\frac {x}{2}}\right )+\frac {x^{2}}{2} + c_3 \end{align*}

\begin{align*} y&= \frac {2 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {c_1}{2}-\frac {x}{2}}\right )^{3}}{3}+3 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {c_1}{2}-\frac {x}{2}}\right )^{2}+4 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {c_1}{2}-\frac {x}{2}}\right )+\frac {x^{2}}{2}-x +c_3 \end{align*}

In summary, these are the solution found for \((y)\)

\begin{align*} y &= \frac {2 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {c_1}{2}-\frac {x}{2}}\right )^{3}}{3}+3 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {c_1}{2}-\frac {x}{2}}\right )^{2}+4 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {c_1}{2}-\frac {x}{2}}\right )+\frac {x^{2}}{2}-x +c_3 \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {1}{2} x^{2}-x +c_2 \\ y &= \frac {2 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {c_1}{2}-\frac {x}{2}}\right )^{3}}{3}+3 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {c_1}{2}-\frac {x}{2}}\right )^{2}+4 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {c_1}{2}-\frac {x}{2}}\right )+\frac {x^{2}}{2}-x +c_3 \\ \end{align*}

2.1.19.2 ✓ Maple. Time used: 0.030 (sec). Leaf size: 122

ode:=diff(diff(y(x),x),x)^2+diff(y(x),x) = x; 
dsolve(ode,y(x), singsol=all);
 

\begin{align*} y &= \int \left (-{\mathrm e}^{2 \operatorname {RootOf}\left (\textit {\_Z} -x -2 \,{\mathrm e}^{\textit {\_Z}}+2+c_1 -\ln \left ({\mathrm e}^{\textit {\_Z}} \left ({\mathrm e}^{\textit {\_Z}}-2\right )^{2}\right )\right )}+2 \,{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z} -x -2 \,{\mathrm e}^{\textit {\_Z}}+2+c_1 -\ln \left ({\mathrm e}^{\textit {\_Z}} \left ({\mathrm e}^{\textit {\_Z}}-2\right )^{2}\right )\right )}+x \right )d x -x +c_2 \\ y &= \frac {2 \operatorname {LambertW}\left (-c_1 \,{\mathrm e}^{-1-\frac {x}{2}}\right )^{3}}{3}+3 \operatorname {LambertW}\left (-c_1 \,{\mathrm e}^{-1-\frac {x}{2}}\right )^{2}+4 \operatorname {LambertW}\left (-c_1 \,{\mathrm e}^{-1-\frac {x}{2}}\right )+\frac {x^{2}}{2}-x +c_2 \\ \end{align*}

Maple trace

Methods for second order ODEs: 
Successful isolation of d^2y/dx^2: 2 solutions were found. Trying to solve each\ 
 resulting ODE. 
   *** Sublevel 2 *** 
   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) = (_a-_b(_a))^(1/2), _b(_a) 
, HINT = [[1, 1]] 
      *** Sublevel 3 *** 
      symmetry methods on request 
      1st order, trying reduction of order with given symmetries: 
[1, 1] 
      1st order, trying the canonical coordinates of the invariance group 
         -> Calling odsolve with the ODE, diff(y(x),x) = 1, y(x) 
            *** Sublevel 4 *** 
            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 
------------------- 
* Tackling next ODE. 
   *** Sublevel 2 *** 
   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) = -(_a-_b(_a))^(1/2), _b(_a 
), HINT = [[1, 1]] 
      *** Sublevel 3 *** 
      symmetry methods on request 
      1st order, trying reduction of order with given symmetries: 
[1, 1] 
      1st order, trying the canonical coordinates of the invariance group 
      <- 1st order, canonical coordinates successful 
   <- differential order: 2; canonical coordinates successful 
   <- differential order 2; missing variables successful
 

2.1.19.3 ✓ Mathematica. Time used: 60.087 (sec). Leaf size: 117

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

\begin{align*} y(x)&\to \int _1^x\left (-W\left (e^{-\frac {c_1}{2}-\frac {K[1]}{2}-1}\right ){}^2-2 W\left (e^{-\frac {c_1}{2}-\frac {K[1]}{2}-1}\right )+K[1]-1\right )dK[1]+c_2\\ y(x)&\to \int _1^x\left (-W\left (-e^{\frac {1}{2} (c_1-K[2]-2)}\right ){}^2-2 W\left (-e^{\frac {1}{2} (c_1-K[2]-2)}\right )+K[2]-1\right )dK[2]+c_2 \end{align*}

2.1.19.4 ✗ Sympy

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

NotImplementedError : multiple generators [log(sqrt(-_X0 + x) + 1), sqrt(-_X0 + x)] 
No algorithms ar
 

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)) 
 
('nth_order_reducible',)