2.1.19 Problem 19

Internal problem ID [10378]
Book : Second order enumerated odes
Section : section 1
Problem number : 19
Date solved : Monday, December 08, 2025 at 08:23:48 PM
CAS classification : [[_2nd_order, _missing_y]]

2.1.19.1 second order ode missing y

1.043 (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}{p -1}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 {p -1}{2 p}&= 0 \end{align*}

for \(p \left (x \right )\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have 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 {x}{2}-\frac {c_1}{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 {x}{2}-\frac {c_1}{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 {x}{2}-\frac {c_1}{2}}\right )+1\right )}^{2}+x \\ \end{align*}

For solution \(u \left (x \right ) = x -1\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which 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*}

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

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

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

\begin{align*} y^{\prime } = -{\left (\operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {x}{2}-\frac {c_1}{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 {-\operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {x}{2}-\frac {c_1}{2}}\right )^{2}-2 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {x}{2}-\frac {c_1}{2}}\right )+x -1\, dx}\\ y &= \frac {2 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {x}{2}-\frac {c_1}{2}}\right )^{3}}{3}+3 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {x}{2}-\frac {c_1}{2}}\right )^{2}+4 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {x}{2}-\frac {c_1}{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 {x}{2}-\frac {c_1}{2}}\right )^{3}}{3}+3 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {x}{2}-\frac {c_1}{2}}\right )^{2}+4 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {x}{2}-\frac {c_1}{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 {x}{2}-\frac {c_1}{2}}\right )^{3}}{3}+3 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {x}{2}-\frac {c_1}{2}}\right )^{2}+4 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {x}{2}-\frac {c_1}{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 are implemented to solve equation C1 + x - 2*sqrt(-_X0 + x) + 2*log(sqrt(-_X0 + x) + 1)