2.1.16 Problem 16

Internal problem ID [10375]
Book : Second order enumerated odes
Section : section 1
Problem number : 16
Date solved : Monday, January 26, 2026 at 09:50:39 PM
CAS classification : [[_2nd_order, _missing_x]]

2.1.16.1 second order ode missing x

4.788 (sec)

\begin{align*} {y^{\prime \prime }}^{2}+y^{\prime }&=1 \\ \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*} p \left (y \right )^{2} \left (\frac {d}{d y}p \left (y \right )\right )^{2}+p \left (y \right ) = 1 \end{align*}

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

Entering first order ode y dy transformation solverApplying the transformation \(h \left (y \right ) = p^{2}\) the ode becomes

\[ \frac {{h^{\prime }\left (y \right )}^{2}}{4}+\sqrt {h \left (y \right )} = 1 \]

Which is now solved for \(h \left (y \right )\).

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

\begin{align*} \frac {p^{2}}{4}+\sqrt {h} = 1 \end{align*}

Solving for \(h \left (y \right )\) from the above results in

\begin{align*} \tag{1} h \left (y \right ) &= \frac {1}{16} p^{4}-\frac {1}{2} p^{2}+1 \\ \end{align*}

This has the form

\begin{align*} h=y f(p)+g(p)\tag {*} \end{align*}

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

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

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

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

\begin{align*} f &= 0\\ g &= \frac {1}{16} p^{4}-\frac {1}{2} p^{2}+1 \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p = \left (\frac {1}{4} p^{3}-p \right ) p^{\prime }\left (y \right ) \end{equation}

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

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

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0 \end{align*}

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

\begin{align*} h \left (y \right ) = 1 \end{align*}

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

\begin{equation} \tag{3} p^{\prime }\left (y \right ) = \frac {p \left (y \right )}{\frac {p \left (y \right )^{3}}{4}-p \left (y \right )} \end{equation}

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

Integrating gives

\begin{align*} \int \left (\frac {p^{2}}{4}-1\right )d p &= dy\\ \frac {1}{12} p^{3}-p&= y +c_1 \end{align*}

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

\begin{align*} h \left (y \right ) &= \frac {{\left (\left (6 c_1 +6 y +2 \sqrt {9 c_1^{2}+18 y c_1 +9 y^{2}-16}\right )^{{1}/{3}}+\frac {4}{\left (6 c_1 +6 y +2 \sqrt {9 c_1^{2}+18 y c_1 +9 y^{2}-16}\right )^{{1}/{3}}}\right )}^{4}}{16}-\frac {{\left (\left (6 c_1 +6 y +2 \sqrt {9 c_1^{2}+18 y c_1 +9 y^{2}-16}\right )^{{1}/{3}}+\frac {4}{\left (6 c_1 +6 y +2 \sqrt {9 c_1^{2}+18 y c_1 +9 y^{2}-16}\right )^{{1}/{3}}}\right )}^{2}}{2}+1 \\ \end{align*}

Simplifying the above gives

\begin{align*} h \left (y \right ) &= 1 \\ h \left (y \right ) &= \frac {-256+128 \left (6 c_1 +6 y +2 \sqrt {9 c_1^{2}+18 y c_1 +9 y^{2}-16}\right )^{{2}/{3}}+48 \left (6 c_1 +6 y +2 \sqrt {9 c_1^{2}+18 y c_1 +9 y^{2}-16}\right )^{{4}/{3}}+192 \left (y +c_1 \right ) \sqrt {9 c_1^{2}+18 y c_1 +9 y^{2}-16}+\left (6 c_1 +6 y +2 \sqrt {9 c_1^{2}+18 y c_1 +9 y^{2}-16}\right )^{{8}/{3}}+576 y^{2}+1152 y c_1 +576 c_1^{2}}{16 \left (6 c_1 +6 y +2 \sqrt {9 c_1^{2}+18 y c_1 +9 y^{2}-16}\right )^{{4}/{3}}} \\ \end{align*}

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

Converting the solution above back using \(h \left (y \right ) = p^{2}\) gives

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

Solving for \(p\) gives

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

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

\begin{align*} y^{\prime } = 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 {1\, dx}\\ y &= x + c_2 \end{align*}

\begin{align*} y&= x +c_2 \end{align*}

Summary of solutions found

\begin{align*} y &= x +c_2 \\ \end{align*}

2.1.16.2 second order ode missing y

0.467 (sec)

\begin{align*} {y^{\prime \prime }}^{2}+y^{\prime }&=1 \\ \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 )-1 = 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 -1 = 0 \end{align*}

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

\begin{align*} \tag{1} u \left (x \right ) &= -p^{2}+1 \\ \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 &= 0\\ g &= -p^{2}+1 \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p = -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 = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0 \end{align*}

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

\begin{align*} u \left (x \right ) = 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 {1}{2}} \end{equation}

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

Since the ode has the form \(p^{\prime }\left (x \right )=f(x)\), then we only need to integrate \(f(x)\).

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

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

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

\begin{align*} u \left (x \right ) &= -\frac {1}{4} x^{2}+c_1 x -c_1^{2}+1 \\ \end{align*}

Simplifying the above gives

\begin{align*} u \left (x \right ) &= 1 \\ u \left (x \right ) &= -\frac {\left (x -2 c_1 +2\right ) \left (x -2 c_1 -2\right )}{4} \\ \end{align*}

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

\begin{align*} u \left (x \right ) &= 1 \\ u \left (x \right ) &= -\frac {\left (x -2 c_1 +2\right ) \left (x -2 c_1 -2\right )}{4} \\ \end{align*}

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

\begin{align*} y^{\prime } = 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 {1\, dx}\\ y &= x + c_2 \end{align*}

\begin{align*} y&= x +c_2 \end{align*}

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

\begin{align*} y &= x +c_2 \\ \end{align*}

For solution \(u \left (x \right ) = -\frac {\left (x -2 c_1 +2\right ) \left (x -2 c_1 -2\right )}{4}\), since \(u=y^{\prime }\) then the new first order ode to solve is

\begin{align*} y^{\prime } = -\frac {\left (x -2 c_1 +2\right ) \left (x -2 c_1 -2\right )}{4} \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 {-\frac {\left (2 c_1 -x +2\right ) \left (2 c_1 -x -2\right )}{4}\, dx}\\ y &= -\frac {x^{3}}{12}+\frac {c_1 \,x^{2}}{2}-\frac {\left (2 c_1 +2\right ) \left (2 c_1 -2\right ) x}{4} + c_3 \end{align*}

\begin{align*} y&= -\frac {1}{12} x^{3}+\frac {1}{2} c_1 \,x^{2}-c_1^{2} x +x +c_3 \end{align*}

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

\begin{align*} y &= -\frac {1}{12} x^{3}+\frac {1}{2} c_1 \,x^{2}-c_1^{2} x +x +c_3 \\ \end{align*}

Summary of solutions found

\begin{align*} y &= x +c_2 \\ y &= -\frac {1}{12} x^{3}+\frac {1}{2} c_1 \,x^{2}-c_1^{2} x +x +c_3 \\ \end{align*}

2.1.16.3 ✓ Maple. Time used: 0.055 (sec). Leaf size: 30

ode:=diff(diff(y(x),x),x)^2+diff(y(x),x) = 1; 
dsolve(ode,y(x), singsol=all);
 

\begin{align*} y &= x +c_{1} \\ y &= -\frac {1}{12} x^{3}+\frac {1}{2} c_{1} x^{2}-x \,c_{1}^{2}+x +c_{2} \\ \end{align*}

Maple trace

Methods for second order ODEs: 
   *** Sublevel 2 *** 
   Methods for second order ODEs: 
   Successful isolation of d^2y/dx^2: 2 solutions were found. Trying to solve e\ 
ach resulting ODE. 
      *** Sublevel 3 *** 
      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 
      -> Calling odsolve with the ODE, diff(diff(diff(y(x),x),x),x)+1/2, y(x) 
         *** Sublevel 4 *** 
         Methods for third order ODEs: 
         --- Trying classification methods --- 
         trying a quadrature 
         <- quadrature successful 
      <- 2nd order ODE linearizable_by_differentiation successful 
   ------------------- 
   * Tackling next ODE. 
      *** Sublevel 3 *** 
      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 
      <- 2nd order ODE linearizable_by_differentiation successful 
-> Calling odsolve with the ODE, diff(y(x),x) = 1, y(x), singsol = none 
   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   <- 1st order linear successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )^{2}+\frac {d}{d x}y \left (x \right )=1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =\frac {d}{d x}y \left (x \right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}u \left (x \right )\right )^{2}+u \left (x \right )=1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}u \left (x \right )=\sqrt {1-u \left (x \right )}, \frac {d}{d x}u \left (x \right )=-\sqrt {1-u \left (x \right )}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}u \left (x \right )=\sqrt {1-u \left (x \right )} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}u \left (x \right )}{\sqrt {1-u \left (x \right )}}=1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}u \left (x \right )}{\sqrt {1-u \left (x \right )}}d x =\int 1d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -2 \sqrt {1-u \left (x \right )}=x +\textit {\_C1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{4} \textit {\_C1}^{2}-\frac {1}{2} \textit {\_C1} x -\frac {1}{4} x^{2}+1 \\ {} & \circ & \textrm {Simplify}\hspace {3pt} \\ {} & {} & u \left (x \right )=-\frac {\left (x +\textit {\_C1} +2\right ) \left (x +\textit {\_C1} -2\right )}{4} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}u \left (x \right )=-\sqrt {1-u \left (x \right )} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}u \left (x \right )}{\sqrt {1-u \left (x \right )}}=-1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}u \left (x \right )}{\sqrt {1-u \left (x \right )}}d x =\int \left (-1\right )d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -2 \sqrt {1-u \left (x \right )}=-x +\textit {\_C1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{4} \textit {\_C1}^{2}+\frac {1}{2} \textit {\_C1} x -\frac {1}{4} x^{2}+1 \\ {} & \circ & \textrm {Simplify}\hspace {3pt} \\ {} & {} & u \left (x \right )=-\frac {\left (x -\textit {\_C1} +2\right ) \left (x -\textit {\_C1} -2\right )}{4} \\ {} & \circ & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & u \left (x \right )=-\frac {\left (x +\textit {\_C1} +2\right ) \left (x +\textit {\_C1} -2\right )}{4} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{u \left (x \right )=-\frac {\left (x +\mathit {C1} +2\right ) \left (x +\mathit {C1} -2\right )}{4}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {\left (x +\mathit {C1} +2\right ) \left (x +\mathit {C1} -2\right )}{4} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =\frac {d}{d x}y \left (x \right ) \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\frac {\left (x +\mathit {C1} +2\right ) \left (x +\mathit {C1} -2\right )}{4} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & \int \left (\frac {d}{d x}y \left (x \right )\right )d x =\int -\frac {\left (x +\mathit {C1} +2\right ) \left (x +\mathit {C1} -2\right )}{4}d x +\mathit {C2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y \left (x \right )=-\frac {x^{3}}{12}-\frac {\mathit {C1} \,x^{2}}{4}-\frac {\left (\mathit {C1} +2\right ) \left (\mathit {C1} -2\right ) x}{4}+\mathit {C2} \end {array} \]

2.1.16.4 ✓ Mathematica. Time used: 0.016 (sec). Leaf size: 67

ode=(D[y[x],{x,2}])^2+D[y[x],x]==1; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to -\frac {x^3}{12}-\frac {c_1 x^2}{4}+x-\frac {c_1{}^2 x}{4}+c_2\\ y(x)&\to -\frac {x^3}{12}+\frac {c_1 x^2}{4}+x-\frac {c_1{}^2 x}{4}+c_2 \end{align*}

2.1.16.5 ✓ Sympy. Time used: 0.658 (sec). Leaf size: 24

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(Derivative(y(x), x) + Derivative(y(x), (x, 2))**2 - 1,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

\[ y{\left (x \right )} = C_{1} - \frac {C_{2}^{2} x}{4} + \frac {C_{2} x^{2}}{4} - \frac {x^{3}}{12} + x \]

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',)