2.1.16 Problem 16

Internal problem ID [10375]
Book : Second order enumerated odes
Section : section 1
Problem number : 16
Date solved : Monday, December 08, 2025 at 08:15:23 PM
CAS classification : [[_2nd_order, _missing_x]]

2.1.16.1 second order ode missing y

0.578 (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*}

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 we now have a new first order ode to solve which 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*}

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 we now have a new first order ode to solve which 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*}

2.1.16.2 ✓ 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} \]

2.1.16.3 ✓ 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*}

2.1.16.4 ✓ Sympy. Time used: 0.620 (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)
 

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