2.1.17 Problem 17

Internal problem ID [10376]
Book : Second order enumerated odes
Section : section 1
Problem number : 17
Date solved : Sunday, March 01, 2026 at 07:54:39 AM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_xy]]

2.1.17.1 second order ode missing x

3.051 (sec)

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

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

Entering first order ode autonomous solverIntegrating gives

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

Singular solutions are found by solving

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

for \(p\). 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 = -1\\ p = 1 \end{align*}

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

\begin{align*} -\frac {\ln \left (y^{\prime }-1\right )}{2}-\frac {\ln \left (y^{\prime }+1\right )}{2} = y+c_1 \end{align*}

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

\begin{align*} -\frac {\ln \left (p -1\right )}{2}-\frac {\ln \left (p +1\right )}{2} = y +c_1 \end{align*}

Solving for \(y\) from the above results in

\begin{align*} \tag{1} y &= -c_1 -\frac {\ln \left (p -1\right )}{2}-\frac {\ln \left (p +1\right )}{2} \\ \end{align*}

This has the form

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

Where \(f,g\) are functions of \(p=y'(x)\). 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 \(y=x f + g\) to (1A) shows that

\begin{align*} f &= 0\\ g &= -c_1 -\frac {\ln \left (p -1\right )}{2}-\frac {\ln \left (p +1\right )}{2} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p = \left (-\frac {1}{2 \left (p -1\right )}-\frac {1}{2 \left (p +1\right )}\right ) 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*} y = -c_1 -\frac {i \pi }{2} \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 )}{-\frac {1}{2 \left (p \left (x \right )-1\right )}-\frac {1}{2 \left (p \left (x \right )+1\right )}} \end{equation}

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

Integrating gives

\begin{align*} \int \frac {1}{-p^{2}+1}d p &= dx\\ \operatorname {arctanh}\left (p \right )&= x +c_2 \end{align*}

Singular solutions are found by solving

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

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

\begin{align*} y &= -c_1 -\frac {\ln \left (\tanh \left (x +c_2 \right )-1\right )}{2}-\frac {\ln \left (\tanh \left (x +c_2 \right )+1\right )}{2} \\ \end{align*}

For solution (2) 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_3 \end{align*}

\begin{align*} y&= c_3 -x \end{align*}

For solution (3) 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_4 \end{align*}

\begin{align*} y&= x +c_4 \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*} y &= c_3 -x \\ y &= x +c_4 \\ y &= -c_1 -\frac {\ln \left (\tanh \left (x +c_2 \right )-1\right )}{2}-\frac {\ln \left (\tanh \left (x +c_2 \right )+1\right )}{2} \\ \end{align*}

2.1.17.2 second order ode missing y

0.472 (sec)

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

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

Entering first order ode autonomous solverIntegrating gives

\begin{align*} \int \frac {1}{-u^{2}+1}d u &= dx\\ \operatorname {arctanh}\left (u \right )&= x +c_1 \end{align*}

Singular solutions are found by solving

\begin{align*} -u^{2}+1&= 0 \end{align*}

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

Solving for \(u \left (x \right )\) gives

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

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

\begin{align*} u \left (x \right ) &= -1 \\ u \left (x \right ) &= 1 \\ u \left (x \right ) &= \tanh \left (x +c_1 \right ) \\ \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 \(u \left (x \right ) = 1\), 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_3 \end{align*}

\begin{align*} y&= x +c_3 \end{align*}

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

\begin{align*} y &= x +c_3 \\ \end{align*}

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

\begin{align*} y^{\prime } = \tanh \left (x +c_1 \right ) \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 {\tanh \left (x +c_1 \right )\, dx}\\ y &= \ln \left (\cosh \left (x +c_1 \right )\right ) + c_4 \end{align*}

\begin{align*} y&= \ln \left (\cosh \left (x +c_1 \right )\right )+c_4 \end{align*}

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

\begin{align*} y &= \ln \left (\cosh \left (x +c_1 \right )\right )+c_4 \\ \end{align*}

Summary of solutions found

\begin{align*} y &= -x +c_2 \\ y &= x +c_3 \\ y &= \ln \left (\cosh \left (x +c_1 \right )\right )+c_4 \\ \end{align*}

2.1.17.3 ✓ Maple. Time used: 0.008 (sec). Leaf size: 22

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

\[ y = -x -\ln \left (2\right )+\ln \left (-c_1 \,{\mathrm e}^{2 x}+c_2 \right ) \]

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 a quadrature 
checking if the LODE has constant coefficients 
<- constant coefficients successful 
<- 2nd order, 2 integrating factors of the form mu(x,y) successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )+\left (\frac {d}{d x}y \left (x \right )\right )^{2}=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} \\ {} & {} & \frac {d}{d x}u \left (x \right )+u \left (x \right )^{2}=1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}u \left (x \right )=-u \left (x \right )^{2}+1 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}u \left (x \right )}{-u \left (x \right )^{2}+1}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}u \left (x \right )}{-u \left (x \right )^{2}+1}d x =\int 1d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \mathrm {arctanh}\left (u \left (x \right )\right )=x +\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\tanh \left (x +\mathit {C1} \right ) \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\tanh \left (x +\mathit {C1} \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =\frac {d}{d x}y \left (x \right ) \\ {} & {} & \frac {d}{d x}y \left (x \right )=\tanh \left (x +\mathit {C1} \right ) \\ \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 \tanh \left (x +\mathit {C1} \right )d x +\mathit {C2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y \left (x \right )=\ln \left (\cosh \left (x +\mathit {C1} \right )\right )+\mathit {C2} \end {array} \]

2.1.17.4 ✓ Mathematica. Time used: 0.426 (sec). Leaf size: 45

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

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

2.1.17.5 ✗ Sympy

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

Timed Out
 

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0