2.1.14 Problem 14

Internal problem ID [10373]
Book : Second order enumerated odes
Section : section 1
Problem number : 14
Date solved : Monday, January 26, 2026 at 09:50:26 PM
CAS classification : [[_2nd_order, _missing_x], _Liouville, [_2nd_order, _reducible, _mu_xy]]

2.1.14.1 second order ode missing x

0.707 (sec)

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

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

2.1.14.2 Solved by factoring the differential equation

Time used: 0.425 (sec)

\begin{align*} p p^{\prime }+p^{2}&=0 \\ \end{align*}

Writing the ode as

\begin{align*} \left (p\right )\left (p^{\prime }+p\right )&=0 \end{align*}

Therefore we need to solve the following equations

\begin{align*} \tag{1} p &= 0 \\ \tag{2} p^{\prime }+p &= 0 \\ \end{align*}

Now each of the above equations is solved in turn.

Solving equation (1)

Entering zero order ode solverSolving for \(p\) from

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

Solving gives

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

Solving equation (2)

Entering first order ode autonomous solverIntegrating gives

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

Singular solutions are found by solving

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

Solving for \(p\) gives

\begin{align*} p &= 0 \\ p &= {\mathrm e}^{-y -c_1} \\ \end{align*}

Summary of solutions found

\begin{align*} p &= 0 \\ p &= {\mathrm e}^{-y -c_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 } = 0 \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 {0\, dx} + c_2 \\ y &= c_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 } = {\mathrm e}^{-y-c_1} \end{align*}

Entering first order ode autonomous solverIntegrating gives

\begin{align*} \int {\mathrm e}^{y +c_1}d y &= dx\\ {\mathrm e}^{y +c_1}&= x +c_3 \end{align*}

Simplifying the above gives

\begin{align*} {\mathrm e}^{y+c_1} &= x +c_3 \\ \end{align*}

Solving for \(y\) gives

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

Summary of solutions found

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

2.1.14.3 second order ode missing y

0.369 (sec)

\begin{align*} y^{\prime \prime }+{y^{\prime }}^{2}&=0 \\ \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} = 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}}d u &= dx\\ \frac {1}{u}&= x +c_1 \end{align*}

Singular solutions are found by solving

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

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

\begin{align*} u \left (x \right ) &= 0 \\ u \left (x \right ) &= \frac {1}{x +c_1} \\ \end{align*}

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

\begin{align*} u \left (x \right ) &= 0 \\ u \left (x \right ) &= \frac {1}{x +c_1} \\ \end{align*}

For solution \(u \left (x \right ) = 0\), since \(u=y^{\prime }\) then the new first order ode to solve is

\begin{align*} y^{\prime } = 0 \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 {0\, dx} + c_2 \\ y &= c_2 \end{align*}

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

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

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

\begin{align*} y^{\prime } = \frac {1}{x +c_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 {\frac {1}{x +c_1}\, dx}\\ y &= \ln \left (x +c_1 \right ) + c_3 \end{align*}

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

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

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

Summary of solutions found

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

2.1.14.4 ✓ Maple. Time used: 0.007 (sec). Leaf size: 10

ode:=diff(diff(y(x),x),x)+diff(y(x),x)^2 = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ y = \ln \left (c_1 x +c_2 \right ) \]

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying 2nd order Liouville 
<- 2nd_order Liouville 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}=0 \\ \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}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}u \left (x \right )=-u \left (x \right )^{2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}u \left (x \right )}{u \left (x \right )^{2}}=-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}}d x =\int \left (-1\right )d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{u \left (x \right )}=-x +\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{-x +\mathit {C1}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & u \left (x \right )=\frac {1}{-\mathit {C1} +x} \\ \bullet & {} & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & u \left (x \right )=\frac {1}{\mathit {C1} +x} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {1}{\mathit {C1} +x} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =\frac {d}{d x}y \left (x \right ) \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {1}{\mathit {C1} +x} \\ \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 {1}{\mathit {C1} +x}d x +\mathit {C2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y \left (x \right )=\ln \left (\mathit {C1} +x \right )+\mathit {C2} \end {array} \]

2.1.14.5 ✓ Mathematica. Time used: 0.123 (sec). Leaf size: 15

ode=D[y[x],{x,2}]+(D[y[x],x])^2==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to \log (x-c_1)+c_2 \end{align*}

2.1.14.6 ✓ Sympy. Time used: 0.357 (sec). Leaf size: 8

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

\[ y{\left (x \right )} = C_{1} + \log {\left (C_{2} + x \right )} \]

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)) 
 
('factorable', 'Liouville', 'nth_order_reducible', 'Liouville_Integral')