2.1.14 Problem 14
Internal
problem
ID
[10373]
Book
:
Second
order
enumerated
odes
Section
:
section
1
Problem
number
:
14
Date
solved
:
Monday, December 08, 2025 at 08:15:12 PM
CAS
classification
:
[[_2nd_order, _missing_x], _Liouville, [_2nd_order, _reducible, _mu_xy]]
2.1.14.1 second order ode missing x
0.778 (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: 0.049 (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 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 = 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 we now have a new first
order ode to solve which 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 we now have a new first order ode to solve which 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
0.333 (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 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*} 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 we
now have a new first order ode to solve which 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 we now have a new first
order ode to solve which 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*}
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*}
0.625 (sec)
\begin{align*}
y^{\prime \prime }+{y^{\prime }}^{2}&=0 \\
\end{align*}
Applying change of variable \(x = \arccos \left (\tau \right )\) to the above ode results in the following new ode
\[
\left (-\tau ^{2}+1\right ) \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right )-\left (\frac {d}{d \tau }y \left (\tau \right )\right ) \tau +\left (\frac {d}{d \tau }y \left (\tau \right )\right )^{2} \left (-\tau ^{2}+1\right ) = 0
\]
Which is now
solved for \(y \left (\tau \right )\). Entering second order ode missing \(y\) solverThis is second order ode with missing
dependent variable \(y \left (\tau \right )\). Let \begin{align*} u(\tau ) &= \frac {d}{d \tau }y \left (\tau \right ) \end{align*}
Then
\begin{align*} u'(\tau ) &= \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right ) \end{align*}
Hence the ode becomes
\begin{align*} \left (-\tau ^{2}+1\right ) \left (\frac {d}{d \tau }u \left (\tau \right )\right )-u \left (\tau \right ) \tau +u \left (\tau \right )^{2} \left (-\tau ^{2}+1\right ) = 0 \end{align*}
Which is now solved for \(u(\tau )\) as first order ode.
Entering first order ode bernoulli solver In canonical form, the ODE is
\begin{align*} u' &= F(\tau ,u)\\ &= -\frac {u \left (\tau ^{2} u +\tau -u \right )}{\tau ^{2}-1} \end{align*}
This is a Bernoulli ODE.
\[ u' = \left (-\frac {\tau }{\tau ^{2}-1}\right ) u \left (\tau \right ) + \left (-1\right )u^{2} \tag {1} \]
The standard Bernoulli ODE has the form \[ u' = f_0(\tau )u+f_1(\tau )u^n \tag {2} \]
Comparing this to (1)
shows that \begin{align*} f_0 &=-\frac {\tau }{\tau ^{2}-1}\\ f_1 &=-1 \end{align*}
The first step is to divide the above equation by \(u^n \) which gives
\[ \frac {u'}{u^n} = f_0(\tau ) u^{1-n} +f_1(\tau ) \tag {3} \]
The next step is use the
substitution \(v = u^{1-n}\) in equation (3) which generates a new ODE in \(v \left (\tau \right )\) which will be linear and can be
easily solved using an integrating factor. Backsubstitution then gives the solution \(u(\tau )\) which is what
we want.
This method is now applied to the ODE at hand. Comparing the ODE (1) With (2) Shows that
\begin{align*} f_0(\tau )&=-\frac {\tau }{\tau ^{2}-1}\\ f_1(\tau )&=-1\\ n &=2 \end{align*}
Dividing both sides of ODE (1) by \(u^n=u^{2}\) gives
\begin{align*} u'\frac {1}{u^{2}} &= -\frac {\tau }{\left (\tau ^{2}-1\right ) u} -1 \tag {4} \end{align*}
Let
\begin{align*} v &= u^{1-n} \\ &= \frac {1}{u} \tag {5} \end{align*}
Taking derivative of equation (5) w.r.t \(\tau \) gives
\begin{align*} v' &= -\frac {1}{u^{2}}u' \tag {6} \end{align*}
Substituting equations (5) and (6) into equation (4) gives
\begin{align*} -\frac {d}{d \tau }v \left (\tau \right )&= -\frac {\tau v \left (\tau \right )}{\tau ^{2}-1}-1\\ v' &= \frac {\tau v}{\tau ^{2}-1}+1 \tag {7} \end{align*}
The above now is a linear ODE in \(v \left (\tau \right )\) which is now solved.
In canonical form a linear first order is
\begin{align*} \frac {d}{d \tau }v \left (\tau \right ) + q(\tau )v \left (\tau \right ) &= p(\tau ) \end{align*}
Comparing the above to the given ode shows that
\begin{align*} q(\tau ) &=-\frac {\tau }{\tau ^{2}-1}\\ p(\tau ) &=1 \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,d\tau }}\\ &= {\mathrm e}^{\int -\frac {\tau }{\tau ^{2}-1}d \tau }\\ &= \frac {1}{\sqrt {\tau ^{2}-1}} \end{align*}
The ode becomes
\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}\tau }}\left ( \mu v\right ) &= \mu \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}\tau }} \left (\frac {v}{\sqrt {\tau ^{2}-1}}\right ) &= \frac {1}{\sqrt {\tau ^{2}-1}}\\ \mathrm {d} \left (\frac {v}{\sqrt {\tau ^{2}-1}}\right ) &= \frac {1}{\sqrt {\tau ^{2}-1}}\mathrm {d} \tau \end{align*}
Integrating gives
\begin{align*} \frac {v}{\sqrt {\tau ^{2}-1}}&= \int {\frac {1}{\sqrt {\tau ^{2}-1}} \,d\tau } \\ &=\ln \left (\tau +\sqrt {\tau ^{2}-1}\right ) + c_1 \end{align*}
Dividing throughout by the integrating factor \(\frac {1}{\sqrt {\tau ^{2}-1}}\) gives the final solution
\[ v \left (\tau \right ) = \sqrt {\tau ^{2}-1}\, \left (\ln \left (\tau +\sqrt {\tau ^{2}-1}\right )+c_1 \right ) \]
The substitution \(v = u^{1-n}\) is now
used to convert the above solution back to \(u \left (\tau \right )\) which results in \[
\frac {1}{u \left (\tau \right )} = \sqrt {\tau ^{2}-1}\, \left (\ln \left (\tau +\sqrt {\tau ^{2}-1}\right )+c_1 \right )
\]
Solving for \(u \left (\tau \right )\) gives \begin{align*}
u \left (\tau \right ) &= \frac {1}{\sqrt {\tau ^{2}-1}\, \left (\ln \left (\tau +\sqrt {\tau ^{2}-1}\right )+c_1 \right )} \\
\end{align*}
In summary, these
are the solution found for \(y \left (\tau \right )\) \begin{align*}
u \left (\tau \right ) &= \frac {1}{\sqrt {\tau ^{2}-1}\, \left (\ln \left (\tau +\sqrt {\tau ^{2}-1}\right )+c_1 \right )} \\
\end{align*}
For solution \(u \left (\tau \right ) = \frac {1}{\sqrt {\tau ^{2}-1}\, \left (\ln \left (\tau +\sqrt {\tau ^{2}-1}\right )+c_1 \right )}\), since \(u=\frac {d}{d \tau }y \left (\tau \right )\) then we now have a new first order ode to solve
which is \begin{align*} \frac {d}{d \tau }y \left (\tau \right ) = \frac {1}{\sqrt {\tau ^{2}-1}\, \left (\ln \left (\tau +\sqrt {\tau ^{2}-1}\right )+c_1 \right )} \end{align*}
Entering first order ode quadrature solverSince the ode has the form \(\frac {d}{d \tau }y \left (\tau \right )=f(\tau )\), then we only need to
integrate \(f(\tau )\).
\begin{align*} \int {dy} &= \int {\frac {1}{\sqrt {\tau ^{2}-1}\, \left (\ln \left (\tau +\sqrt {\tau ^{2}-1}\right )+c_1 \right )}\, d\tau }\\ y \left (\tau \right ) &= \ln \left (\ln \left (\tau +\sqrt {\tau ^{2}-1}\right )+c_1 \right ) + c_2 \end{align*}
In summary, these are the solution found for \((y \left (\tau \right ))\)
\begin{align*}
y \left (\tau \right ) &= \ln \left (\ln \left (\tau +\sqrt {\tau ^{2}-1}\right )+c_1 \right )+c_2 \\
\end{align*}
Applying change of variable \(\tau = \cos \left (x \right )\) to the solutions
above gives \begin{align*}
y &= \ln \left (\ln \left (\cos \left (x \right )+\sqrt {\cos \left (x \right )^{2}-1}\right )+c_1 \right )+c_2 \\
\end{align*}
2.1.14.5 ✓ 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} \]
2.1.14.6 ✓ 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*}
2.1.14.7 ✓ Sympy. Time used: 0.312 (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)
\[
y{\left (x \right )} = C_{1} + \log {\left (C_{2} + x \right )}
\]