2.1.20 Problem 20
Internal
problem
ID
[10379]
Book
:
Second
order
enumerated
odes
Section
:
section
1
Problem
number
:
20
Date
solved
:
Monday, December 08, 2025 at 08:34:46 PM
CAS
classification
:
[[_2nd_order, _missing_y], [_2nd_order, _reducible, _mu_xy]]
2.1.20.1 second order ode missing y
2.027 (sec)
\begin{align*}
y^{\prime \prime }+{y^{\prime }}^{2}&=x \\
\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}-x = 0 \end{align*}
Which is now solved for \(u(x)\) as first order ode.
Entering first order ode reduced riccati solverThis is reduced Riccati ode of the form
\begin{align*} u^{\prime }\left (x \right )&=a \,x^{n}+b u \left (x \right )^{2} \end{align*}
Comparing the given ode to the above shows that
\begin{align*} a &= 1\\ b &= -1\\ n &= 1 \end{align*}
Since \(n\neq -2\) then the solution of the reduced Riccati ode is given by
\begin{align*} w & =\sqrt {x}\left \{ \begin {array}[c]{cc} c_{1}\operatorname {BesselJ}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {ab} x^{k}\right ) +c_{2}\operatorname {BesselY}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {ab}x^{k}\right ) & ab>0\\ c_{1}\operatorname {BesselI}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {-ab}x^{k}\right ) +c_{2}\operatorname {BesselK}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {-ab}x^{k}\right ) & ab<0 \end {array} \right . \tag {1}\\ u \left (x \right ) & =-\frac {1}{b}\frac {w^{\prime }}{w}\nonumber \\ k &=1+\frac {n}{2}\nonumber \end{align*}
Since \(ab<0\) then EQ(1) gives
\begin{align*} k &= {\frac {3}{2}}\\ w &= \sqrt {x}\, \left (c_1 \operatorname {BesselI}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )+c_2 \operatorname {BesselK}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )\right ) \end{align*}
Therefore the solution becomes
\begin{align*} u \left (x \right ) & =-\frac {1}{b}\frac {w^{\prime }}{w} \end{align*}
Substituting the value of \(b,w\) found above and simplifying gives
\[
u \left (x \right ) = \frac {\left (-\operatorname {BesselK}\left (\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right ) c_2 +\operatorname {BesselI}\left (-\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right ) c_1 \right ) \sqrt {x}}{c_1 \operatorname {BesselI}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )+c_2 \operatorname {BesselK}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )}
\]
Letting \(c_2 = 1\) the above becomes \[
u \left (x \right ) = \frac {\sqrt {x}\, \left (\operatorname {BesselI}\left (-\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right ) c_1 -\operatorname {BesselK}\left (\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right )\right )}{c_1 \operatorname {BesselI}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )+\operatorname {BesselK}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )}
\]
In
summary, these are the solution found for \(y\) \begin{align*}
u \left (x \right ) &= \frac {\sqrt {x}\, \left (\operatorname {BesselI}\left (-\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right ) c_1 -\operatorname {BesselK}\left (\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right )\right )}{c_1 \operatorname {BesselI}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )+\operatorname {BesselK}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )} \\
\end{align*}
For solution \(u \left (x \right ) = \frac {\sqrt {x}\, \left (\operatorname {BesselI}\left (-\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right ) c_1 -\operatorname {BesselK}\left (\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right )\right )}{c_1 \operatorname {BesselI}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )+\operatorname {BesselK}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )}\), since \(u=y^{\prime }\) then we now have a new first
order ode to solve which is \begin{align*} y^{\prime } = \frac {\sqrt {x}\, \left (\operatorname {BesselI}\left (-\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right ) c_1 -\operatorname {BesselK}\left (\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right )\right )}{c_1 \operatorname {BesselI}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )+\operatorname {BesselK}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\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 {-\frac {\sqrt {x}\, \left (-\operatorname {BesselI}\left (-\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right ) c_1 +\operatorname {BesselK}\left (\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right )\right )}{c_1 \operatorname {BesselI}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )+\operatorname {BesselK}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )}\, dx}\\ y &= \int -\frac {\sqrt {x}\, \left (-\operatorname {BesselI}\left (-\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right ) c_1 +\operatorname {BesselK}\left (\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right )\right )}{c_1 \operatorname {BesselI}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )+\operatorname {BesselK}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )}d x + c_3 \end{align*}
\begin{align*} y&= \int -\frac {\sqrt {x}\, \left (-\operatorname {BesselI}\left (-\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right ) c_1 +\operatorname {BesselK}\left (\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right )\right )}{c_1 \operatorname {BesselI}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )+\operatorname {BesselK}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )}d x +c_3 \end{align*}
In summary, these are the solution found for \((y)\)
\begin{align*}
y &= \int -\frac {\sqrt {x}\, \left (-\operatorname {BesselI}\left (-\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right ) c_1 +\operatorname {BesselK}\left (\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right )\right )}{c_1 \operatorname {BesselI}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )+\operatorname {BesselK}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )}d x +c_3 \\
\end{align*}
Summary of solutions found
\begin{align*}
y &= \int -\frac {\sqrt {x}\, \left (-\operatorname {BesselI}\left (-\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right ) c_1 +\operatorname {BesselK}\left (\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right )\right )}{c_1 \operatorname {BesselI}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )+\operatorname {BesselK}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )}d x +c_3 \\
\end{align*}
2.1.20.2 ✓ Maple. Time used: 0.009 (sec). Leaf size: 18
ode:=diff(diff(y(x),x),x)+diff(y(x),x)^2 = x;
dsolve(ode,y(x), singsol=all);
\[
y = \ln \left (\pi \right )+\ln \left (c_1 \operatorname {AiryAi}\left (x \right )-c_2 \operatorname {AiryBi}\left (x \right )\right )
\]
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
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful
<- 2nd order, 2 integrating factors of the form mu(x,y) successful
2.1.20.3 ✓ Mathematica. Time used: 0.122 (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*}
2.1.20.4 ✗ Sympy
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(-x + Derivative(y(x), x)**2 + Derivative(y(x), (x, 2)),0)
ics = {}
dsolve(ode,func=y(x),ics=ics)
TypeError : bad operand type for unary -: list