2.1.20 problem 20
Internal
problem
ID
[8767]
Book
:
Second
order
enumerated
odes
Section
:
section
1
Problem
number
:
20
Date
solved
:
Thursday, December 12, 2024 at 09:43:09 AM
CAS
classification
:
[[_2nd_order, _missing_y], [_2nd_order, _reducible, _mu_xy]]
Solve
\begin{align*} y^{\prime \prime }+{y^{\prime }}^{2}&=x \end{align*}
Solved as second order missing y ode
Time used: 0.741 (sec)
This is second order ode with missing dependent variable \(y\). Let
\begin{align*} p(x) &= y^{\prime } \end{align*}
Then
\begin{align*} p'(x) &= y^{\prime \prime } \end{align*}
Hence the ode becomes
\begin{align*} p^{\prime }\left (x \right )+p \left (x \right )^{2}-x = 0 \end{align*}
Which is now solve for \(p(x)\) as first order ode. This is reduced Riccati ode of the form
\begin{align*} p^{\prime }\left (x \right )&=a \,x^{n}+b p \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}\\ p \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*} p \left (x \right ) & =-\frac {1}{b}\frac {w^{\prime }}{w} \end{align*}
Substituting the value of \(b,w\) found above and simplyfing gives
\[
p \left (x \right ) = \frac {\sqrt {x}\, \left (\operatorname {BesselI}\left (-\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right ) c_1 -\operatorname {BesselK}\left (\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right ) c_2 \right )}{c_1 \operatorname {BesselI}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )+c_2 \operatorname {BesselK}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )}
\]
Letting \(c_2 = 1\) the above becomes
\[
p \left (x \right ) = \frac {\sqrt {x}\, \left (\operatorname {BesselI}\left (-\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right ) c_1 -\operatorname {BesselK}\left (\frac {2}{3}, \frac {2 x^{{3}/{2}}}{3}\right )\right )}{c_1 \operatorname {BesselI}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )+\operatorname {BesselK}\left (\frac {1}{3}, \frac {2 x^{{3}/{2}}}{3}\right )}
\]
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 } = \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*}
Since 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*}
Will add steps showing solving for IC soon.
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*}
Maple step by step solution
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`
Maple dsolve solution
Solving time : 0.016 (sec)
Leaf size : 18
dsolve(diff(diff(y(x),x),x)+diff(y(x),x)^2 = x,
y(x),singsol=all)
\[
y = \ln \left (\pi \right )+\ln \left (c_{1} \operatorname {AiryAi}\left (x \right )-c_{2} \operatorname {AiryBi}\left (x \right )\right )
\]
Mathematica DSolve solution
Solving time : 0.106 (sec)
Leaf size : 15
DSolve[{D[y[x],{x,2}]+(D[y[x],x])^2==0,{}},
y[x],x,IncludeSingularSolutions->True]
\[
y(x)\to \log (x-c_1)+c_2
\]