2.1.20 Problem 20

Internal problem ID [9090]
Book : Second order enumerated odes
Section : section 1
Problem number : 20
Date solved : Friday, April 25, 2025 at 05:41:27 PM
CAS classification : [[_2nd_order, _missing_y], [_2nd_order, _reducible, _mu_xy]]

Solved as second order missing y ode

Time used: 0.971 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }+{y^{\prime }}^2& =x\end{array}$$

This is second order ode with missing dependent variable $y$. Let

$$\begin{array}{rl}u(x)& =y^{\prime }\end{array}$$

Then

$$\begin{array}{rl}{u}^{\prime }(x)& =y^{\prime \prime }\end{array}$$

Hence the ode becomes

$$\begin{array}{r}{u}^{\prime }\left(x\right)+u{\left(x\right)}^2-x=0\end{array}$$

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

This is reduced Riccati ode of the form

$$\begin{array}{rl}{u}^{\prime }& =a{x}^n+b{u}^2\end{array}$$

Comparing the given ode to the above shows that

$$\begin{array}{rl}a& =1\\ b& =-1\\ n& =1\end{array}$$

Since $n\ne -2$ then the solution of the reduced Riccati ode is given by

$$\begin{array}{rl}\text{(1)}& w& =\sqrt{x}\{\begin{array}{cc}{c}_1\mathrm{BesselJ}(\frac{1}{2k},\frac{1}{k}\sqrt{ab}{x}^k)+c_2\mathrm{BesselY}(\frac{1}{2k},\frac{1}{k}\sqrt{ab}{x}^k)& ab>0\\ c_1\mathrm{BesselI}(\frac{1}{2k},\frac{1}{k}\sqrt{-ab}{x}^k)+c_2\mathrm{BesselK}(\frac{1}{2k},\frac{1}{k}\sqrt{-ab}{x}^k)& ab<0\end{array}u& =-\frac{1}{b}\frac{w^{\prime }}{w}\\ k& =1+\frac{n}{2}\end{array}$$

Since $ab<0$ then EQ(1) gives

$$\begin{array}{rl}k& =\frac{3}{2}\\ w& =\sqrt{x}(c_1\mathrm{BesselI}(\frac{1}{3},\frac{2x^{3/2}}{3})+c_2\mathrm{BesselK}(\frac{1}{3},\frac{2x^{3/2}}{3}))\end{array}$$

Therefore the solution becomes

$$\begin{array}{rl}u& =-\frac{1}{b}\frac{w^{\prime }}{w}\end{array}$$

Substituting the value of $b,w$ found above and simplyfing gives

$$u=\frac{\sqrt{x}(\mathrm{BesselI}(-\frac{2}{3},\frac{2x^{3/2}}{3})c_1-\mathrm{BesselK}(\frac{2}{3},\frac{2x^{3/2}}{3})c_2)}{c_1\mathrm{BesselI}(\frac{1}{3},\frac{2x^{3/2}}{3})+c_2\mathrm{BesselK}(\frac{1}{3},\frac{2x^{3/2}}{3})}$$

Letting $c_2=1$ the above becomes

$$u=\frac{\sqrt{x}(\mathrm{BesselI}(-\frac{2}{3},\frac{2x^{3/2}}{3})c_1-\mathrm{BesselK}(\frac{2}{3},\frac{2x^{3/2}}{3}))}{c_1\mathrm{BesselI}(\frac{1}{3},\frac{2x^{3/2}}{3})+\mathrm{BesselK}(\frac{1}{3},\frac{2x^{3/2}}{3})}$$

In summary, these are the solution found for $y\left(x\right)$

$$\begin{array}{rl}u& =\frac{\sqrt{x}(\mathrm{BesselI}(-\frac{2}{3},\frac{2x^{3/2}}{3})c_1-\mathrm{BesselK}(\frac{2}{3},\frac{2x^{3/2}}{3}))}{c_1\mathrm{BesselI}(\frac{1}{3},\frac{2x^{3/2}}{3})+\mathrm{BesselK}(\frac{1}{3},\frac{2x^{3/2}}{3})}\end{array}$$

For solution $u=\frac{\sqrt{x}(\mathrm{BesselI}(-\frac{2}{3},\frac{2x^{3/2}}{3})c_1-\mathrm{BesselK}(\frac{2}{3},\frac{2x^{3/2}}{3}))}{c_1\mathrm{BesselI}(\frac{1}{3},\frac{2x^{3/2}}{3})+\mathrm{BesselK}(\frac{1}{3},\frac{2x^{3/2}}{3})}$, since $u=y^{\prime }\left(x\right)$ then we now have a new first order ode to solve which is

$$\begin{array}{r}{y}^{\prime }\left(x\right)=\frac{\sqrt{x}(\mathrm{BesselI}(-\frac{2}{3},\frac{2x^{3/2}}{3})c_1-\mathrm{BesselK}(\frac{2}{3},\frac{2x^{3/2}}{3}))}{c_1\mathrm{BesselI}(\frac{1}{3},\frac{2x^{3/2}}{3})+\mathrm{BesselK}(\frac{1}{3},\frac{2x^{3/2}}{3})}\end{array}$$

Since the ode has the form $y^{\prime }=f(x)$, then we only need to integrate $f(x)$.

$$\begin{array}{rl}\int dy& =\int -\frac{\sqrt{x}(-\mathrm{BesselI}(-\frac{2}{3},\frac{2x^{3/2}}{3})c_1+\mathrm{BesselK}(\frac{2}{3},\frac{2x^{3/2}}{3}))}{c_1\mathrm{BesselI}(\frac{1}{3},\frac{2x^{3/2}}{3})+\mathrm{BesselK}(\frac{1}{3},\frac{2x^{3/2}}{3})}dx\\ y& =\int -\frac{\sqrt{x}(-\mathrm{BesselI}(-\frac{2}{3},\frac{2x^{3/2}}{3})c_1+\mathrm{BesselK}(\frac{2}{3},\frac{2x^{3/2}}{3}))}{c_1\mathrm{BesselI}(\frac{1}{3},\frac{2x^{3/2}}{3})+\mathrm{BesselK}(\frac{1}{3},\frac{2x^{3/2}}{3})}dx+c_3\end{array}$$

$$\begin{array}{rl}y& =\int -\frac{\sqrt{x}(-\mathrm{BesselI}(-\frac{2}{3},\frac{2x^{3/2}}{3})c_1+\mathrm{BesselK}(\frac{2}{3},\frac{2x^{3/2}}{3}))}{c_1\mathrm{BesselI}(\frac{1}{3},\frac{2x^{3/2}}{3})+\mathrm{BesselK}(\frac{1}{3},\frac{2x^{3/2}}{3})}dx+c_3\end{array}$$

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

$$\begin{array}{rl}y& =\int -\frac{\sqrt{x}(-\mathrm{BesselI}(-\frac{2}{3},\frac{2x^{3/2}}{3})c_1+\mathrm{BesselK}(\frac{2}{3},\frac{2x^{3/2}}{3}))}{c_1\mathrm{BesselI}(\frac{1}{3},\frac{2x^{3/2}}{3})+\mathrm{BesselK}(\frac{1}{3},\frac{2x^{3/2}}{3})}dx+c_3\end{array}$$

Summary of solutions found

$$\begin{array}{rl}y& =\int -\frac{\sqrt{x}(-\mathrm{BesselI}(-\frac{2}{3},\frac{2x^{3/2}}{3})c_1+\mathrm{BesselK}(\frac{2}{3},\frac{2x^{3/2}}{3}))}{c_1\mathrm{BesselI}(\frac{1}{3},\frac{2x^{3/2}}{3})+\mathrm{BesselK}(\frac{1}{3},\frac{2x^{3/2}}{3})}dx+c_3\end{array}$$

✓ Maple. Time used: 0.018 (sec). Leaf size: 18

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

$$y=\ln \left(\pi \right)+\ln (c_1\mathrm{AiryAi}\left(x\right)-c_2\mathrm{AiryBi}\left(x\right))$$

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
 

✓ Mathematica. Time used: 0.202 (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]
 

$$y(x)\to \log (x-c_1)+c_2$$

✗ 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