2.1.20 problem 20

Internal problem ID [8503]
Book : Second order enumerated odes
Section : section 1
Problem number : 20
Date solved : Sunday, November 10, 2024 at 03:55:14 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: 1.155 (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. In canonical form the ODE is

\begin{align*} p' &= F(x,p)\\ &= -p^{2}+x \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ p' = -p^{2}+x \]

With Riccati ODE standard form

\[ p' = f_0(x)+ f_1(x)p+f_2(x)p^{2} \]

Shows that \(f_0(x)=x\), \(f_1(x)=0\) and \(f_2(x)=-1\). Let

\begin{align*} p &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-u} \tag {1} \end{align*}

Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for \(u(x)\) which is

\begin{align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end{align*}

But

\begin{align*} f_2' &=0\\ f_1 f_2 &=0\\ f_2^2 f_0 &=x \end{align*}

Substituting the above terms back in equation (2) gives

\begin{align*} -u^{\prime \prime }\left (x \right )+x u \left (x \right ) = 0 \end{align*}

This is Airy ODE. It has the general form

\[ a \frac {d^{2}u}{d x^{2}} + b \frac {d u}{d x} + c x u = F(x) \]

Where in this case

\begin{align*} a &= -1\\ b &= 0\\ c &= 1\\ F &= 0 \end{align*}

Therefore the solution to the homogeneous Airy ODE becomes

\[ u = c_1 \operatorname {AiryAi}\left (-x \left (-1\right )^{{1}/{3}}\right )+c_2 \operatorname {AiryBi}\left (-x \left (-1\right )^{{1}/{3}}\right ) \]

Will add steps showing solving for IC soon.

Taking derivative gives

\[ u^{\prime }\left (x \right ) = -c_1 \left (-1\right )^{{1}/{3}} \operatorname {AiryAi}\left (1, -x \left (-1\right )^{{1}/{3}}\right )-c_2 \left (-1\right )^{{1}/{3}} \operatorname {AiryBi}\left (1, -x \left (-1\right )^{{1}/{3}}\right ) \]

Doing change of constants, the solution becomes

\[ p = \frac {-c_3 \left (-1\right )^{{1}/{3}} \operatorname {AiryAi}\left (1, -x \left (-1\right )^{{1}/{3}}\right )-\left (-1\right )^{{1}/{3}} \operatorname {AiryBi}\left (1, -x \left (-1\right )^{{1}/{3}}\right )}{c_3 \operatorname {AiryAi}\left (-x \left (-1\right )^{{1}/{3}}\right )+\operatorname {AiryBi}\left (-x \left (-1\right )^{{1}/{3}}\right )} \]

For solution (1) found earlier, since \(p=y^{\prime }\left (x \right )\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime }\left (x \right ) = -\frac {\left (1+i \sqrt {3}\right ) \left (\operatorname {AiryAi}\left (1, -\frac {x \left (1+i \sqrt {3}\right )}{2}\right ) c_3 +\operatorname {AiryBi}\left (1, -\frac {x \left (1+i \sqrt {3}\right )}{2}\right )\right )}{2 c_3 \operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+2 \operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )} \end{align*}

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

\begin{align*} \int {dy} &= \int {-\frac {\left (1+i \sqrt {3}\right ) \left (\operatorname {AiryAi}\left (1, -\frac {x \left (1+i \sqrt {3}\right )}{2}\right ) c_3 +\operatorname {AiryBi}\left (1, -\frac {x \left (1+i \sqrt {3}\right )}{2}\right )\right )}{2 \left (c_3 \operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+\operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )\right )}\, dx}\\ y \left (x \right ) &= -\frac {\left (1+i \sqrt {3}\right ) \ln \left (c_3 \operatorname {AiryAi}\left (\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x \right )+\operatorname {AiryBi}\left (\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x \right )\right )}{2 \left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )} + c_4 \end{align*}

\begin{align*} y \left (x \right )&= \ln \left (c_3 \operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+\operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )\right )+c_4 \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y \left (x \right ) &= \ln \left (c_3 \operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+\operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )\right )+c_4 \\ \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]
 

\[ y(x)\to \log (x-c_1)+c_2 \]