Internal problem ID [7409]
Internal file name [OUTPUT/6376_Sunday_June_05_2022_04_42_16_PM_97106534/index.tex]
Book: Second order enumerated odes
Section: section 1
Problem number: 20.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_y], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {y^{\prime \prime }+{y^{\prime }}^{2}=x} \]
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*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} \operatorname {AiryAi}\left (x \right )+c_{2} \operatorname {AiryBi}\left (x \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = c_{1} \operatorname {AiryAi}\left (1, x\right )+c_{2} \operatorname {AiryBi}\left (1, x\right ) \] Using the above in (1) gives the solution \[ p \left (x \right ) = \frac {c_{1} \operatorname {AiryAi}\left (1, x\right )+c_{2} \operatorname {AiryBi}\left (1, x\right )}{c_{1} \operatorname {AiryAi}\left (x \right )+c_{2} \operatorname {AiryBi}\left (x \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ p \left (x \right ) = \frac {c_{3} \operatorname {AiryAi}\left (1, x\right )+\operatorname {AiryBi}\left (1, x\right )}{c_{3} \operatorname {AiryAi}\left (x \right )+\operatorname {AiryBi}\left (x \right )} \] Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = \frac {c_{3} \operatorname {AiryAi}\left (1, x\right )+\operatorname {AiryBi}\left (1, x\right )}{c_{3} \operatorname {AiryAi}\left (x \right )+\operatorname {AiryBi}\left (x \right )} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { \frac {c_{3} \operatorname {AiryAi}\left (1, x\right )+\operatorname {AiryBi}\left (1, x\right )}{c_{3} \operatorname {AiryAi}\left (x \right )+\operatorname {AiryBi}\left (x \right )}\,\mathop {\mathrm {d}x}}\\ &= \ln \left (c_{3} \operatorname {AiryAi}\left (x \right )+\operatorname {AiryBi}\left (x \right )\right )+c_{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \ln \left (c_{3} \operatorname {AiryAi}\left (x \right )+\operatorname {AiryBi}\left (x \right )\right )+c_{4} \\ \end{align*}
Verification of solutions
\[ y = \ln \left (c_{3} \operatorname {AiryAi}\left (x \right )+\operatorname {AiryBi}\left (x \right )\right )+c_{4} \] Verified OK.
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`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 18
dsolve(diff(y(x),x$2)+diff(y(x),x)^2=x,y(x), singsol=all)
\[ y \left (x \right ) = \ln \left (\pi \right )+\ln \left (c_{1} \operatorname {AiryAi}\left (x \right )-c_{2} \operatorname {AiryBi}\left (x \right )\right ) \]
✓ Solution by Mathematica
Time used: 0.114 (sec). Leaf size: 15
DSolve[y''[x]+(y'[x])^2==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \log (x-c_1)+c_2 \]