Internal problem ID [13482]
Internal file name [OUTPUT/12654_Friday_February_16_2024_12_04_42_AM_96321983/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 13. Higher order equations: Extending first order concepts. Additional exercises
page 259
Problem number: 13.2 (e).
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]]
\[ \boxed {x y^{\prime \prime }-{y^{\prime }}^{2}=6 x^{5}} \]
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 ) x -p \left (x \right )^{2}-6 x^{5} = 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)\\ &= \frac {6 x^{5}+p^{2}}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ p' = 6 x^{4}+\frac {p^{2}}{x} \] With Riccati ODE standard form \[ p' = f_0(x)+ f_1(x)p+f_2(x)p^{2} \] Shows that \(f_0(x)=6 x^{4}\), \(f_1(x)=0\) and \(f_2(x)=\frac {1}{x}\). Let \begin {align*} p &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u}{x}} \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' &=-\frac {1}{x^{2}}\\ f_1 f_2 &=0\\ f_2^2 f_0 &=6 x^{2} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {u^{\prime \prime }\left (x \right )}{x}+\frac {u^{\prime }\left (x \right )}{x^{2}}+6 x^{2} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} \operatorname {BesselJ}\left (0, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )+c_{2} \operatorname {BesselY}\left (0, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \sqrt {6}\, x^{\frac {3}{2}} \left (-c_{1} \operatorname {BesselJ}\left (1, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )-c_{2} \operatorname {BesselY}\left (1, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )\right ) \] Using the above in (1) gives the solution \[ p \left (x \right ) = -\frac {\sqrt {6}\, x^{\frac {5}{2}} \left (-c_{1} \operatorname {BesselJ}\left (1, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )-c_{2} \operatorname {BesselY}\left (1, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )\right )}{c_{1} \operatorname {BesselJ}\left (0, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )+c_{2} \operatorname {BesselY}\left (0, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ p \left (x \right ) = \frac {\sqrt {6}\, x^{\frac {5}{2}} \left (c_{3} \operatorname {BesselJ}\left (1, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )+\operatorname {BesselY}\left (1, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )\right )}{c_{3} \operatorname {BesselJ}\left (0, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )+\operatorname {BesselY}\left (0, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )} \] Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = \frac {\sqrt {6}\, x^{\frac {5}{2}} \left (c_{3} \operatorname {BesselJ}\left (1, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )+\operatorname {BesselY}\left (1, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )\right )}{c_{3} \operatorname {BesselJ}\left (0, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )+\operatorname {BesselY}\left (0, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )} \end {align*}
Integrating both sides gives \begin {align*} y = \int \frac {\sqrt {6}\, x^{\frac {5}{2}} \left (c_{3} \operatorname {BesselJ}\left (1, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )+\operatorname {BesselY}\left (1, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )\right )}{c_{3} \operatorname {BesselJ}\left (0, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )+\operatorname {BesselY}\left (0, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )}d x +c_{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \int \frac {\sqrt {6}\, x^{\frac {5}{2}} \left (c_{3} \operatorname {BesselJ}\left (1, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )+\operatorname {BesselY}\left (1, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )\right )}{c_{3} \operatorname {BesselJ}\left (0, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )+\operatorname {BesselY}\left (0, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )}d x +c_{4} \\ \end{align*}
Verification of solutions
\[ y = \int \frac {\sqrt {6}\, x^{\frac {5}{2}} \left (c_{3} \operatorname {BesselJ}\left (1, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )+\operatorname {BesselY}\left (1, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )\right )}{c_{3} \operatorname {BesselJ}\left (0, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )+\operatorname {BesselY}\left (0, \frac {2 \sqrt {6}\, x^{\frac {5}{2}}}{5}\right )}d x +c_{4} \] 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 differential order: 2; missing variables `, `-> Computing symmetries using: way = 3 `, `-> Computing symmetries using: way = exp_sym -> Calling odsolve with the ODE`, diff(_b(_a), _a) = (6*_a^5+_b(_a)^2)/_a, _b(_a)` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying Chini differential order: 1; looking for linear symmetries trying exact Looking for potential symmetries trying Riccati trying Riccati sub-methods: trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = -(diff(y(x), x))/x-6*y(x)*x^3, y(x)` *** Sublevel 3 *** Methods for second order ODEs: --- Trying classification methods --- 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 <- Riccati to 2nd Order successful <- differential order: 2; canonical coordinates successful <- differential order 2; missing variables successful`
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 64
dsolve(x*diff(y(x),x$2)-diff(y(x),x)^2=6*x^5,y(x), singsol=all)
\[ y \left (x \right ) = \sqrt {6}\, \left (\int \frac {x^{\frac {5}{2}} \left (\operatorname {BesselY}\left (1, \frac {2 x^{\frac {5}{2}} \sqrt {6}}{5}\right ) c_{1} +\operatorname {BesselJ}\left (1, \frac {2 x^{\frac {5}{2}} \sqrt {6}}{5}\right )\right )}{c_{1} \operatorname {BesselY}\left (0, \frac {2 x^{\frac {5}{2}} \sqrt {6}}{5}\right )+\operatorname {BesselJ}\left (0, \frac {2 x^{\frac {5}{2}} \sqrt {6}}{5}\right )}d x \right )+c_{2} \]
✓ Solution by Mathematica
Time used: 60.384 (sec). Leaf size: 109
DSolve[x*y''[x]-y'[x]^2==6*x^5,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \int _1^x\frac {\sqrt {6} \left (2 \operatorname {BesselY}\left (1,\frac {2}{5} \sqrt {6} K[1]^{5/2}\right )+\operatorname {BesselJ}\left (1,\frac {2}{5} \sqrt {6} K[1]^{5/2}\right ) c_1\right ) K[1]^{5/2}}{2 \operatorname {BesselY}\left (0,\frac {2}{5} \sqrt {6} K[1]^{5/2}\right )+\operatorname {BesselJ}\left (0,\frac {2}{5} \sqrt {6} K[1]^{5/2}\right ) c_1}dK[1]+c_2 \]