problem 19

Internal problem ID [7408]
Internal ﬁle name [OUTPUT/6375_Sunday_June_05_2022_04_42_10_PM_194809/index.tex]

Book: Second order enumerated odes
Section: section 1
Problem number: 19.
ODE order: 2.
ODE degree: 2.

1.19.1 Solving as second order ode missing y ode

This is second order ode with missing dependent variable \(y\). Let \begin {align*} p(x) &= y^{\prime } \end {align*}

Hence the ode becomes \begin {align*} {p^{\prime }\left (x \right )}^{2}+p \left (x \right )-x = 0 \end {align*}

Which is now solve for \(p(x)\) as ﬁrst order ode. The ode has the form \begin {align*} (p')^{\frac {n}{m}} &= a x + b p + c \tag {1} \end {align*}

Where \(n=2, m=1, a=1 , b=-1 , c=0\). Hence the ode is \begin {align*} (p')^{2} &= -p \left (x \right )+x \end {align*}

Substituting the above in (1) gives \begin {align*} \left (\frac {u'-a}{b} \right )^{\frac {n}{m}} &= u \\ \left (\frac {u'-a}{b} \right )^n &= u^m \end {align*}

Plugging in the above the values for \(n,m,a,b,c\) gives \begin {align*} \left (-u^{\prime }\left (x \right )+1\right )^{2} &= u \end {align*}

Therefore the solutions are \begin {align*} -u^{\prime }\left (x \right )+1&=\sqrt {u}\\ -u^{\prime }\left (x \right )+1&=-\sqrt {u} \end {align*}

Rewriting the above gives \begin {align*} u^{\prime }\left (x \right ) &= -\sqrt {u}+1\\ u^{\prime }\left (x \right ) &= \sqrt {u}+1 \end {align*}

Each of the above is a separable ODE in \(u \left (x \right )\). This results in \begin {align*} \frac {du}{-\sqrt {u}+1} &= dx\\ \frac {du}{\sqrt {u}+1} &= dx \end {align*}

Integrating each of the above solutions gives \begin {align*} \int \frac {du}{-\sqrt {u}+1} &= x + c_{1}\\ \int \frac {du}{\sqrt {u}+1} &= x + c_{1} \end {align*}

But since \begin {align*} u &= a x + b p + c \\ &= -p \left (x \right )+x \end {align*}

Then the solutions can be written as \begin {align*} \int ^{-p \left (x \right )+x} \frac {1}{-\sqrt {\tau }+1} \, d\tau &= x + c_{1}\\ \int ^{-p \left (x \right )+x} \frac {1}{\sqrt {\tau }+1} \, d\tau &= x + c_{1} \end {align*}

For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new ﬁrst order ode to solve which is \begin {align*} \int _{}^{-y^{\prime }+x}\frac {1}{-\sqrt {\tau }+1}d \tau = x +c_{1} \end {align*}

Integrating both sides gives \begin {align*} y = \int \operatorname {RootOf}\left (-\left (\int _{}^{-\textit {\_Z} +x}-\frac {1}{\sqrt {\tau }-1}d \tau \right )+x +c_{1} \right )d x +c_{2} \end {align*}

For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new ﬁrst order ode to solve which is \begin {align*} \int _{}^{-y^{\prime }+x}\frac {1}{\sqrt {\tau }+1}d \tau = x +c_{1} \end {align*}

Integrating both sides gives \begin {align*} y = \int \operatorname {RootOf}\left (-\left (\int _{}^{-\textit {\_Z} +x}\frac {1}{\sqrt {\tau }+1}d \tau \right )+x +c_{1} \right )d x +c_{3} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \int \operatorname {RootOf}\left (-\left (\int _{}^{-\textit {\_Z} +x}-\frac {1}{\sqrt {\tau }-1}d \tau \right )+x +c_{1} \right )d x +c_{2} \\ \tag{2} y &= \int \operatorname {RootOf}\left (-\left (\int _{}^{-\textit {\_Z} +x}\frac {1}{\sqrt {\tau }+1}d \tau \right )+x +c_{1} \right )d x +c_{3} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \int \operatorname {RootOf}\left (-\left (\int _{}^{-\textit {\_Z} +x}-\frac {1}{\sqrt {\tau }-1}d \tau \right )+x +c_{1} \right )d x +c_{2} \] Veriﬁed OK.

\[ y = \int \operatorname {RootOf}\left (-\left (\int _{}^{-\textit {\_Z} +x}\frac {1}{\sqrt {\tau }+1}d \tau \right )+x +c_{1} \right )d x +c_{3} \] Veriﬁed OK.

Maple trace

`Methods for second order ODEs: 
Successful isolation of d^2y/dx^2: 2 solutions were found. Trying to solve each resulting ODE. 
   *** Sublevel 2 *** 
   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 
   -> Calling odsolve with the ODE`, diff(_b(_a), _a) = (-_b(_a)+_a)^(1/2), _b(_a), HINT = [[1, 1]]`      *** Sublevel 3 *** 
      symmetry methods on request 
   `, `1st order, trying reduction of order with given symmetries:`[1, 1]

\begin{align*} y \left (x \right ) &= \int \left (-{\mathrm e}^{2 \operatorname {RootOf}\left (\textit {\_Z} -x -2 \,{\mathrm e}^{\textit {\_Z}}+2+c_{1} -\ln \left ({\mathrm e}^{\textit {\_Z}} \left ({\mathrm e}^{\textit {\_Z}}-2\right )^{2}\right )\right )}+2 \,{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z} -x -2 \,{\mathrm e}^{\textit {\_Z}}+2+c_{1} -\ln \left ({\mathrm e}^{\textit {\_Z}} \left ({\mathrm e}^{\textit {\_Z}}-2\right )^{2}\right )\right )}+x \right )d x -x +c_{2} \\ y \left (x \right ) &= \frac {2 \operatorname {LambertW}\left (-c_{1} {\mathrm e}^{-\frac {x}{2}-1}\right )^{3}}{3}+3 \operatorname {LambertW}\left (-c_{1} {\mathrm e}^{-\frac {x}{2}-1}\right )^{2}+4 \operatorname {LambertW}\left (-c_{1} {\mathrm e}^{-\frac {x}{2}-1}\right )+\frac {x^{2}}{2}-x +c_{2} \\ \end{align*}

\begin{align*} y(x)\to \frac {2}{3} W\left (e^{-\frac {x}{2}-1-\frac {c_1}{2}}\right ){}^3+3 W\left (e^{-\frac {x}{2}-1-\frac {c_1}{2}}\right ){}^2+4 W\left (e^{-\frac {x}{2}-1-\frac {c_1}{2}}\right )+\frac {x^2}{2}-x+c_2 \\ y(x)\to \frac {2}{3} W\left (-e^{\frac {1}{2} (-x-2+c_1)}\right ){}^3+3 W\left (-e^{\frac {1}{2} (-x-2+c_1)}\right ){}^2+4 W\left (-e^{\frac {1}{2} (-x-2+c_1)}\right )+\frac {x^2}{2}-x+c_2 \\ y(x)\to \frac {x^2}{2}-x+c_2 \\ y(x)\to \frac {2}{3} W\left (-e^{-\frac {x}{2}-1}\right )^3+3 W\left (-e^{-\frac {x}{2}-1}\right )^2+4 W\left (-e^{-\frac {x}{2}-1}\right )+\frac {x^2}{2}-x+c_2 \\ \end{align*}

1.19 problem 19

1.19.1 Solving as second order ode missing y ode