Internal problem ID [11315]
Internal file name [OUTPUT/10301_Tuesday_December_27_2022_04_05_59_AM_47236562/index.tex]
Book: An elementary treatise on differential equations by Abraham Cohen. DC heath publishers.
1906
Section: Chapter IX, Miscellaneous methods for solving equations of higher order than first.
Article 57. Dependent variable absent. Page 132
Problem number: Ex 5.
ODE order: 2.
ODE degree: 2.
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]]
\[ \boxed {\left (y^{\prime }-x y^{\prime \prime }\right )^{2}-{y^{\prime \prime }}^{2}=1} \]
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*} \left (x^{2} p^{\prime }\left (x \right )-2 p \left (x \right ) x -p^{\prime }\left (x \right )\right ) p^{\prime }\left (x \right )+p \left (x \right )^{2}-1 = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. This is Clairaut ODE. It has the form \[ p=x p^{\prime }\left (x \right )+g\left (p^{\prime }\left (x \right )\right ) \] Where \(g\) is function of \(p'(x)\). Let \(p=p^{\prime }\left (x \right )\) the ode becomes \begin {align*} \left (x^{2} p -2 p x -p \right ) p +p^{2} = 1 \end {align*}
Solving for \(p \left (x \right )\) from the above results in \begin {align*} p \left (x \right ) &= p x +\sqrt {p^{2}+1}\tag {1A} \\ p \left (x \right ) &= p x -\sqrt {p^{2}+1}\tag {2A} \end {align*}
Each of the above ode’s is a Clairaut ode which is now solved. Solving ode 1A We start by replacing \(p^{\prime }\left (x \right )\) by \(p\) which gives \begin {align*} p \left (x \right )&=p x +\sqrt {p^{2}+1}\\ &=p x +\sqrt {p^{2}+1} \end {align*}
Writing the ode as \begin {align*} p \left (x \right )&= p x +g \left (p \right ) \end {align*}
We now write \(g\equiv g\left ( p\right ) \) to make notation simpler but we should always remember that \(g\) is function of \(p\) which in turn is function of \(x\). Hence the above becomes \begin {align*} p = p x +g\tag {1} \end {align*}
Then we see that \begin {align*} g&=\sqrt {p^{2}+1} \end {align*}
Taking derivative of (1) w.r.t. \(x\) gives \begin {align*} p &=\frac {d}{dx}\left (x p+g\right ) \\ p & =\left ( p+x\frac {dp}{dx}\right ) +\left ( g' \frac {dp}{dx}\right ) \\ p & =p+\left ( x+g'\right ) \frac {dp}{dx}\\ 0 & =\left ( x+g'\right ) \frac {dp}{dx} \end {align*}
Where \(g'\) is derivative of \(g\left ( p\right ) \) w.r.t. \(p\). The general solution is given by \begin {align*} \frac {dp}{dx} & =0\\ p &=c_{1} \end {align*}
Substituting this in (1) gives the general solution as \begin {align*} p \left (x \right ) = c_{1} x +\sqrt {c_{1}^{2}+1} \end {align*}
The singular solution is found from solving for \(p\) from \begin {align*} x+g'\left ( p\right ) &=0 \end {align*}
And substituting the result back in (1). Since we found above that \(g=\sqrt {p^{2}+1}\), then the above equation becomes \begin {align*} x+g'\left ( p\right ) &= x +\frac {p}{\sqrt {p^{2}+1}}\\ &= 0 \end {align*}
Solving the above for \(p\) results in \begin {align*} p_{1} &=-x \sqrt {-\frac {1}{x^{2}-1}} \end {align*}
Substituting the above back in (1) results in \begin {align*} p \left (x \right )_{1} &=\left (-x^{2}+1\right ) \sqrt {-\frac {1}{x^{2}-1}} \end {align*}
Solving ode 2A We start by replacing \(p^{\prime }\left (x \right )\) by \(p\) which gives \begin {align*} p \left (x \right )&=p x -\sqrt {p^{2}+1}\\ &=p x -\sqrt {p^{2}+1} \end {align*}
Writing the ode as \begin {align*} p \left (x \right )&= p x +g \left (p \right ) \end {align*}
We now write \(g\equiv g\left ( p\right ) \) to make notation simpler but we should always remember that \(g\) is function of \(p\) which in turn is function of \(x\). Hence the above becomes \begin {align*} p = p x +g\tag {1} \end {align*}
Then we see that \begin {align*} g&=-\sqrt {p^{2}+1} \end {align*}
Taking derivative of (1) w.r.t. \(x\) gives \begin {align*} p &=\frac {d}{dx}\left (x p+g\right ) \\ p & =\left ( p+x\frac {dp}{dx}\right ) +\left ( g' \frac {dp}{dx}\right ) \\ p & =p+\left ( x+g'\right ) \frac {dp}{dx}\\ 0 & =\left ( x+g'\right ) \frac {dp}{dx} \end {align*}
Where \(g'\) is derivative of \(g\left ( p\right ) \) w.r.t. \(p\). The general solution is given by \begin {align*} \frac {dp}{dx} & =0\\ p &=c_{1} \end {align*}
Substituting this in (1) gives the general solution as \begin {align*} p \left (x \right ) = c_{2} x -\sqrt {c_{2}^{2}+1} \end {align*}
The singular solution is found from solving for \(p\) from \begin {align*} x+g'\left ( p\right ) &=0 \end {align*}
And substituting the result back in (1). Since we found above that \(g=-\sqrt {p^{2}+1}\), then the above equation becomes \begin {align*} x+g'\left ( p\right ) &= x -\frac {p}{\sqrt {p^{2}+1}}\\ &= 0 \end {align*}
Solving the above for \(p\) results in \begin {align*} p_{1} &=x \sqrt {-\frac {1}{x^{2}-1}} \end {align*}
Substituting the above back in (1) results in \begin {align*} p \left (x \right )_{1} &=\sqrt {-\frac {1}{x^{2}-1}}\, \left (x^{2}-1\right ) \end {align*}
For solution (1) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = c_{1} x +\sqrt {c_{1}^{2}+1} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { c_{1} x +\sqrt {c_{1}^{2}+1}\,\mathop {\mathrm {d}x}}\\ &= \frac {x \left (c_{1} x +2 \sqrt {c_{1}^{2}+1}\right )}{2}+c_{3} \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = \left (-x^{2}+1\right ) \sqrt {-\frac {1}{x^{2}-1}} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -\sqrt {-\frac {1}{x^{2}-1}}\, \left (x^{2}-1\right )\,\mathop {\mathrm {d}x}}\\ &= -\frac {\sqrt {-\frac {1}{x^{2}-1}}\, \sqrt {x^{2}-1}\, \left (x \sqrt {x^{2}-1}-\ln \left (x +\sqrt {x^{2}-1}\right )\right )}{2}+c_{4} \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = c_{2} x -\sqrt {c_{2}^{2}+1} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { c_{2} x -\sqrt {c_{2}^{2}+1}\,\mathop {\mathrm {d}x}}\\ &= \frac {c_{2} x^{2}}{2}-\sqrt {c_{2}^{2}+1}\, x +c_{5} \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = \sqrt {-\frac {1}{x^{2}-1}}\, \left (x^{2}-1\right ) \end {align*}
Integrating both sides gives \begin {align*} y &= \int { \sqrt {-\frac {1}{x^{2}-1}}\, \left (x^{2}-1\right )\,\mathop {\mathrm {d}x}}\\ &= \frac {\sqrt {-\frac {1}{x^{2}-1}}\, \sqrt {x^{2}-1}\, \left (x \sqrt {x^{2}-1}-\ln \left (x +\sqrt {x^{2}-1}\right )\right )}{2}+c_{6} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x \left (c_{1} x +2 \sqrt {c_{1}^{2}+1}\right )}{2}+c_{3} \\ \tag{2} y &= -\frac {\sqrt {-\frac {1}{x^{2}-1}}\, \sqrt {x^{2}-1}\, \left (x \sqrt {x^{2}-1}-\ln \left (x +\sqrt {x^{2}-1}\right )\right )}{2}+c_{4} \\ \tag{3} y &= \frac {c_{2} x^{2}}{2}-\sqrt {c_{2}^{2}+1}\, x +c_{5} \\ \tag{4} y &= \frac {\sqrt {-\frac {1}{x^{2}-1}}\, \sqrt {x^{2}-1}\, \left (x \sqrt {x^{2}-1}-\ln \left (x +\sqrt {x^{2}-1}\right )\right )}{2}+c_{6} \\ \end{align*}
Verification of solutions
\[ y = \frac {x \left (c_{1} x +2 \sqrt {c_{1}^{2}+1}\right )}{2}+c_{3} \] Verified OK.
\[ y = -\frac {\sqrt {-\frac {1}{x^{2}-1}}\, \sqrt {x^{2}-1}\, \left (x \sqrt {x^{2}-1}-\ln \left (x +\sqrt {x^{2}-1}\right )\right )}{2}+c_{4} \] Verified OK.
\[ y = \frac {c_{2} x^{2}}{2}-\sqrt {c_{2}^{2}+1}\, x +c_{5} \] Verified OK.
\[ y = \frac {\sqrt {-\frac {1}{x^{2}-1}}\, \sqrt {x^{2}-1}\, \left (x \sqrt {x^{2}-1}-\ln \left (x +\sqrt {x^{2}-1}\right )\right )}{2}+c_{6} \] Verified OK.
Maple trace
`Methods for second order ODEs: *** Sublevel 2 *** Methods for second order ODEs: Successful isolation of d^2y/dx^2: 2 solutions were found. Trying to solve each resulting ODE. *** Sublevel 3 *** 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 <- 2nd order ODE linearizable_by_differentiation successful ------------------- * Tackling next ODE. *** Sublevel 3 *** 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 <- 2nd order ODE linearizable_by_differentiation successful -> Calling odsolve with the ODE`, (diff(y(x), x))^2 = -x^2+1, y(x), singsol = none` *** Sublevel 2 *** Methods for first order ODEs: -> Solving 1st order ODE of high degree, 1st attempt trying 1st order WeierstrassP solution for high degree ODE trying 1st order WeierstrassPPrime solution for high degree ODE trying 1st order JacobiSN solution for high degree ODE trying 1st order ODE linearizable_by_differentiation trying differential order: 1; missing variables <- differential order: 1; missing y(x) successful`
✓ Solution by Maple
Time used: 0.094 (sec). Leaf size: 63
dsolve((diff(y(x),x)-x*diff(y(x),x$2))^2=1+diff(y(x),x$2)^2,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {x \sqrt {-x^{2}+1}}{2}+\frac {\arcsin \left (x \right )}{2}+c_{1} \\ y \left (x \right ) &= -\frac {x \sqrt {-x^{2}+1}}{2}-\frac {\arcsin \left (x \right )}{2}+c_{1} \\ y \left (x \right ) &= \frac {\sqrt {c_{1}^{2}-1}\, x^{2}}{2}+c_{1} x +c_{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.215 (sec). Leaf size: 58
DSolve[(y'[x]-x*y''[x])^2==1+(y''[x])^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {c_1 x^2}{2}-\sqrt {1+c_1{}^2} x+c_2 \\ y(x)\to \frac {c_1 x^2}{2}+\sqrt {1+c_1{}^2} x+c_2 \\ \end{align*}