Internal problem ID [10022]
Internal file name [OUTPUT/8969_Monday_June_06_2022_06_05_37_AM_12026490/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1701 (book 6.110).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]
\[ \boxed {y^{\prime \prime } y-{y^{\prime }}^{2}=-1} \]
This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using \begin {align*} y' &= p(y) \end {align*}
Then \begin {align*} y'' &= \frac {dp}{dx}\\ &= \frac {dy}{dx} \frac {dp}{dy}\\ &= p \frac {dp}{dy} \end {align*}
Hence the ode becomes \begin {align*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) y -p \left (y \right )^{2} = -1 \end {align*}
Which is now solved as first order ode for \(p(y)\). In canonical form the ODE is \begin {align*} p' &= F(y,p)\\ &= f( y) g(p)\\ &= \frac {p^{2}-1}{p y} \end {align*}
Where \(f(y)=\frac {1}{y}\) and \(g(p)=\frac {p^{2}-1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {p^{2}-1}{p}} \,dp &= \frac {1}{y} \,d y \\ \int { \frac {1}{\frac {p^{2}-1}{p}} \,dp} &= \int {\frac {1}{y} \,d y} \\ \frac {\ln \left (p^{2}-1\right )}{2}&=\ln \left (y \right )+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {p^{2}-1} &= {\mathrm e}^{\ln \left (y \right )+c_{1}} \end {align*}
Which simplifies to \begin {align*} \sqrt {p^{2}-1} &= c_{2} y \end {align*}
Which simplifies to \[ \sqrt {p \left (y \right )^{2}-1} = c_{2} y \,{\mathrm e}^{c_{1}} \] The solution is \[ \sqrt {p \left (y \right )^{2}-1} = c_{2} y \,{\mathrm e}^{c_{1}} \] For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} \sqrt {{y^{\prime }}^{2}-1} = c_{2} y \,{\mathrm e}^{c_{1}} \end {align*}
Solving the given ode for \(y^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\sqrt {1+c_{2}^{2} y^{2} {\mathrm e}^{2 c_{1}}} \tag {1} \\ y^{\prime }&=-\sqrt {1+c_{2}^{2} y^{2} {\mathrm e}^{2 c_{1}}} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} \int \frac {1}{\sqrt {1+c_{2}^{2} y^{2} {\mathrm e}^{2 c_{1}}}}d y &= c_{3} +x\\ \frac {{\mathrm e}^{-c_{1}} \operatorname {arcsinh}\left (c_{2} y \,{\mathrm e}^{c_{1}}\right )}{c_{2}}&=c_{3} +x \end {align*}
Solving for \(y\) gives these solutions \begin {align*} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} \int -\frac {1}{\sqrt {1+c_{2}^{2} y^{2} {\mathrm e}^{2 c_{1}}}}d y &= x +c_{4}\\ -\frac {{\mathrm e}^{-c_{1}} \operatorname {arcsinh}\left (c_{2} y \,{\mathrm e}^{c_{1}}\right )}{c_{2}}&=x +c_{4} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\sinh \left (c_{2} \left (c_{3} +x \right ) {\mathrm e}^{c_{1}}\right ) {\mathrm e}^{-c_{1}}}{c_{2}} \\ \tag{2} y &= -\frac {\sinh \left (c_{2} \left (x +c_{4} \right ) {\mathrm e}^{c_{1}}\right ) {\mathrm e}^{-c_{1}}}{c_{2}} \\ \end{align*}
Verification of solutions
\[ y = \frac {\sinh \left (c_{2} \left (c_{3} +x \right ) {\mathrm e}^{c_{1}}\right ) {\mathrm e}^{-c_{1}}}{c_{2}} \] Verified OK.
\[ y = -\frac {\sinh \left (c_{2} \left (x +c_{4} \right ) {\mathrm e}^{c_{1}}\right ) {\mathrm e}^{-c_{1}}}{c_{2}} \] 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 -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)-(_b(_a)^2-1)/_a = 0, _b(_a), HINT = [[_a, 0]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, 0]
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 59
dsolve(diff(diff(y(x),x),x)*y(x)-diff(y(x),x)^2+1=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {c_{1} \left (-{\mathrm e}^{\frac {c_{2} +x}{c_{1}}}+{\mathrm e}^{\frac {-c_{2} -x}{c_{1}}}\right )}{2} \\ y \left (x \right ) &= -\frac {c_{1} \left (-{\mathrm e}^{\frac {c_{2} +x}{c_{1}}}+{\mathrm e}^{\frac {-c_{2} -x}{c_{1}}}\right )}{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 60.331 (sec). Leaf size: 85
DSolve[y''[x]*y[x]-y'[x]^2+1==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {i e^{-c_1} \tanh \left (e^{c_1} (x+c_2)\right )}{\sqrt {-\text {sech}^2\left (e^{c_1} (x+c_2)\right )}} \\ y(x)\to \frac {i e^{-c_1} \tanh \left (e^{c_1} (x+c_2)\right )}{\sqrt {-\text {sech}^2\left (e^{c_1} (x+c_2)\right )}} \\ \end{align*}