Internal problem ID [9973]
Internal file name [OUTPUT/8921_Monday_June_06_2022_05_52_19_AM_5479834/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1652 (book 6.61).
ODE order: 2.
ODE degree: 2.
The type(s) of ODE detected by this program : "second_order_ode_missing_x", "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_x]]
\[ \boxed {y^{\prime \prime }-a \sqrt {{y^{\prime }}^{2}+1}=b} \]
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 )-a \sqrt {p \left (x \right )^{2}+1}-b = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. Integrating both sides gives \begin {align*} \int \frac {1}{a \sqrt {p^{2}+1}+b}d p &= \int {dx}\\ \int _{}^{p \left (x \right )}\frac {1}{a \sqrt {\textit {\_a}^{2}+1}+b}d \textit {\_a}&= 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 }}\frac {1}{a \sqrt {\textit {\_a}^{2}+1}+b}d \textit {\_a} = x +c_{1} \end {align*}
Integrating both sides gives \begin {align*} y = \int \operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {1}{a \sqrt {\textit {\_a}^{2}+1}+b}d \textit {\_a} \right )+x +c_{1} \right )d x +c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \int \operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {1}{a \sqrt {\textit {\_a}^{2}+1}+b}d \textit {\_a} \right )+x +c_{1} \right )d x +c_{2} \\ \end{align*}
Verification of solutions
\[ y = \int \operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {1}{a \sqrt {\textit {\_a}^{2}+1}+b}d \textit {\_a} \right )+x +c_{1} \right )d x +c_{2} \] Verified OK.
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 ) = a \sqrt {p \left (y \right )^{2}+1}+b \end {align*}
Which is now solved as first order ode for \(p(y)\). Integrating both sides gives \begin{align*} \int \frac {p}{a \sqrt {p^{2}+1}+b}d p &= \int d y \\ \frac {\sqrt {p \left (y \right )^{2}+1}}{a}+\frac {b \ln \left (a \sqrt {p \left (y \right )^{2}+1}-b \right )}{2 a^{2}}-\frac {b \ln \left (a \sqrt {p \left (y \right )^{2}+1}+b \right )}{2 a^{2}}-\frac {b \ln \left (a^{2} \left (p \left (y \right )^{2}+1\right )-b^{2}\right )}{2 a^{2}}&=y +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*} \frac {\sqrt {{y^{\prime }}^{2}+1}}{a}+\frac {b \ln \left (a \sqrt {{y^{\prime }}^{2}+1}-b \right )}{2 a^{2}}-\frac {b \ln \left (a \sqrt {{y^{\prime }}^{2}+1}+b \right )}{2 a^{2}}-\frac {b \ln \left (a^{2} \left ({y^{\prime }}^{2}+1\right )-b^{2}\right )}{2 a^{2}} = y+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 }&=\frac {\sqrt {\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right )^{2} b^{2}+2 \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}-a^{2}+b^{2}}}{a} \tag {1} \\ y^{\prime }&=-\frac {\sqrt {\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right )^{2} b^{2}+2 \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}-a^{2}+b^{2}}}{a} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin{align*} \int \frac {a}{\sqrt {\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {c_{1} a^{2}+a^{2} y +b}{b}}}{b}\right )^{2} b^{2}+2 \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {c_{1} a^{2}+a^{2} y +b}{b}}}{b}\right ) b^{2}-a^{2}+b^{2}}}d y &= \int d x \\ -\frac {b \left (\frac {\ln \left (\frac {\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}+b^{2}}{\sqrt {b^{2}}}+\sqrt {\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right )^{2} b^{2}+2 \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}-a^{2}+b^{2}}\right )}{\sqrt {b^{2}}}-\frac {\ln \left (\frac {-2 a^{2}+2 b^{2}+2 \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}+2 \sqrt {-a^{2}+b^{2}}\, \sqrt {\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right )^{2} b^{2}+2 \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}-a^{2}+b^{2}}}{\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right )}\right )}{\sqrt {-a^{2}+b^{2}}}\right )}{a}&=x +c_{2} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {a}{\sqrt {\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {c_{1} a^{2}+a^{2} y +b}{b}}}{b}\right )^{2} b^{2}+2 \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {c_{1} a^{2}+a^{2} y +b}{b}}}{b}\right ) b^{2}-a^{2}+b^{2}}}d y &= \int d x \\ \frac {b \left (\frac {\ln \left (\frac {\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}+b^{2}}{\sqrt {b^{2}}}+\sqrt {\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right )^{2} b^{2}+2 \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}-a^{2}+b^{2}}\right )}{\sqrt {b^{2}}}-\frac {\ln \left (\frac {-2 a^{2}+2 b^{2}+2 \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}+2 \sqrt {-a^{2}+b^{2}}\, \sqrt {\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right )^{2} b^{2}+2 \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}-a^{2}+b^{2}}}{\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right )}\right )}{\sqrt {-a^{2}+b^{2}}}\right )}{a}&=x +c_{3} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} -\frac {b \left (\frac {\ln \left (\frac {\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}+b^{2}}{\sqrt {b^{2}}}+\sqrt {\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right )^{2} b^{2}+2 \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}-a^{2}+b^{2}}\right )}{\sqrt {b^{2}}}-\frac {\ln \left (\frac {-2 a^{2}+2 b^{2}+2 \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}+2 \sqrt {-a^{2}+b^{2}}\, \sqrt {\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right )^{2} b^{2}+2 \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}-a^{2}+b^{2}}}{\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right )}\right )}{\sqrt {-a^{2}+b^{2}}}\right )}{a} &= x +c_{2} \\ \tag{2} \frac {b \left (\frac {\ln \left (\frac {\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}+b^{2}}{\sqrt {b^{2}}}+\sqrt {\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right )^{2} b^{2}+2 \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}-a^{2}+b^{2}}\right )}{\sqrt {b^{2}}}-\frac {\ln \left (\frac {-2 a^{2}+2 b^{2}+2 \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}+2 \sqrt {-a^{2}+b^{2}}\, \sqrt {\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right )^{2} b^{2}+2 \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}-a^{2}+b^{2}}}{\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right )}\right )}{\sqrt {-a^{2}+b^{2}}}\right )}{a} &= x +c_{3} \\ \end{align*}
Verification of solutions
\[ -\frac {b \left (\frac {\ln \left (\frac {\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}+b^{2}}{\sqrt {b^{2}}}+\sqrt {\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right )^{2} b^{2}+2 \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}-a^{2}+b^{2}}\right )}{\sqrt {b^{2}}}-\frac {\ln \left (\frac {-2 a^{2}+2 b^{2}+2 \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}+2 \sqrt {-a^{2}+b^{2}}\, \sqrt {\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right )^{2} b^{2}+2 \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}-a^{2}+b^{2}}}{\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right )}\right )}{\sqrt {-a^{2}+b^{2}}}\right )}{a} = x +c_{2} \] Warning, solution could not be verified
\[ \frac {b \left (\frac {\ln \left (\frac {\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}+b^{2}}{\sqrt {b^{2}}}+\sqrt {\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right )^{2} b^{2}+2 \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}-a^{2}+b^{2}}\right )}{\sqrt {b^{2}}}-\frac {\ln \left (\frac {-2 a^{2}+2 b^{2}+2 \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}+2 \sqrt {-a^{2}+b^{2}}\, \sqrt {\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right )^{2} b^{2}+2 \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right ) b^{2}-a^{2}+b^{2}}}{\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {a^{2} y+c_{1} a^{2}+b}{b}}}{b}\right )}\right )}{\sqrt {-a^{2}+b^{2}}}\right )}{a} = x +c_{3} \] Warning, solution could not be verified
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) = a*(_b(_a)^2+1)^(1/2)+b, _b(_a), HINT = [[1, 0]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[1, 0]
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 31
dsolve(diff(diff(y(x),x),x)=a*sqrt(1+diff(y(x),x)^2)+b,y(x), singsol=all)
\[ y \left (x \right ) = \int \operatorname {RootOf}\left (x -\left (\int _{}^{\textit {\_Z}}\frac {1}{a \sqrt {\textit {\_f}^{2}+1}+b}d \textit {\_f} \right )+c_{1} \right )d x +c_{2} \]
✓ Solution by Mathematica
Time used: 60.646 (sec). Leaf size: 972
DSolve[y''[x]==a*Sqrt[1+y'[x]^2]+b,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2-\frac {-\frac {2 a \text {InverseFunction}\left [\frac {\frac {2 b \arctan \left (\frac {b+a \left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\log \left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right )}{a}\&\right ][x+c_1]{}^2}{\sqrt {\text {InverseFunction}\left [\frac {\frac {2 b \arctan \left (\frac {b+a \left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\log \left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right )}{a}\&\right ][x+c_1]{}^2+1}}+b \log \left (\text {InverseFunction}\left [\frac {\frac {2 b \arctan \left (\frac {b+a \left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\log \left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right )}{a}\&\right ][x+c_1]{}^2 a^2+a^2-b^2\right )+b \log \left (\frac {2 a^2 \left (a-i \sqrt {a^2-b^2} \text {InverseFunction}\left [\frac {\frac {2 b \arctan \left (\frac {b+a \left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\log \left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right )}{a}\&\right ][x+c_1]+b \sqrt {\text {InverseFunction}\left [\frac {\frac {2 b \arctan \left (\frac {b+a \left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\log \left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right )}{a}\&\right ][x+c_1]{}^2+1}\right )}{b^3 \left (a \text {InverseFunction}\left [\frac {\frac {2 b \arctan \left (\frac {b+a \left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\log \left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right )}{a}\&\right ][x+c_1]+i \sqrt {a^2-b^2}\right )}\right )+b \log \left (\frac {2 a^2 \left (a+i \sqrt {a^2-b^2} \text {InverseFunction}\left [\frac {\frac {2 b \arctan \left (\frac {b+a \left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\log \left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right )}{a}\&\right ][x+c_1]+b \sqrt {\text {InverseFunction}\left [\frac {\frac {2 b \arctan \left (\frac {b+a \left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\log \left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right )}{a}\&\right ][x+c_1]{}^2+1}\right )}{b^3 \left (a \text {InverseFunction}\left [\frac {\frac {2 b \arctan \left (\frac {b+a \left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\log \left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right )}{a}\&\right ][x+c_1]-i \sqrt {a^2-b^2}\right )}\right )-\frac {2 a}{\sqrt {\text {InverseFunction}\left [\frac {\frac {2 b \arctan \left (\frac {b+a \left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\log \left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right )}{a}\&\right ][x+c_1]{}^2+1}}}{2 a^2} \]