Internal problem ID [9972]
Internal file name [OUTPUT/8919_Monday_June_06_2022_05_51_12_AM_93774806/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1650 (book 6.60).
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]]
\[ \boxed {y^{\prime \prime }-a \sqrt {{y^{\prime }}^{2}+1}=0} \]
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} = 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}}d p &= x +c_{1}\\ \frac {\operatorname {arcsinh}\left (p \right )}{a}&=x +c_{1} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&=\sinh \left (a c_{1} +x a \right ) \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = \sinh \left (a c_{1} +x a \right ) \end {align*}
Integrating both sides gives \begin {align*} y &= \int { \sinh \left (a c_{1} +x a \right )\,\mathop {\mathrm {d}x}}\\ &= \frac {\cosh \left (a c_{1} +x a \right )}{a}+c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\cosh \left (a c_{1} +x a \right )}{a}+c_{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {\cosh \left (a c_{1} +x a \right )}{a}+c_{2} \] 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} \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}}d p &= \int d y \\ \frac {\sqrt {p \left (y \right )^{2}+1}}{a}&=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} = 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 }&=\sqrt {-1+c_{1}^{2} a^{2}+2 y c_{1} a^{2}+y^{2} a^{2}} \tag {1} \\ y^{\prime }&=-\sqrt {-1+c_{1}^{2} a^{2}+2 y c_{1} a^{2}+y^{2} a^{2}} \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 {c_{1}^{2} a^{2}+2 y c_{1} a^{2}+y^{2} a^{2}-1}}d y &= \int d x \\ \frac {\ln \left (\frac {c_{1} a^{2}+a^{2} y}{\sqrt {a^{2}}}+\sqrt {-1+c_{1}^{2} a^{2}+2 y c_{1} a^{2}+y^{2} a^{2}}\right )}{\sqrt {a^{2}}}&=x +c_{2} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {1}{\sqrt {c_{1}^{2} a^{2}+2 y c_{1} a^{2}+y^{2} a^{2}-1}}d y &= \int d x \\ -\frac {\ln \left (\frac {c_{1} a^{2}+a^{2} y}{\sqrt {a^{2}}}+\sqrt {-1+c_{1}^{2} a^{2}+2 y c_{1} a^{2}+y^{2} a^{2}}\right )}{\sqrt {a^{2}}}&=x +c_{3} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-2 c_{1} a^{2} {\mathrm e}^{c_{2} \sqrt {a^{2}}+x \sqrt {a^{2}}}+{\mathrm e}^{2 c_{2} \sqrt {a^{2}}+2 x \sqrt {a^{2}}} \sqrt {a^{2}}+\sqrt {a^{2}}\right ) {\mathrm e}^{-c_{2} \sqrt {a^{2}}-x \sqrt {a^{2}}}}{2 a^{2}} \\ \tag{2} y &= \frac {\left (-2 c_{1} a^{2} {\mathrm e}^{-c_{3} \sqrt {a^{2}}-x \sqrt {a^{2}}}+{\mathrm e}^{-2 c_{3} \sqrt {a^{2}}-2 x \sqrt {a^{2}}} \sqrt {a^{2}}+\sqrt {a^{2}}\right ) {\mathrm e}^{c_{3} \sqrt {a^{2}}+x \sqrt {a^{2}}}}{2 a^{2}} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (-2 c_{1} a^{2} {\mathrm e}^{c_{2} \sqrt {a^{2}}+x \sqrt {a^{2}}}+{\mathrm e}^{2 c_{2} \sqrt {a^{2}}+2 x \sqrt {a^{2}}} \sqrt {a^{2}}+\sqrt {a^{2}}\right ) {\mathrm e}^{-c_{2} \sqrt {a^{2}}-x \sqrt {a^{2}}}}{2 a^{2}} \] Verified OK.
\[ y = \frac {\left (-2 c_{1} a^{2} {\mathrm e}^{-c_{3} \sqrt {a^{2}}-x \sqrt {a^{2}}}+{\mathrm e}^{-2 c_{3} \sqrt {a^{2}}-2 x \sqrt {a^{2}}} \sqrt {a^{2}}+\sqrt {a^{2}}\right ) {\mathrm e}^{c_{3} \sqrt {a^{2}}+x \sqrt {a^{2}}}}{2 a^{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=a \sqrt {{y^{\prime }}^{2}+1} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=a \sqrt {u \left (x \right )^{2}+1} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=a \sqrt {u \left (x \right )^{2}+1} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{\sqrt {u \left (x \right )^{2}+1}}=a \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{\sqrt {u \left (x \right )^{2}+1}}d x =\int a d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \mathrm {arcsinh}\left (u \left (x \right )\right )=x a +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\sinh \left (x a +c_{1} \right ) \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\sinh \left (x a +c_{1} \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\sinh \left (x a +c_{1} \right ) \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \sinh \left (x a +c_{1} \right )d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\frac {\cosh \left (x a +c_{1} \right )}{a}+c_{2} \end {array} \]
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 -> Calling odsolve with the ODE`, diff(diff(diff(y(x), x), x), x)-a^2*(diff(y(x), x)), y(x)` *** Sublevel 2 *** Methods for third order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients <- constant coefficients successful 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(_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: 0.219 (sec). Leaf size: 34
dsolve(diff(diff(y(x),x),x)=a*(diff(y(x),x)^2+1)^(1/2),y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -i x +c_{1} \\ y \left (x \right ) &= i x +c_{1} \\ y \left (x \right ) &= c_{2} +\frac {\cosh \left (a \left (c_{1} +x \right )\right )}{a} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.728 (sec). Leaf size: 35
DSolve[-(a*Sqrt[1 + y'[x]^2]) + y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {e^{a x+c_1} \left (1+e^{-2 (a x+c_1)}\right )}{2 a}+c_2 \]