Internal problem ID [4846]
Internal file name [OUTPUT/4339_Sunday_June_05_2022_01_03_09_PM_8720060/index.tex]
Book: Mathematical Methods in the Physical Sciences. third edition. Mary L. Boas. John Wiley.
2006
Section: Chapter 8, Ordinary differential equations. Section 7. Other second-Order equations.
page 435
Problem number: 5.
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]]
\[ \boxed {{y^{\prime \prime }}^{2}-k^{2} \left (1+{y^{\prime }}^{2}\right )=0} \]
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 \left (x \right )^{2} k^{2}+{p^{\prime }\left (x \right )}^{2}-k^{2} = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. Solving the given ode for \(p^{\prime }\left (x \right )\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} p^{\prime }\left (x \right )&=\sqrt {p \left (x \right )^{2}+1}\, k \tag {1} \\ p^{\prime }\left (x \right )&=-\sqrt {p \left (x \right )^{2}+1}\, k \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 {p^{2}+1}\, k}d p &= x +c_{1}\\ \frac {\operatorname {arcsinh}\left (p \right )}{k}&=x +c_{1} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&=\sinh \left (c_{1} k +x k \right ) \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} \int -\frac {1}{\sqrt {p^{2}+1}\, k}d p &= x +c_{2}\\ -\frac {\operatorname {arcsinh}\left (p \right )}{k}&=x +c_{2} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&=-\sinh \left (c_{2} k +x k \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 } = \sinh \left (c_{1} k +x k \right ) \end {align*}
Integrating both sides gives \begin {align*} y &= \int { \sinh \left (c_{1} k +x k \right )\,\mathop {\mathrm {d}x}}\\ &= \frac {\cosh \left (c_{1} k +x k \right )}{k}+c_{3} \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = -\sinh \left (c_{2} k +x k \right ) \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -\sinh \left (c_{2} k +x k \right )\,\mathop {\mathrm {d}x}}\\ &= -\frac {\cosh \left (c_{2} k +x k \right )}{k}+c_{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\cosh \left (c_{1} k +x k \right )}{k}+c_{3} \\ \tag{2} y &= -\frac {\cosh \left (c_{2} k +x k \right )}{k}+c_{4} \\ \end{align*}
Verification of solutions
\[ y = \frac {\cosh \left (c_{1} k +x k \right )}{k}+c_{3} \] Verified OK.
\[ y = -\frac {\cosh \left (c_{2} k +x k \right )}{k}+c_{4} \] 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 )^{2} \left (\frac {d}{d y}p \left (y \right )\right )^{2}-p \left (y \right )^{2} k^{2} = k^{2} \end {align*}
Which is now solved as first order ode for \(p(y)\). Solving the given ode for \(\frac {d}{d y}p \left (y \right )\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} \frac {d}{d y}p \left (y \right )&=\frac {\sqrt {p \left (y \right )^{2}+1}\, k}{p \left (y \right )} \tag {1} \\ \frac {d}{d y}p \left (y \right )&=-\frac {\sqrt {p \left (y \right )^{2}+1}\, k}{p \left (y \right )} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin{align*} \int \frac {p}{\sqrt {p^{2}+1}\, k}d p &= \int d y \\ \frac {\sqrt {p \left (y \right )^{2}+1}}{k}&=y +c_{1} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {p}{\sqrt {p^{2}+1}\, k}d p &= \int d y \\ -\frac {\sqrt {p \left (y \right )^{2}+1}}{k}&=y +c_{2} \\ \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 {1+{y^{\prime }}^{2}}}{k} = 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} k^{2}+2 y c_{1} k^{2}+y^{2} k^{2}} \tag {1} \\ y^{\prime }&=-\sqrt {-1+c_{1}^{2} k^{2}+2 y c_{1} k^{2}+y^{2} k^{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} k^{2}+2 y c_{1} k^{2}+y^{2} k^{2}-1}}d y &= \int d x \\ \frac {\ln \left (\frac {c_{1} k^{2}+k^{2} y}{\sqrt {k^{2}}}+\sqrt {-1+c_{1}^{2} k^{2}+2 y c_{1} k^{2}+y^{2} k^{2}}\right )}{\sqrt {k^{2}}}&=x +c_{3} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {1}{\sqrt {c_{1}^{2} k^{2}+2 y c_{1} k^{2}+y^{2} k^{2}-1}}d y &= \int d x \\ -\frac {\ln \left (\frac {c_{1} k^{2}+k^{2} y}{\sqrt {k^{2}}}+\sqrt {-1+c_{1}^{2} k^{2}+2 y c_{1} k^{2}+y^{2} k^{2}}\right )}{\sqrt {k^{2}}}&=x +c_{4} \\ \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*} -\frac {\sqrt {1+{y^{\prime }}^{2}}}{k} = y+c_{2} \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} k^{2}+2 y c_{2} k^{2}+y^{2} k^{2}} \tag {1} \\ y^{\prime }&=-\sqrt {-1+c_{2}^{2} k^{2}+2 y c_{2} k^{2}+y^{2} k^{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_{2}^{2} k^{2}+2 y c_{2} k^{2}+y^{2} k^{2}-1}}d y &= \int d x \\ \frac {\ln \left (\frac {c_{2} k^{2}+k^{2} y}{\sqrt {k^{2}}}+\sqrt {-1+c_{2}^{2} k^{2}+2 y c_{2} k^{2}+y^{2} k^{2}}\right )}{\sqrt {k^{2}}}&=x +c_{5} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {1}{\sqrt {c_{2}^{2} k^{2}+2 y c_{2} k^{2}+y^{2} k^{2}-1}}d y &= \int d x \\ -\frac {\ln \left (\frac {c_{2} k^{2}+k^{2} y}{\sqrt {k^{2}}}+\sqrt {-1+c_{2}^{2} k^{2}+2 y c_{2} k^{2}+y^{2} k^{2}}\right )}{\sqrt {k^{2}}}&=x +c_{6} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-2 c_{1} k^{2} {\mathrm e}^{c_{3} \sqrt {k^{2}}+x \sqrt {k^{2}}}+{\mathrm e}^{2 c_{3} \sqrt {k^{2}}+2 x \sqrt {k^{2}}} \sqrt {k^{2}}+\sqrt {k^{2}}\right ) {\mathrm e}^{-c_{3} \sqrt {k^{2}}-x \sqrt {k^{2}}}}{2 k^{2}} \\ \tag{2} y &= \frac {\left (-2 c_{1} k^{2} {\mathrm e}^{-c_{4} \sqrt {k^{2}}-x \sqrt {k^{2}}}+{\mathrm e}^{-2 c_{4} \sqrt {k^{2}}-2 x \sqrt {k^{2}}} \sqrt {k^{2}}+\sqrt {k^{2}}\right ) {\mathrm e}^{c_{4} \sqrt {k^{2}}+x \sqrt {k^{2}}}}{2 k^{2}} \\ \tag{3} y &= \frac {\left (-2 c_{2} k^{2} {\mathrm e}^{c_{5} \sqrt {k^{2}}+x \sqrt {k^{2}}}+{\mathrm e}^{2 c_{5} \sqrt {k^{2}}+2 x \sqrt {k^{2}}} \sqrt {k^{2}}+\sqrt {k^{2}}\right ) {\mathrm e}^{-c_{5} \sqrt {k^{2}}-x \sqrt {k^{2}}}}{2 k^{2}} \\ \tag{4} y &= \frac {\left (-2 c_{2} k^{2} {\mathrm e}^{-c_{6} \sqrt {k^{2}}-x \sqrt {k^{2}}}+{\mathrm e}^{-2 c_{6} \sqrt {k^{2}}-2 x \sqrt {k^{2}}} \sqrt {k^{2}}+\sqrt {k^{2}}\right ) {\mathrm e}^{c_{6} \sqrt {k^{2}}+x \sqrt {k^{2}}}}{2 k^{2}} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (-2 c_{1} k^{2} {\mathrm e}^{c_{3} \sqrt {k^{2}}+x \sqrt {k^{2}}}+{\mathrm e}^{2 c_{3} \sqrt {k^{2}}+2 x \sqrt {k^{2}}} \sqrt {k^{2}}+\sqrt {k^{2}}\right ) {\mathrm e}^{-c_{3} \sqrt {k^{2}}-x \sqrt {k^{2}}}}{2 k^{2}} \] Verified OK.
\[ y = \frac {\left (-2 c_{1} k^{2} {\mathrm e}^{-c_{4} \sqrt {k^{2}}-x \sqrt {k^{2}}}+{\mathrm e}^{-2 c_{4} \sqrt {k^{2}}-2 x \sqrt {k^{2}}} \sqrt {k^{2}}+\sqrt {k^{2}}\right ) {\mathrm e}^{c_{4} \sqrt {k^{2}}+x \sqrt {k^{2}}}}{2 k^{2}} \] Verified OK.
\[ y = \frac {\left (-2 c_{2} k^{2} {\mathrm e}^{c_{5} \sqrt {k^{2}}+x \sqrt {k^{2}}}+{\mathrm e}^{2 c_{5} \sqrt {k^{2}}+2 x \sqrt {k^{2}}} \sqrt {k^{2}}+\sqrt {k^{2}}\right ) {\mathrm e}^{-c_{5} \sqrt {k^{2}}-x \sqrt {k^{2}}}}{2 k^{2}} \] Verified OK.
\[ y = \frac {\left (-2 c_{2} k^{2} {\mathrm e}^{-c_{6} \sqrt {k^{2}}-x \sqrt {k^{2}}}+{\mathrm e}^{-2 c_{6} \sqrt {k^{2}}-2 x \sqrt {k^{2}}} \sqrt {k^{2}}+\sqrt {k^{2}}\right ) {\mathrm e}^{c_{6} \sqrt {k^{2}}+x \sqrt {k^{2}}}}{2 k^{2}} \] 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 -> Calling odsolve with the ODE`, diff(diff(diff(y(x), x), x), x)-k^2*(diff(y(x), x)), y(x)` *** Sublevel 4 *** Methods for third order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients <- constant coefficients successful <- 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`
✓ Solution by Maple
Time used: 0.5 (sec). Leaf size: 55
dsolve((diff(y(x),x$2))^2=k^2*(1+ (diff(y(x),x))^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 ) &= \frac {4 c_{2}^{2} {\mathrm e}^{k x} k^{2}+4 c_{1} c_{2} k^{2}+{\mathrm e}^{-k x}}{4 c_{2} k^{2}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.451 (sec). Leaf size: 71
DSolve[(y''[x])^2==k^2*(1+ (y'[x])^2),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {e^{k x-c_1}+e^{-k x+c_1}-2 c_2 k}{2 k} \\ y(x)\to \frac {e^{k x+c_1} \left (1+e^{-2 (k x+c_1)}\right )}{2 k}+c_2 \\ \end{align*}