Internal problem ID [2287]
Internal file name [OUTPUT/2287_Tuesday_February_27_2024_08_24_06_AM_57231313/index.tex]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 35, page 157
Problem number: 15.
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]]
\[ \boxed {y^{\prime \prime }-\sqrt {{y^{\prime }}^{2}+1}=0} \]
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 )-\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}{\sqrt {p^{2}+1}}d p &= x +c_{1}\\ \operatorname {arcsinh}\left (p \right )&=x +c_{1} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&=\sinh \left (x +c_{1} \right ) \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = \sinh \left (x +c_{1} \right ) \end {align*}
Integrating both sides gives \begin {align*} y &= \int { \sinh \left (x +c_{1} \right )\,\mathop {\mathrm {d}x}}\\ &= \cosh \left (x +c_{1} \right )+c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \cosh \left (x +c_{1} \right )+c_{2} \\ \end{align*}
Verification of solutions
\[ y = \cosh \left (x +c_{1} \right )+c_{2} \] 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 ) = \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}{\sqrt {p^{2}+1}}d p &= \int d y \\ \sqrt {p \left (y \right )^{2}+1}&=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*} \sqrt {{y^{\prime }}^{2}+1} = 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}+2 y c_{1} +y^{2}} \tag {1} \\ y^{\prime }&=-\sqrt {-1+c_{1}^{2}+2 y c_{1} +y^{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}+2 c_{1} y +y^{2}-1}}d y &= \int d x \\ \ln \left (y+c_{1} +\sqrt {-1+c_{1}^{2}+2 y c_{1} +y^{2}}\right )&=x +c_{2} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {1}{\sqrt {c_{1}^{2}+2 c_{1} y +y^{2}-1}}d y &= \int d x \\ -\ln \left (y+c_{1} +\sqrt {-1+c_{1}^{2}+2 y c_{1} +y^{2}}\right )&=x +c_{3} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left ({\mathrm e}^{2 x +2 c_{2}}-2 \,{\mathrm e}^{x +c_{2}} c_{1} +1\right ) {\mathrm e}^{-x -c_{2}}}{2} \\ \tag{2} y &= \frac {\left ({\mathrm e}^{-2 x -2 c_{3}}-2 \,{\mathrm e}^{-x -c_{3}} c_{1} +1\right ) {\mathrm e}^{x +c_{3}}}{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left ({\mathrm e}^{2 x +2 c_{2}}-2 \,{\mathrm e}^{x +c_{2}} c_{1} +1\right ) {\mathrm e}^{-x -c_{2}}}{2} \] Verified OK.
\[ y = \frac {\left ({\mathrm e}^{-2 x -2 c_{3}}-2 \,{\mathrm e}^{-x -c_{3}} c_{1} +1\right ) {\mathrm e}^{x +c_{3}}}{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\sqrt {{y^{\prime }}^{2}+1} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=\sqrt {u \left (x \right )^{2}+1} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=\sqrt {u \left (x \right )^{2}+1} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{\sqrt {u \left (x \right )^{2}+1}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{\sqrt {u \left (x \right )^{2}+1}}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \mathrm {arcsinh}\left (u \left (x \right )\right )=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\sinh \left (x +c_{1} \right ) \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\sinh \left (x +c_{1} \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\sinh \left (x +c_{1} \right ) \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \sinh \left (x +c_{1} \right )d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\cosh \left (x +c_{1} \right )+c_{2} \end {array} \]
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)-(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) = (_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: 0.39 (sec). Leaf size: 28
dsolve(diff(y(x),x$2)=sqrt(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 ) &= \cosh \left (c_{1} +x \right )+c_{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.34 (sec). Leaf size: 29
DSolve[y''[x]==Sqrt[1+y'[x]^2],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{2} \left (e^{-x-c_1}+e^{x+c_1}\right )+c_2 \]