Internal problem ID [15200]
Internal file name [OUTPUT/15201_Tuesday_April_23_2024_04_53_59_PM_61062009/index.tex]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 14. Differential equations admitting of
depression of their order. Exercises page 107
Problem number: 343.
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}\\ \arcsin \left (p \right )&=x +c_{1} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&=\sin \left (x +c_{1} \right ) \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = \sin \left (x +c_{1} \right ) \end {align*}
Integrating both sides gives \begin {align*} y &= \int { \sin \left (x +c_{1} \right )\,\mathop {\mathrm {d}x}}\\ &= -\cos \left (x +c_{1} \right )+c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\cos \left (x +c_{1} \right )+c_{2} \\ \end{align*}
Verification of solutions
\[ y = -\cos \left (x +c_{1} \right )+c_{2} \] 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 {-y^{2}-2 y c_{1} -c_{1}^{2}+1} \tag {1} \\ y^{\prime }&=-\sqrt {-y^{2}-2 y c_{1} -c_{1}^{2}+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 {-c_{1}^{2}-2 c_{1} y -y^{2}+1}}d y &= x +c_{2}\\ \arcsin \left (y +c_{1} \right )&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=-c_{1} +\sin \left (x +c_{2} \right ) \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 &= c_{3} +x\\ -\arcsin \left (y +c_{1} \right )&=c_{3} +x \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=-c_{1} -\sin \left (c_{3} +x \right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -c_{1} +\sin \left (x +c_{2} \right ) \\ \tag{2} y &= -c_{1} -\sin \left (c_{3} +x \right ) \\ \end{align*}
Verification of solutions
\[ y = -c_{1} +\sin \left (x +c_{2} \right ) \] Verified OK.
\[ y = -c_{1} -\sin \left (c_{3} +x \right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\sqrt {-{y^{\prime }}^{2}+1} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=\sqrt {-u \left (x \right )^{2}+1} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=\sqrt {-u \left (x \right )^{2}+1} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{\sqrt {-u \left (x \right )^{2}+1}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{\sqrt {-u \left (x \right )^{2}+1}}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \arcsin \left (u \left (x \right )\right )=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\sin \left (x +c_{1} \right ) \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\sin \left (x +c_{1} \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\sin \left (x +c_{1} \right ) \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \sin \left (x +c_{1} \right )d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=-\cos \left (x +c_{1} \right )+c_{2} \end {array} \]
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: 2.297 (sec). Leaf size: 26
dsolve(diff(y(x),x$2)=sqrt(1-diff(y(x),x)^2),y(x), singsol=all)
\begin{align*} y &= -x +c_{1} \\ y &= x +c_{1} \\ y &= -\cos \left (x +c_{1} \right )+c_{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.166 (sec). Leaf size: 24
DSolve[y''[x]==Sqrt[1-y'[x]^2],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \sin (x+c_1)+c_2 \\ y(x)\to \text {Interval}[\{-1,1\}]+c_2 \\ \end{align*}