Internal problem ID [12496]
Internal file name [OUTPUT/11149_Monday_October_16_2023_09_52_02_PM_17799216/index.tex]
Book: DIFFERENTIAL and INTEGRAL CALCULUS. VOL I. by N. PISKUNOV. MIR
PUBLISHERS, Moscow 1969.
Section: Chapter 8. Differential equations. Exercises page 595
Problem number: 125.
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]]
\[ \boxed {{y^{\prime \prime }}^{2}+{y^{\prime }}^{2}=a^{2}} \] With initial conditions \begin {align*} [y \left (0\right ) = -1, y^{\prime }\left (0\right ) = 0] \end {align*}
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 )}^{2}+p \left (x \right )^{2}-a^{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}+a^{2}} \tag {1} \\ p^{\prime }\left (x \right )&=-\sqrt {-p \left (x \right )^{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 {a^{2}-p^{2}}}d p &= \int {dx}\\ \arctan \left (\frac {p}{\sqrt {a^{2}-p^{2}}}\right )&= x +c_{1} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(p=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = 0 \end {align*}
Trying the constant \begin {align*} c_{1} = 0 \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \arctan \left (\frac {p}{\sqrt {a^{2}-p^{2}}}\right ) = x \end {align*}
The constant \(c_{1} = 0\) gives valid solution.
Solving equation (2)
Integrating both sides gives \begin {align*} \int -\frac {1}{\sqrt {a^{2}-p^{2}}}d p &= \int {dx}\\ \arctan \left (\frac {\sqrt {a^{2}-p^{2}}}{p}\right )&= x +c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=0\) and \(p=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \textit {undefined} = c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = \textit {undefined} \end {align*}
Trying the constant \begin {align*} c_{2} = \textit {undefined} \end {align*}
The constant \(c_{2} = \textit {undefined}\) does not give valid solution.
Warning: Unable to solve for constant of integration. Unable to determine ODE type.
For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} \arctan \left (\frac {y^{\prime }}{\sqrt {-{y^{\prime }}^{2}+a^{2}}}\right ) = x \end {align*}
Integrating both sides gives \begin {align*} y &= \int { \tan \left (x \right ) \sqrt {\frac {a^{2}}{1+\tan \left (x \right )^{2}}}\,\mathop {\mathrm {d}x}}\\ &= -\sqrt {\frac {a^{2}}{1+\tan \left (x \right )^{2}}}+c_{3} \end {align*}
Initial conditions are used to solve for \(c_{3}\). Substituting \(x=0\) and \(y=-1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -1 = -\operatorname {csgn}\left (a \right ) a +c_{3} \end {align*}
The solutions are \begin {align*} c_{3} = a -1 \end {align*}
Trying the constant \begin {align*} c_{3} = a -1 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=-\sqrt {\cos \left (x \right )^{2} a^{2}}+a -1 \end {align*}
But this does not satisfy the initial conditions. Hence no solution can be found. The constant \(c_{3} = a -1\) does not give valid solution.
Which is valid for any constant of integration. Therefore keeping the constant in place. Initial conditions are used to solve for the constants of integration.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\sqrt {\cos \left (x \right )^{2} a^{2}}+a -1 \\ \end{align*}
Verification of solutions
\[ y = -\sqrt {\cos \left (x \right )^{2} a^{2}}+a -1 \] 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} = a^{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}+a^{2}}}{p \left (y \right )} \tag {1} \\ \frac {d}{d y}p \left (y \right )&=-\frac {\sqrt {-p \left (y \right )^{2}+a^{2}}}{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 {a^{2}-p^{2}}}d p &= \int {dy}\\ -\sqrt {a^{2}-p^{2}}&= y +c_{1} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(y=-1\) and \(p=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -\operatorname {csgn}\left (a \right ) a = c_{1} -1 \end {align*}
The solutions are \begin {align*} c_{1} = 1-a \end {align*}
Trying the constant \begin {align*} c_{1} = 1-a \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} -\sqrt {a^{2}-p^{2}} = y +1-a \end {align*}
The constant \(c_{1} = 1-a\) gives valid solution.
Solving equation (2)
Integrating both sides gives \begin {align*} \int -\frac {p}{\sqrt {a^{2}-p^{2}}}d p &= \int {dy}\\ \sqrt {a^{2}-p^{2}}&= y +c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(y=-1\) and \(p=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \operatorname {csgn}\left (a \right ) a = -1+c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = a +1 \end {align*}
Trying the constant \begin {align*} c_{2} = a +1 \end {align*}
Substituting \(c_{2}\) found above in the general solution gives \begin {align*} \sqrt {a^{2}-p^{2}} = y +a +1 \end {align*}
The constant \(c_{2} = a +1\) gives valid solution.
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}+a^{2}} = y+1-a \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+2 a y-y^{2}+2 a -2 y} \tag {1} \\ y^{\prime }&=-\sqrt {-1+2 a y-y^{2}+2 a -2 y} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Since ode has form \(y^{\prime }= f(y)\) and initial conditions \(y = -1\) is verified to satisfy the ode, then the solution is \begin {align*} y&=y_0 \\ &=-1 \end {align*}
Solving equation (2)
Since ode has form \(y^{\prime }= f(y)\) and initial conditions \(y = -1\) is verified to satisfy the ode, then the solution is \begin {align*} y&=y_0 \\ &=-1 \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*} \sqrt {-{y^{\prime }}^{2}+a^{2}} = y+a +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 {-1-2 a y-y^{2}-2 a -2 y} \tag {1} \\ y^{\prime }&=-\sqrt {-1-2 a y-y^{2}-2 a -2 y} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Since ode has form \(y^{\prime }= f(y)\) and initial conditions \(y = -1\) is verified to satisfy the ode, then the solution is \begin {align*} y&=y_0 \\ &=-1 \end {align*}
Solving equation (2)
Since ode has form \(y^{\prime }= f(y)\) and initial conditions \(y = -1\) is verified to satisfy the ode, then the solution is \begin {align*} y&=y_0 \\ &=-1 \end {align*}
Initial conditions are used to solve for the constants of integration.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -1 \\ \tag{2} y &= -1 \\ \end{align*}
Verification of solutions
\[ y = -1 \] Verified OK.
\[ y = -1 \] 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)+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 -> Calling odsolve with the ODE`, diff(y(x), x) = -a, y(x), singsol = none` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful -> Calling odsolve with the ODE`, diff(y(x), x) = a, y(x), singsol = none` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful`
✓ Solution by Maple
Time used: 0.657 (sec). Leaf size: 24
dsolve([diff(y(x),x$2)^2+diff(y(x),x)^2=a^2,y(0) = -1, D(y)(0) = 0],y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -a -1+a \cos \left (x \right ) \\ y \left (x \right ) &= a -1-a \cos \left (x \right ) \\ \end{align*}
✓ Solution by Mathematica
Time used: 15.637 (sec). Leaf size: 37
DSolve[{(y''[x])^2+(y'[x])^2==a^2,{y[0]==-1,y'[0]==0}},y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to a \left (\frac {1}{\sqrt {\sec ^2(x)}}-1\right )-1 \\ y(x)\to -\frac {a}{\sqrt {\sec ^2(x)}}+a-1 \\ \end{align*}