Internal problem ID [6840]
Internal file name [OUTPUT/6087_Thursday_July_28_2022_04_29_58_AM_68133921/index.tex]
Book: Elementary differential equations. By Earl D. Rainville, Phillip E. Bedient. Macmilliam
Publishing Co. NY. 6th edition. 1981.
Section: CHAPTER 16. Nonlinear equations. Section 101. Independent variable missing.
EXERCISES Page 324
Problem number: 21.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x", "second_order_ode_can_be_made_integrable"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]
\[ \boxed {2 y^{\prime \prime }-\sin \left (2 y\right )=0} \] With initial conditions \begin {align*} \left [y \left (0\right ) = -\frac {\pi }{2}, y^{\prime }\left (0\right ) = 1\right ] \end {align*}
Multiplying the ode by \(y^{\prime }\) gives \[ 2 y^{\prime } y^{\prime \prime }-y^{\prime } \sin \left (2 y\right ) = 0 \] Integrating the above w.r.t \(x\) gives \begin {align*} \int \left (2 y^{\prime } y^{\prime \prime }-y^{\prime } \sin \left (2 y\right )\right )d x &= 0 \\ {y^{\prime }}^{2}+\frac {\cos \left (2 y\right )}{2} = c_2 \end {align*}
Which is now solved for \(y\). 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 }&=\frac {\sqrt {-2 \cos \left (2 y\right )+4 c_{1}}}{2} \tag {1} \\ y^{\prime }&=-\frac {\sqrt {-2 \cos \left (2 y\right )+4 c_{1}}}{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 {2}{\sqrt {-2 \cos \left (2 y \right )+4 c_{1}}}d y &= \int d x \\ \frac {2 \sqrt {-\frac {\cos \left (2 y\right )-2 c_{1}}{2 c_{1} -1}}\, \operatorname {InverseJacobiAM}\left (y, \sqrt {2}\, \sqrt {-\frac {1}{2 c_{1} -1}}\right )}{\sqrt {-2 \cos \left (2 y\right )+4 c_{1}}}&=x +c_{2} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {2}{\sqrt {-2 \cos \left (2 y \right )+4 c_{1}}}d y &= \int d x \\ -\frac {2 \sqrt {-\frac {\cos \left (2 y\right )-2 c_{1}}{2 c_{1} -1}}\, \operatorname {InverseJacobiAM}\left (y, \sqrt {2}\, \sqrt {-\frac {1}{2 c_{1} -1}}\right )}{\sqrt {-2 \cos \left (2 y\right )+4 c_{1}}}&=x +c_{3} \\ \end{align*} Initial conditions are used to solve for the constants of integration.
Looking at the First solution \begin {align*} \frac {2 \sqrt {-\frac {\cos \left (2 y\right )-2 c_{1}}{2 c_{1} -1}}\, \operatorname {InverseJacobiAM}\left (y, \sqrt {2}\, \sqrt {-\frac {1}{2 c_{1} -1}}\right )}{\sqrt {-2 \cos \left (2 y\right )+4 c_{1}}} = x +c_{2} \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = -\frac {\pi }{2}\) and \(x = 0\) in the above gives \begin {align*} -\frac {2 \sqrt {\frac {1+2 c_{1}}{2 c_{1} -1}}\, \operatorname {EllipticK}\left (\sqrt {2}\, \sqrt {-\frac {1}{2 c_{1} -1}}\right )}{\sqrt {2+4 c_{1}}} = c_{2}\tag {1A} \end {align*}
Unable to solve for \(y\) to solve for constant of integration
Looking at the Second solution \begin {align*} -\frac {2 \sqrt {-\frac {\cos \left (2 y\right )-2 c_{1}}{2 c_{1} -1}}\, \operatorname {InverseJacobiAM}\left (y, \sqrt {2}\, \sqrt {-\frac {1}{2 c_{1} -1}}\right )}{\sqrt {-2 \cos \left (2 y\right )+4 c_{1}}} = x +c_{3} \tag {2} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = -\frac {\pi }{2}\) and \(x = 0\) in the above gives \begin {align*} \frac {2 \sqrt {\frac {1+2 c_{1}}{2 c_{1} -1}}\, \operatorname {EllipticK}\left (\sqrt {2}\, \sqrt {-\frac {1}{2 c_{1} -1}}\right )}{\sqrt {2+4 c_{1}}} = c_{3}\tag {1A} \end {align*}
Unable to solve for \(y\) to solve for constant of integration
Verification of solutions N/A
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*} 2 p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) = \sin \left (2 y \right ) \end {align*}
Which is now solved as first order ode for \(p(y)\). In canonical form the ODE is \begin {align*} p' &= F(y,p)\\ &= f( y) g(p)\\ &= \frac {\sin \left (2 y \right )}{2 p} \end {align*}
Where \(f(y)=\frac {\sin \left (2 y \right )}{2}\) and \(g(p)=\frac {1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p}} \,dp &= \frac {\sin \left (2 y \right )}{2} \,d y \\ \int { \frac {1}{\frac {1}{p}} \,dp} &= \int {\frac {\sin \left (2 y \right )}{2} \,d y} \\ \frac {p^{2}}{2}&=-\frac {\cos \left (2 y \right )}{4}+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (y \right )^{2}}{2}+\frac {\cos \left (2 y \right )}{4}-c_{1} = 0 \] Initial conditions are used to solve for \(c_{1}\). Substituting \(y=-\frac {\pi }{2}\) and \(p=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -c_{1} +\frac {1}{4} = 0 \end {align*}
The solutions are \begin {align*} c_{1} = {\frac {1}{4}} \end {align*}
Trying the constant \begin {align*} c_{1} = {\frac {1}{4}} \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {p^{2}}{2}+\frac {\cos \left (2 y \right )}{4}-\frac {1}{4} = 0 \end {align*}
The constant \(c_{1} = {\frac {1}{4}}\) 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*} \frac {{y^{\prime }}^{2}}{2}+\frac {\cos \left (2 y\right )}{4}-\frac {1}{4} = 0 \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 }&=\frac {\sqrt {2-2 \cos \left (2 y\right )}}{2} \tag {1} \\ y^{\prime }&=-\frac {\sqrt {2-2 \cos \left (2 y\right )}}{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 {2}{\sqrt {2-2 \cos \left (2 y \right )}}d y &= \int {dx}\\ -\frac {2 \sin \left (y \right ) \operatorname {arctanh}\left (\cos \left (y \right )\right )}{\sqrt {2-2 \cos \left (2 y \right )}}&= x +c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=0\) and \(y=-\frac {\pi }{2}\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = 0 \end {align*}
Trying the constant \begin {align*} c_{2} = 0 \end {align*}
Substituting \(c_{2}\) found above in the general solution gives \begin {align*} -\frac {2 \sin \left (y \right ) \operatorname {arctanh}\left (\cos \left (y \right )\right )}{\sqrt {2-2 \cos \left (2 y \right )}} = x \end {align*}
The constant \(c_{2} = 0\) does not give valid solution.
Which is valid for any constant of integration. Therefore keeping the constant in place. Solving equation (2)
Integrating both sides gives \begin {align*} \int -\frac {2}{\sqrt {2-2 \cos \left (2 y \right )}}d y &= \int {dx}\\ \frac {2 \sin \left (y \right ) \operatorname {arctanh}\left (\cos \left (y \right )\right )}{\sqrt {2-2 \cos \left (2 y \right )}}&= x +c_{3} \end {align*}
Initial conditions are used to solve for \(c_{3}\). Substituting \(x=0\) and \(y=-\frac {\pi }{2}\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = c_{3} \end {align*}
The solutions are \begin {align*} c_{3} = 0 \end {align*}
Trying the constant \begin {align*} c_{3} = 0 \end {align*}
Substituting \(c_{3}\) found above in the general solution gives \begin {align*} \frac {2 \sin \left (y \right ) \operatorname {arctanh}\left (\cos \left (y \right )\right )}{\sqrt {2-2 \cos \left (2 y \right )}} = x \end {align*}
The constant \(c_{3} = 0\) does not give valid solution.
Which is valid for any constant of integration. Therefore keeping the constant in place. Simplifying the solution \(-\operatorname {arctanh}\left (\cos \left (y\right )\right ) \operatorname {csgn}\left (\sin \left (y\right )\right ) = x\) to \(-\operatorname {arctanh}\left (\cos \left (y\right )\right ) = x\) Simplifying the solution \(\operatorname {arctanh}\left (\cos \left (y\right )\right ) \operatorname {csgn}\left (\sin \left (y\right )\right ) = x\) to \(\operatorname {arctanh}\left (\cos \left (y\right )\right ) = x\) Initial conditions are used to solve for the constants of integration.
Summary
The solution(s) found are the following \begin{align*} \tag{1} -\operatorname {arctanh}\left (\cos \left (y\right )\right ) &= x \\ \tag{2} \operatorname {arctanh}\left (\cos \left (y\right )\right ) &= x \\ \end{align*}
Verification of solutions
\[ -\operatorname {arctanh}\left (\cos \left (y\right )\right ) = x \] Verified OK.
\[ \operatorname {arctanh}\left (\cos \left (y\right )\right ) = x \] Verified OK.
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 trying 2nd order, 2 integrating factors of the form mu(x,y) trying differential order: 2; missing variables `, `-> Computing symmetries using: way = 3 `, `-> Computing symmetries using: way = exp_sym -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)-(1/2)*sin(2*_a) = 0, _b(_a)` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli <- Bernoulli successful <- differential order: 2; canonical coordinates successful <- differential order 2; missing variables successful`
✓ Solution by Maple
Time used: 107.406 (sec). Leaf size: 1490
dsolve([2*diff(y(x),x$2)=sin(2*y(x)),y(0) = -1/2*Pi, D(y)(0) = 1],y(x), singsol=all)
\[ \text {Expression too large to display} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[{2*y''[x]==Sin[2*y[x]],{y[0]==-Pi/2,y'[0]==1}},y[x],x,IncludeSingularSolutions -> True]
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