Internal problem ID [9929]
Internal file name [OUTPUT/8876_Monday_June_06_2022_05_44_09_AM_93873351/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1607 (6.17).
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]]
\[ \boxed {y^{\prime \prime }+a \sin \left (y\right )=0} \]
Multiplying the ode by \(y^{\prime }\) gives \[ y^{\prime } y^{\prime \prime }+y^{\prime } a \sin \left (y\right ) = 0 \] Integrating the above w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime } y^{\prime \prime }+y^{\prime } a \sin \left (y\right )\right )d x &= 0 \\ \frac {{y^{\prime }}^{2}}{2}-a \cos \left (y\right ) = 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 }&=\sqrt {2 a \cos \left (y\right )+2 c_{1}} \tag {1} \\ y^{\prime }&=-\sqrt {2 a \cos \left (y\right )+2 c_{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 {2 a \cos \left (y \right )+2 c_{1}}}d y &= \int d x \\ \frac {2 \sqrt {\frac {a \cos \left (y\right )+c_{1}}{c_{1} +a}}\, \operatorname {InverseJacobiAM}\left (\frac {y}{2}, \frac {\sqrt {2}\, \sqrt {a}}{\sqrt {c_{1} +a}}\right )}{\sqrt {2 a \cos \left (y\right )+2 c_{1}}}&=x +c_{2} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {1}{\sqrt {2 a \cos \left (y \right )+2 c_{1}}}d y &= \int d x \\ -\frac {2 \sqrt {\frac {a \cos \left (y\right )+c_{1}}{c_{1} +a}}\, \operatorname {InverseJacobiAM}\left (\frac {y}{2}, \frac {\sqrt {2}\, \sqrt {a}}{\sqrt {c_{1} +a}}\right )}{\sqrt {2 a \cos \left (y\right )+2 c_{1}}}&=x +c_{3} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} \frac {2 \sqrt {\frac {a \cos \left (y\right )+c_{1}}{c_{1} +a}}\, \operatorname {InverseJacobiAM}\left (\frac {y}{2}, \frac {\sqrt {2}\, \sqrt {a}}{\sqrt {c_{1} +a}}\right )}{\sqrt {2 a \cos \left (y\right )+2 c_{1}}} &= x +c_{2} \\ \tag{2} -\frac {2 \sqrt {\frac {a \cos \left (y\right )+c_{1}}{c_{1} +a}}\, \operatorname {InverseJacobiAM}\left (\frac {y}{2}, \frac {\sqrt {2}\, \sqrt {a}}{\sqrt {c_{1} +a}}\right )}{\sqrt {2 a \cos \left (y\right )+2 c_{1}}} &= x +c_{3} \\ \end{align*}
Verification of solutions
\[ \frac {2 \sqrt {\frac {a \cos \left (y\right )+c_{1}}{c_{1} +a}}\, \operatorname {InverseJacobiAM}\left (\frac {y}{2}, \frac {\sqrt {2}\, \sqrt {a}}{\sqrt {c_{1} +a}}\right )}{\sqrt {2 a \cos \left (y\right )+2 c_{1}}} = x +c_{2} \] Verified OK.
\[ -\frac {2 \sqrt {\frac {a \cos \left (y\right )+c_{1}}{c_{1} +a}}\, \operatorname {InverseJacobiAM}\left (\frac {y}{2}, \frac {\sqrt {2}\, \sqrt {a}}{\sqrt {c_{1} +a}}\right )}{\sqrt {2 a \cos \left (y\right )+2 c_{1}}} = x +c_{3} \] 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 ) = -a \sin \left (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 {a \sin \left (y \right )}{p} \end {align*}
Where \(f(y)=-a \sin \left (y \right )\) and \(g(p)=\frac {1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p}} \,dp &= -a \sin \left (y \right ) \,d y \\ \int { \frac {1}{\frac {1}{p}} \,dp} &= \int {-a \sin \left (y \right ) \,d y} \\ \frac {p^{2}}{2}&=a \cos \left (y \right )+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (y \right )^{2}}{2}-a \cos \left (y \right )-c_{1} = 0 \] 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}-a \cos \left (y\right )-c_{1} = 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 }&=\sqrt {2 a \cos \left (y\right )+2 c_{1}} \tag {1} \\ y^{\prime }&=-\sqrt {2 a \cos \left (y\right )+2 c_{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 {2 a \cos \left (y \right )+2 c_{1}}}d y &= \int d x \\ \frac {2 \sqrt {\frac {a \cos \left (y\right )+c_{1}}{c_{1} +a}}\, \operatorname {InverseJacobiAM}\left (\frac {y}{2}, \frac {\sqrt {2}\, \sqrt {a}}{\sqrt {c_{1} +a}}\right )}{\sqrt {2 a \cos \left (y\right )+2 c_{1}}}&=x +c_{2} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {1}{\sqrt {2 a \cos \left (y \right )+2 c_{1}}}d y &= \int d x \\ -\frac {2 \sqrt {\frac {a \cos \left (y\right )+c_{1}}{c_{1} +a}}\, \operatorname {InverseJacobiAM}\left (\frac {y}{2}, \frac {\sqrt {2}\, \sqrt {a}}{\sqrt {c_{1} +a}}\right )}{\sqrt {2 a \cos \left (y\right )+2 c_{1}}}&=x +c_{3} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} \frac {2 \sqrt {\frac {a \cos \left (y\right )+c_{1}}{c_{1} +a}}\, \operatorname {InverseJacobiAM}\left (\frac {y}{2}, \frac {\sqrt {2}\, \sqrt {a}}{\sqrt {c_{1} +a}}\right )}{\sqrt {2 a \cos \left (y\right )+2 c_{1}}} &= x +c_{2} \\ \tag{2} -\frac {2 \sqrt {\frac {a \cos \left (y\right )+c_{1}}{c_{1} +a}}\, \operatorname {InverseJacobiAM}\left (\frac {y}{2}, \frac {\sqrt {2}\, \sqrt {a}}{\sqrt {c_{1} +a}}\right )}{\sqrt {2 a \cos \left (y\right )+2 c_{1}}} &= x +c_{3} \\ \end{align*}
Verification of solutions
\[ \frac {2 \sqrt {\frac {a \cos \left (y\right )+c_{1}}{c_{1} +a}}\, \operatorname {InverseJacobiAM}\left (\frac {y}{2}, \frac {\sqrt {2}\, \sqrt {a}}{\sqrt {c_{1} +a}}\right )}{\sqrt {2 a \cos \left (y\right )+2 c_{1}}} = x +c_{2} \] Verified OK.
\[ -\frac {2 \sqrt {\frac {a \cos \left (y\right )+c_{1}}{c_{1} +a}}\, \operatorname {InverseJacobiAM}\left (\frac {y}{2}, \frac {\sqrt {2}\, \sqrt {a}}{\sqrt {c_{1} +a}}\right )}{\sqrt {2 a \cos \left (y\right )+2 c_{1}}} = x +c_{3} \] 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)+sin(_a)*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: 0.0 (sec). Leaf size: 49
dsolve(diff(diff(y(x),x),x)+a*sin(y(x))=0,y(x), singsol=all)
\begin{align*} \int _{}^{y \left (x \right )}\frac {1}{\sqrt {2 a \cos \left (\textit {\_a} \right )+c_{1}}}d \textit {\_a} -x -c_{2} &= 0 \\ -\left (\int _{}^{y \left (x \right )}\frac {1}{\sqrt {2 a \cos \left (\textit {\_a} \right )+c_{1}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 7.619 (sec). Leaf size: 79
DSolve[a*Sin[y[x]] + y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -2 \operatorname {JacobiAmplitude}\left (\frac {1}{2} \sqrt {(2 a+c_1) (x+c_2){}^2},\frac {4 a}{2 a+c_1}\right ) \\ y(x)\to 2 \operatorname {JacobiAmplitude}\left (\frac {1}{2} \sqrt {(2 a+c_1) (x+c_2){}^2},\frac {4 a}{2 a+c_1}\right ) \\ \end{align*}