Internal problem ID [6100]
Internal file name [OUTPUT/5348_Sunday_June_05_2022_03_35_01_PM_69825552/index.tex]
Book: An introduction to Ordinary Differential Equations. Earl A. Coddington. Dover. NY
1961
Section: Chapter 6. Existence and uniqueness of solutions to systems and nth order equations.
Page 238
Problem number: 5(c).
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]]
\[ \boxed {y^{\prime \prime }+\sin \left (y\right )=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 2] \end {align*}
Multiplying the ode by \(y^{\prime }\) gives \[ y^{\prime } y^{\prime \prime }+y^{\prime } \sin \left (y\right ) = 0 \] Integrating the above w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime } y^{\prime \prime }+y^{\prime } \sin \left (y\right )\right )d x &= 0 \\ \frac {{y^{\prime }}^{2}}{2}-\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 \cos \left (y\right )+2 c_{1}} \tag {1} \\ y^{\prime }&=-\sqrt {2 \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 \cos \left (y \right )+2 c_{1}}}d y &= \int d x \\ \frac {2 \sqrt {\frac {\cos \left (y\right )+c_{1}}{1+c_{1}}}\, \operatorname {InverseJacobiAM}\left (\frac {y}{2}, \frac {2}{\sqrt {2 c_{1} +2}}\right )}{\sqrt {2 \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 \cos \left (y \right )+2 c_{1}}}d y &= \int d x \\ -\frac {2 \sqrt {\frac {\cos \left (y\right )+c_{1}}{1+c_{1}}}\, \operatorname {InverseJacobiAM}\left (\frac {y}{2}, \frac {2}{\sqrt {2 c_{1} +2}}\right )}{\sqrt {2 \cos \left (y\right )+2 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 (y\right )+c_{1}}{1+c_{1}}}\, \operatorname {InverseJacobiAM}\left (\frac {y}{2}, \frac {2}{\sqrt {2 c_{1} +2}}\right )}{\sqrt {2 \cos \left (y\right )+2 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 \(y = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} \text {Expression too large to display} \end {align*}
substituting \(y^{\prime } = 2\) and \(x = 0\) in the above gives \begin {align*} \text {Expression too large to display}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). There is no solution for the constants of integrations. This solution is removed.
Looking at the Second solution \begin {align*} -\frac {2 \sqrt {\frac {\cos \left (y\right )+c_{1}}{1+c_{1}}}\, \operatorname {InverseJacobiAM}\left (\frac {y}{2}, \frac {2}{\sqrt {2 c_{1} +2}}\right )}{\sqrt {2 \cos \left (y\right )+2 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 \(y = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = c_{3}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} \text {Expression too large to display} \end {align*}
substituting \(y^{\prime } = 2\) and \(x = 0\) in the above gives \begin {align*} \text {Expression too large to display}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{3}\}\). There is no solution for the constants of integrations. This solution is removed.
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*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) = -\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 {\sin \left (y \right )}{p} \end {align*}
Where \(f(y)=-\sin \left (y \right )\) and \(g(p)=\frac {1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p}} \,dp &= -\sin \left (y \right ) \,d y \\ \int { \frac {1}{\frac {1}{p}} \,dp} &= \int {-\sin \left (y \right ) \,d y} \\ \frac {p^{2}}{2}&=\cos \left (y \right )+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (y \right )^{2}}{2}-\cos \left (y \right )-c_{1} = 0 \] Initial conditions are used to solve for \(c_{1}\). Substituting \(y=0\) and \(p=2\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 1-c_{1} = 0 \end {align*}
The solutions are \begin {align*} c_{1} = 1 \end {align*}
Trying the constant \begin {align*} c_{1} = 1 \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {p^{2}}{2}-\cos \left (y \right )-1 = 0 \end {align*}
The constant \(c_{1} = 1\) 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*} \frac {{y^{\prime }}^{2}}{2}-\cos \left (y\right )-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 \cos \left (y\right )+2} \tag {1} \\ y^{\prime }&=-\sqrt {2 \cos \left (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 {1}{\sqrt {2 \cos \left (y \right )+2}}d y &= \int {dx}\\ \operatorname {InverseJacobiAM}\left (\frac {y}{2}, 1\right )&= x +c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=0\) and \(y=0\) 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*} \operatorname {InverseJacobiAM}\left (\frac {y}{2}, 1\right ) = x \end {align*}
The constant \(c_{2} = 0\) gives valid solution.
Solving equation (2)
Integrating both sides gives \begin {align*} \int -\frac {1}{\sqrt {2 \cos \left (y \right )+2}}d y &= \int {dx}\\ -\operatorname {InverseJacobiAM}\left (\frac {y}{2}, 1\right )&= x +c_{3} \end {align*}
Initial conditions are used to solve for \(c_{3}\). Substituting \(x=0\) and \(y=0\) 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*} -\operatorname {InverseJacobiAM}\left (\frac {y}{2}, 1\right ) = x \end {align*}
The constant \(c_{3} = 0\) gives valid solution.
Initial conditions are used to solve for the constants of integration.
Summary
The solution(s) found are the following \begin{align*} \tag{1} \operatorname {InverseJacobiAM}\left (\frac {y}{2}, 1\right ) &= x \\ \tag{2} -\operatorname {InverseJacobiAM}\left (\frac {y}{2}, 1\right ) &= x \\ \end{align*}
Verification of solutions
\[ \operatorname {InverseJacobiAM}\left (\frac {y}{2}, 1\right ) = x \] Verified OK.
\[ -\operatorname {InverseJacobiAM}\left (\frac {y}{2}, 1\right ) = x \] 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)+sin(_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: 1.296 (sec). Leaf size: 23
dsolve([diff(y(x),x$2)+sin(y(x))=0,y(0) = 0, D(y)(0) = 2],y(x), singsol=all)
\[ y \left (x \right ) = \operatorname {RootOf}\left (-\left (\int _{0}^{\textit {\_Z}}\sec \left (\frac {\textit {\_a}}{2}\right ) \operatorname {csgn}\left (\cos \left (\frac {\textit {\_a}}{2}\right )\right )d \textit {\_a} \right )+2 x \right ) \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[{y''[x]+Sin[y[x]]==0,{y[0]==0,y'[0]==2}},y[x],x,IncludeSingularSolutions -> True]
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