Internal problem ID [2378]
Internal file name [OUTPUT/2378_Tuesday_February_27_2024_08_36_32_AM_36735985/index.tex]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 40, page 186
Problem number: 15.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Taylor series method"
Maple gives the following as the ode type
[NONE]
\[ \boxed {y^{\prime \prime }-\sin \left (y x \right )=0} \] With initial conditions \begin {align*} \left [y \left (\frac {\pi }{2}\right ) = 1, y^{\prime }\left (\frac {\pi }{2}\right ) = 1\right ] \end {align*}
With the expansion point for the power series method at \(x = \frac {\pi }{2}\).
The ode does not have its expansion point at \(x = 0\), therefore to simplify the computation of power series expansion, change of variable is made on the independent variable to shift the initial conditions and the expasion point back to zero. The new ode is then solved more easily since the expansion point is now at zero. The solution converted back to the original independent variable. Let \[ t = x -\frac {\pi }{2} \] The ode is converted to be in terms of the new independent variable \(t\). This results in \[ \frac {d^{2}}{d t^{2}}y \left (t \right ) = \sin \left (y \left (t \right ) \left (t +\frac {\pi }{2}\right )\right ) \] With its expansion point and initial conditions now at \(t = 0\). With initial conditions now becoming \begin {align*} y(0) &= 1\\ y'(0) &= 1 \end {align*}
The transformed ODE is now solved. Solving ode using Taylor series method. This gives review on how the Taylor series method works for solving second order ode.
Let \[ y^{\prime \prime }=f\left ( x,y,y^{\prime }\right ) \] Assuming expansion is at \(x_{0}=0\) (we can always shift the actual expansion point to \(0\) by change of variables) and assuming \(f\left ( x,y,y^{\prime }\right ) \) is analytic at \(x_{0}\) which must be the case for an ordinary point. Let initial conditions be \(y\left ( x_{0}\right ) =y_{0}\) and \(y^{\prime }\left ( x_{0}\right ) =y_{0}^{\prime }\). Using Taylor series gives\begin {align*} y\left ( x\right ) & =y\left ( x_{0}\right ) +\left ( x-x_{0}\right ) y^{\prime }\left ( x_{0}\right ) +\frac {\left ( x-x_{0}\right ) ^{2}}{2}y^{\prime \prime }\left ( x_{0}\right ) +\frac {\left ( x-x_{0}\right ) ^{3}}{3!}y^{\prime \prime \prime }\left ( x_{0}\right ) +\cdots \\ & =y_{0}+xy_{0}^{\prime }+\frac {x^{2}}{2}\left . f\right \vert _{x_{0},y_{0},y_{0}^{\prime }}+\frac {x^{3}}{3!}\left . f^{\prime }\right \vert _{x_{0},y_{0},y_{0}^{\prime }}+\cdots \\ & =y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+2\right ) !}\left . \frac {d^{n}f}{dx^{n}}\right \vert _{x_{0},y_{0},y_{0}^{\prime }} \end {align*}
But \begin {align} \frac {df}{dx} & =\frac {\partial f}{\partial x}\frac {dx}{dx}+\frac {\partial f}{\partial y}\frac {dy}{dx}+\frac {\partial f}{\partial y^{\prime }}\frac {dy^{\prime }}{dx}\tag {1}\\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}y^{\prime \prime }\\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}f\\ \frac {d^{2}f}{dx^{2}} & =\frac {d}{dx}\left ( \frac {df}{dx}\right ) \nonumber \\ & =\frac {\partial }{\partial x}\left ( \frac {df}{dx}\right ) +\frac {\partial }{\partial y}\left ( \frac {df}{dx}\right ) y^{\prime }+\frac {\partial }{\partial y^{\prime }}\left ( \frac {df}{dx}\right ) f\tag {2}\\ \frac {d^{3}f}{dx^{3}} & =\frac {d}{dx}\left ( \frac {d^{2}f}{dx^{2}}\right ) \nonumber \\ & =\frac {\partial }{\partial x}\left ( \frac {d^{2}f}{dx^{2}}\right ) +\left ( \frac {\partial }{\partial y}\frac {d^{2}f}{dx^{2}}\right ) y^{\prime }+\frac {\partial }{\partial y^{\prime }}\left ( \frac {d^{2}f}{dx^{2}}\right ) f\tag {3}\\ & \vdots \nonumber \end {align}
And so on. Hence if we name \(F_{0}=f\left ( x,y,y^{\prime }\right ) \) then the above can be written as \begin {align} F_{0} & =f\left ( x,y,y^{\prime }\right ) \tag {4}\\ F_{1} & =\frac {df}{dx}\nonumber \\ & =\frac {dF_{0}}{dx}\nonumber \\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}y^{\prime \prime }\nonumber \\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}f\tag {5}\\ & =\frac {\partial F_{0}}{\partial x}+\frac {\partial F_{0}}{\partial y}y^{\prime }+\frac {\partial F_{0}}{\partial y^{\prime }}F_{0}\nonumber \\ F_{2} & =\frac {d}{dx}\left ( \frac {d}{dx}f\right ) \nonumber \\ & =\frac {d}{dx}\left ( F_{1}\right ) \nonumber \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\nonumber \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{1}}{\partial y^{\prime }}\right ) F_{0}\nonumber \\ & \vdots \nonumber \\ F_{n} & =\frac {d}{dx}\left ( F_{n-1}\right ) \nonumber \\ & =\frac {\partial }{\partial x}F_{n-1}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{n-1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\nonumber \\ & =\frac {\partial }{\partial x}F_{n-1}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{n-1}}{\partial y^{\prime }}\right ) F_{0} \tag {6} \end {align}
Therefore (6) can be used from now on along with \begin {equation} y\left ( x\right ) =y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+2\right ) !}\left . F_{n}\right \vert _{x_{0},y_{0},y_{0}^{\prime }} \tag {7} \end {equation} To find \(y\left ( x\right ) \) series solution around \(x=0\). Hence \begin {align*} F_0 &= \sin \left (y \left (t \right ) \left (t +\frac {\pi }{2}\right )\right )\\ F_1 &= \frac {d F_0}{dt} \\ &= \frac {\partial F_{0}}{\partial t}+ \frac {\partial F_{0}}{\partial y} \frac {d}{d t}y \left (t \right )+ \frac {\partial F_{0}}{\partial \frac {d}{d t}y \left (t \right )} F_0 \\ &= \left (\left (\frac {d}{d t}y \left (t \right )\right ) \left (t +\frac {\pi }{2}\right )+y \left (t \right )\right ) \cos \left (y \left (t \right ) \left (t +\frac {\pi }{2}\right )\right )\\ F_2 &= \frac {d F_1}{dt} \\ &= \frac {\partial F_{1}}{\partial t}+ \frac {\partial F_{1}}{\partial y} \frac {d}{d t}y \left (t \right )+ \frac {\partial F_{1}}{\partial \frac {d}{d t}y \left (t \right )} F_1 \\ &= \frac {\left (\left (4 t +2 \pi \right ) \cos \left (y \left (t \right ) \left (t +\frac {\pi }{2}\right )\right )-4 \left (\left (\frac {d}{d t}y \left (t \right )\right ) \left (t +\frac {\pi }{2}\right )+y \left (t \right )\right )^{2}\right ) \sin \left (y \left (t \right ) \left (t +\frac {\pi }{2}\right )\right )}{4}+2 \cos \left (y \left (t \right ) \left (t +\frac {\pi }{2}\right )\right ) \left (\frac {d}{d t}y \left (t \right )\right )\\ F_3 &= \frac {d F_2}{dt} \\ &= \frac {\partial F_{2}}{\partial t}+ \frac {\partial F_{2}}{\partial y} \frac {d}{d t}y \left (t \right )+ \frac {\partial F_{2}}{\partial \frac {d}{d t}y \left (t \right )} F_2 \\ &= \frac {\left (24 \sin \left (y \left (t \right ) \left (t +\frac {\pi }{2}\right )\right )+\left (-\pi ^{3}-6 \pi ^{2} t -12 \pi \,t^{2}-8 t^{3}\right ) \left (\frac {d}{d t}y \left (t \right )\right )^{3}+\left (-6 \pi ^{2}-24 \pi t -24 t^{2}\right ) y \left (t \right ) \left (\frac {d}{d t}y \left (t \right )\right )^{2}+\left (-24 t -12 \pi \right ) y \left (t \right )^{2} \left (\frac {d}{d t}y \left (t \right )\right )-8 y \left (t \right )^{3}\right ) \cos \left (y \left (t \right ) \left (t +\frac {\pi }{2}\right )\right )}{8}-3 \left (\left (\frac {d}{d t}y \left (t \right )\right ) \left (t +\frac {\pi }{2}\right )+y \left (t \right )\right ) \left (\left (t +\frac {\pi }{2}\right ) \sin \left (y \left (t \right ) \left (t +\frac {\pi }{2}\right )\right )^{2}+2 \left (\frac {d}{d t}y \left (t \right )\right ) \sin \left (y \left (t \right ) \left (t +\frac {\pi }{2}\right )\right )-\frac {\cos \left (\frac {y \left (t \right ) \left (2 t +\pi \right )}{2}\right )^{2} \left (t +\frac {\pi }{2}\right )}{3}\right ) \end {align*}
And so on. Evaluating all the above at initial conditions \(t = 0\) and \(y \left (0\right ) = 1\) and \(y^{\prime }\left (0\right ) = 1\) gives \begin {align*} F_0 &= 1\\ F_1 &= 0\\ F_2 &= -\frac {1}{4} \pi ^{2}-\pi -1\\ F_3 &= -\frac {3}{4} \pi ^{2}-\frac {9}{2} \pi -6 \end {align*}
Substituting all the above in (7) and simplifying gives the solution as \[ y \left (t \right ) = t +1+\frac {t^{2}}{2}-\frac {t^{4} \pi ^{2}}{96}-\frac {t^{4} \pi }{24}-\frac {t^{4}}{24}-\frac {t^{5} \pi ^{2}}{160}-\frac {3 t^{5} \pi }{80}-\frac {t^{5}}{20}+O\left (t^{5}\right ) \] \[ y \left (t \right ) = t +1+\frac {t^{2}}{2}-\frac {t^{4} \pi ^{2}}{96}-\frac {t^{4} \pi }{24}-\frac {t^{4}}{24}-\frac {t^{5} \pi ^{2}}{160}-\frac {3 t^{5} \pi }{80}-\frac {t^{5}}{20}+O\left (t^{5}\right ) \] Unable to also solve using normal power series since not linear ode. Not currently supported. Replacing \(t\) in the above with the original independent variable \(xs\)using \(t = x -\frac {\pi }{2}\) results in \[ y = x -\frac {\pi }{2}+1+\frac {\left (x -\frac {\pi }{2}\right )^{2}}{2}-\frac {\left (x -\frac {\pi }{2}\right )^{4} \pi ^{2}}{96}-\frac {\left (x -\frac {\pi }{2}\right )^{4} \pi }{24}-\frac {\left (x -\frac {\pi }{2}\right )^{4}}{24}-\frac {\left (x -\frac {\pi }{2}\right )^{5} \pi ^{2}}{160}-\frac {3 \left (x -\frac {\pi }{2}\right )^{5} \pi }{80}-\frac {\left (x -\frac {\pi }{2}\right )^{5}}{20}+O\left (\left (x -\frac {\pi }{2}\right )^{5}\right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= x -\frac {\pi }{2}+1+\frac {\left (x -\frac {\pi }{2}\right )^{2}}{2}-\frac {\left (x -\frac {\pi }{2}\right )^{4} \pi ^{2}}{96}-\frac {\left (x -\frac {\pi }{2}\right )^{4} \pi }{24}-\frac {\left (x -\frac {\pi }{2}\right )^{4}}{24}-\frac {\left (x -\frac {\pi }{2}\right )^{5} \pi ^{2}}{160}-\frac {3 \left (x -\frac {\pi }{2}\right )^{5} \pi }{80}-\frac {\left (x -\frac {\pi }{2}\right )^{5}}{20}+O\left (\left (x -\frac {\pi }{2}\right )^{5}\right ) \\ \end{align*}
Verification of solutions
\[ y = x -\frac {\pi }{2}+1+\frac {\left (x -\frac {\pi }{2}\right )^{2}}{2}-\frac {\left (x -\frac {\pi }{2}\right )^{4} \pi ^{2}}{96}-\frac {\left (x -\frac {\pi }{2}\right )^{4} \pi }{24}-\frac {\left (x -\frac {\pi }{2}\right )^{4}}{24}-\frac {\left (x -\frac {\pi }{2}\right )^{5} \pi ^{2}}{160}-\frac {3 \left (x -\frac {\pi }{2}\right )^{5} \pi }{80}-\frac {\left (x -\frac {\pi }{2}\right )^{5}}{20}+O\left (\left (x -\frac {\pi }{2}\right )^{5}\right ) \] 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 -> trying 2nd order, dynamical_symmetries, fully reducible to Abel through one integrating factor of the form G(x,y)/(1+H(x,y)*y)^2 trying 2nd order, integrating factors of the form mu(x,y)/(y)^n, only the singular cases trying symmetries linear in x and y(x) trying differential order: 2; exact nonlinear trying 2nd order, integrating factor of the form mu(y) trying 2nd order, integrating factor of the form mu(x,y) trying 2nd order, integrating factor of the form mu(x,y)/(y)^n, only the general case trying 2nd order, integrating factor of the form mu(y,y) trying differential order: 2; mu polynomial in y trying 2nd order, integrating factor of the form mu(x,y) differential order: 2; looking for linear symmetries -> trying 2nd order, the S-function method -> trying a change of variables {x -> y(x), y(x) -> x} and re-entering methods for the S-function -> trying 2nd order, the S-function method -> trying 2nd order, No Point Symmetries Class V -> trying 2nd order, No Point Symmetries Class V -> trying 2nd order, No Point Symmetries Class V trying 2nd order, integrating factor of the form mu(x,y)/(y)^n, only the general case -> trying 2nd order, dynamical_symmetries, only a reduction of order through one integrating factor of the form G(x,y)/(1+H(x,y)*y)^ --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 3 `, `-> Computing symmetries using: way = 5 `, `-> Computing symmetries using: way = patterns`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 22
Order:=5; dsolve([diff(y(x),x$2)=sin(x*y(x)),y(1/2*Pi) = 1, D(y)(1/2*Pi) = 1],y(x),type='series',x=1/2*Pi);
\[ y \left (x \right ) = 1+\left (-\frac {\pi }{2}+x \right )+\frac {1}{2} \left (-\frac {\pi }{2}+x \right )^{2}-\frac {1}{96} \left (\pi +2\right )^{2} \left (-\frac {\pi }{2}+x \right )^{4}+\operatorname {O}\left (\left (-\frac {\pi }{2}+x \right )^{5}\right ) \]
✓ Solution by Mathematica
Time used: 0.165 (sec). Leaf size: 47
AsymptoticDSolveValue[{y''[x]==Sin[x*y[x]],{y[Pi/2]==1,y'[Pi/2]==1}},y[x],{x,1/2*Pi,4}]
\[ y(x)\to \frac {1}{96} \left (-4-4 \pi -\pi ^2\right ) \left (x-\frac {\pi }{2}\right )^4+\frac {1}{2} \left (x-\frac {\pi }{2}\right )^2+x-\frac {\pi }{2}+1 \]