Internal problem ID [15471]
Internal file name [OUTPUT/15472_Wednesday_May_08_2024_04_01_03_PM_71973850/index.tex
]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 18.1 Integration of differential equation in
series. Power series. Exercises page 171
Problem number: 728.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_y", "second order series method. Ordinary point", "second order series method. Taylor series method"
Maple gives the following as the ode type
[[_2nd_order, _missing_y]]
\[ \boxed {y^{\prime \prime }-y^{\prime } \sin \left (x \right )=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 1] \end {align*}
With the expansion point for the power series method at \(x = 0\).
This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime \prime } + p(x)y^{\prime } + q(x) y &= F \end {align*}
Where here \begin {align*} p(x) &=-\sin \left (x \right )\\ q(x) &=0\\ F &=0 \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }-y^{\prime } \sin \left (x \right ) = 0 \end {align*}
The domain of \(p(x)=-\sin \left (x \right )\) is \[
\{-\infty 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 &= y^{\prime } \sin \left (x \right )\\ F_1 &= \frac {d F_0}{dx} \\ &= \frac {\partial F_{0}}{\partial x}+ \frac {\partial F_{0}}{\partial y} y^{\prime }+ \frac {\partial F_{0}}{\partial y^{\prime }} F_0 \\ &= y^{\prime } \left (\sin \left (x \right )^{2}+\cos \left (x \right )\right )\\ F_2 &= \frac {d F_1}{dx} \\ &= \frac {\partial F_{1}}{\partial x}+ \frac {\partial F_{1}}{\partial y} y^{\prime }+ \frac {\partial F_{1}}{\partial y^{\prime }} F_1 \\ &= -\sin \left (x \right ) \cos \left (x \right ) \left (\cos \left (x \right )-3\right ) y^{\prime }\\ F_3 &= \frac {d F_2}{dx} \\ &= \frac {\partial F_{2}}{\partial x}+ \frac {\partial F_{2}}{\partial y} y^{\prime }+ \frac {\partial F_{2}}{\partial y^{\prime }} F_2 \\ &= \left (\cos \left (x \right )^{4}-6 \cos \left (x \right )^{3}+5 \cos \left (x \right )^{2}+5 \cos \left (x \right )-3\right ) y^{\prime }\\ F_4 &= \frac {d F_3}{dx} \\ &= \frac {\partial F_{3}}{\partial x}+ \frac {\partial F_{3}}{\partial y} y^{\prime }+ \frac {\partial F_{3}}{\partial y^{\prime }} F_3 \\ &= y^{\prime } \sin \left (x \right ) \left (\cos \left (x \right )^{4}-10 \cos \left (x \right )^{3}+23 \cos \left (x \right )^{2}-5 \cos \left (x \right )-8\right ) \end {align*}
And so on. Evaluating all the above at initial conditions \(x = 0\) and \(y \left (0\right ) = 0\) and \(y^{\prime }\left (0\right ) = 1\) gives \begin {align*} F_0 &= 0\\ F_1 &= 1\\ F_2 &= 0\\ F_3 &= 2\\ F_4 &= 0 \end {align*}
Substituting all the above in (7) and simplifying gives the solution as \[
y = x +\frac {x^{3}}{6}+\frac {x^{5}}{60}+O\left (x^{6}\right )
\] \[
y = x +\frac {x^{3}}{6}+\frac {x^{5}}{60}+O\left (x^{6}\right )
\] Since the expansion
point \(x = 0\) is an ordinary, we can also solve this using standard power series Let the solution be
represented as power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n} \] Then \begin {align*} y^{\prime } &= \moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2} \end {align*}
Substituting the above back into the ode gives \begin {align*} \moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2} = \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right ) \sin \left (x \right )\tag {1} \end {align*}
Expanding \(-\sin \left (x \right )\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} -\sin \left (x \right ) &= -x +\frac {1}{6} x^{3}-\frac {1}{120} x^{5}+\frac {1}{5040} x^{7} + \dots \\ &= -x +\frac {1}{6} x^{3}-\frac {1}{120} x^{5}+\frac {1}{5040} x^{7} \end {align*}
Hence the ODE in Eq (1) becomes \[
\left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+\left (-x +\frac {1}{6} x^{3}-\frac {1}{120} x^{5}+\frac {1}{5040} x^{7}\right ) \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right ) = 0
\] Expanding the second term in (1) gives \[
\left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+-x \cdot \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right )+\frac {x^{3}}{6}\cdot \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right )-\frac {x^{5}}{120}\cdot \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right )+\frac {x^{7}}{5040}\cdot \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right ) = 0
\] Which simplifies
to \begin{equation}
\tag{2} \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-n a_{n} x^{n}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\frac {n \,x^{n +2} a_{n}}{6}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-\frac {n \,x^{n +4} a_{n}}{120}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\frac {n \,x^{n +6} a_{n}}{5040}\right ) = 0
\end{equation} The next step is to make all powers of \(x\) be \(n\) in each summation term. Going over each
summation term above with power of \(x\) in it which is not already \(x^{n}\) and adjusting the
power and the corresponding index gives \begin{align*}
\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2} &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +2\right ) a_{n +2} \left (n +1\right ) x^{n} \\
\moverset {\infty }{\munderset {n =1}{\sum }}\frac {n \,x^{n +2} a_{n}}{6} &= \moverset {\infty }{\munderset {n =3}{\sum }}\frac {\left (n -2\right ) a_{n -2} x^{n}}{6} \\
\moverset {\infty }{\munderset {n =1}{\sum }}\left (-\frac {n \,x^{n +4} a_{n}}{120}\right ) &= \moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {\left (n -4\right ) a_{n -4} x^{n}}{120}\right ) \\
\moverset {\infty }{\munderset {n =1}{\sum }}\frac {n \,x^{n +6} a_{n}}{5040} &= \moverset {\infty }{\munderset {n =7}{\sum }}\frac {\left (n -6\right ) a_{n -6} x^{n}}{5040} \\
\end{align*} Substituting all the above in Eq (2) gives
the following equation where now all powers of \(x\) are the same and equal to \(n\). \begin{equation}
\tag{3} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +2\right ) a_{n +2} \left (n +1\right ) x^{n}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-n a_{n} x^{n}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}\frac {\left (n -2\right ) a_{n -2} x^{n}}{6}\right )+\moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {\left (n -4\right ) a_{n -4} x^{n}}{120}\right )+\left (\moverset {\infty }{\munderset {n =7}{\sum }}\frac {\left (n -6\right ) a_{n -6} x^{n}}{5040}\right ) = 0
\end{equation} \(n=1\)
gives \[
6 a_{3}-a_{1}=0
\] Which after substituting earlier equations, simplifies to \[
a_{3} = \frac {a_{1}}{6}
\] \(n=2\) gives \[
12 a_{4}-2 a_{2} = 0
\] Which after
substituting earlier equations, simplifies to \[
12 a_{4} = 0
\] Or \[
a_{4} = 0
\] \(n=3\) gives \[
20 a_{5}-3 a_{3}+\frac {a_{1}}{6}=0
\] Which after substituting earlier
equations, simplifies to \[
a_{5} = \frac {a_{1}}{60}
\] \(n=4\) gives \[
30 a_{6}-4 a_{4}+\frac {a_{2}}{3} = 0
\] Which after substituting earlier equations, simplifies to
\[
30 a_{6} = 0
\] Or \[
a_{6} = 0
\] \(n=5\) gives \[
42 a_{7}-5 a_{5}+\frac {a_{3}}{2}-\frac {a_{1}}{120}=0
\] Which after substituting earlier equations, simplifies to \[
a_{7} = \frac {a_{1}}{5040}
\] For \(7\le n\), the
recurrence equation is \begin{equation}
\tag{4} \left (n +2\right ) a_{n +2} \left (n +1\right )-n a_{n}+\frac {\left (n -2\right ) a_{n -2}}{6}-\frac {\left (n -4\right ) a_{n -4}}{120}+\frac {\left (n -6\right ) a_{n -6}}{5040} = 0
\end{equation} Solving for \(a_{n +2}\), gives \begin{align*}
\tag{5} a_{n +2}&= \frac {5040 n a_{n}-n a_{n -6}+42 n a_{n -4}-840 n a_{n -2}+6 a_{n -6}-168 a_{n -4}+1680 a_{n -2}}{5040 \left (n +2\right ) \left (n +1\right )} \\
&= \frac {n a_{n}}{\left (n +2\right ) \left (n +1\right )}+\frac {\left (-n +6\right ) a_{n -6}}{5040 \left (n +2\right ) \left (n +1\right )}+\frac {\left (42 n -168\right ) a_{n -4}}{5040 \left (n +2\right ) \left (n +1\right )}+\frac {\left (-840 n +1680\right ) a_{n -2}}{5040 \left (n +2\right ) \left (n +1\right )} \\
\end{align*} And so on. Therefore the solution is
\begin {align*} y &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\\ &= a_{3} x^{3}+a_{2} x^{2}+a_{1} x +a_{0} + \dots \end {align*}
Substituting the values for \(a_{n}\) found above, the solution becomes \[
y = a_{0}+a_{1} x +\frac {1}{6} a_{1} x^{3}+\frac {1}{60} a_{1} x^{5}+\dots
\]
Collecting terms, the solution becomes \begin{equation}
\tag{3} y = a_{0}+\left (x +\frac {1}{6} x^{3}+\frac {1}{60} x^{5}\right ) a_{1}+O\left (x^{6}\right )
\end{equation} At \(x = 0\) the solution above becomes \[
y = c_{1} +\left (x +\frac {1}{6} x^{3}+\frac {1}{60} x^{5}\right ) c_{2} +O\left (x^{6}\right )
\] \[
y = x +\frac {x^{3}}{6}+\frac {x^{5}}{60}+O\left (x^{6}\right )
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= x +\frac {x^{3}}{6}+\frac {x^{5}}{60}+O\left (x^{6}\right ) \\
\tag{2} y &= x +\frac {x^{3}}{6}+\frac {x^{5}}{60}+O\left (x^{6}\right ) \\
\end{align*} Verification of solutions
\[
y = x +\frac {x^{3}}{6}+\frac {x^{5}}{60}+O\left (x^{6}\right )
\] Verified OK.
\[
y = x +\frac {x^{3}}{6}+\frac {x^{5}}{60}+O\left (x^{6}\right )
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }=y^{\prime } \sin \left (x \right ), y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=u \left (x \right ) \sin \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=u \left (x \right ) \sin \left (x \right ) \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{u \left (x \right )}=\sin \left (x \right ) \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{u \left (x \right )}d x =\int \sin \left (x \right )d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (u \left (x \right )\right )=-\cos \left (x \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )={\mathrm e}^{-\cos \left (x \right )+c_{1}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )={\mathrm e}^{-\cos \left (x \right )+c_{1}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }={\mathrm e}^{-\cos \left (x \right )+c_{1}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int {\mathrm e}^{-\cos \left (x \right )+c_{1}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\int {\mathrm e}^{-\cos \left (x \right )+c_{1}}d x +c_{2} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 14
\[
y \left (x \right ) = x +\frac {1}{6} x^{3}+\frac {1}{60} x^{5}+\operatorname {O}\left (x^{6}\right )
\]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 19
\[
y(x)\to \frac {x^5}{30}-\frac {x^3}{6}+x
\]
23.5.2 Maple step by step solution
`Methods for second order ODEs:
--- Trying classification methods ---
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
<- LODE missing y successful`
Order:=6;
dsolve([diff(y(x),x$2)-diff(y(x),x)*sin(x)=0,y(0) = 0, D(y)(0) = 1],y(x),type='series',x=0);
AsymptoticDSolveValue[{y''[x]+Sin[x]*y'[x]==0,{y[0]==0,y'[0]==1}},y[x],{x,0,5}]