2.4.47 Problem 44
Internal
problem
ID
[10210]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
4.0
Problem
number
:
44
Date
solved
:
Thursday, November 27, 2025 at 10:26:30 AM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
\begin{align*}
y^{\prime \prime } \cos \left (x \right )+2 y^{\prime } x -x y&=0 \\
\end{align*}
Series expansion around
\(x=0\).
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 &= -\frac {x \left (2 y^{\prime }-y\right )}{\cos \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 \\ &= \sec \left (x \right ) \left (2 x^{2} \left (2 y^{\prime }-y\right ) \sec \left (x \right )+\left (-2 x \tan \left (x \right )+x -2\right ) y^{\prime }+y \left (x \tan \left (x \right )+1\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 \\ &= 4 \left (\left (\frac {\left (x +1\right ) \cos \left (x \right )^{2}}{2}+\left (\left (-1+\frac {x}{2}\right ) \sin \left (x \right )-x^{2}+3 x \right ) \cos \left (x \right )-2 x^{3}+3 \sin \left (x \right ) x^{2}-x \right ) y^{\prime }+\left (-\frac {\cos \left (x \right )^{2} x}{4}+\frac {\left (\frac {x^{2}}{2}-3 x +\sin \left (x \right )\right ) \cos \left (x \right )}{2}+x^{3}-\frac {3 \sin \left (x \right ) x^{2}}{2}+\frac {x}{2}\right ) y\right ) \sec \left (x \right )^{3}\\ 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 \\ &= -8 \sec \left (x \right )^{4} \left (\left (\left (-\frac {3}{4}+\frac {3 x}{8}\right ) \cos \left (x \right )^{3}+\left (\left (-\frac {x}{4}-\frac {3}{4}\right ) \sin \left (x \right )-\frac {3}{2}+\frac {27 x^{2}}{8}+\frac {9 x}{4}\right ) \cos \left (x \right )^{2}+\left (\left (-7 x +\frac {9}{4} x^{2}\right ) \sin \left (x \right )+6 x^{2}+\frac {3}{2}-\frac {3 x}{4}-\frac {3 x^{3}}{2}\right ) \cos \left (x \right )-2 x \left (\left (-3 x^{2}-\frac {3}{4}\right ) \sin \left (x \right )+x^{3}+\frac {11 x}{4}\right )\right ) y^{\prime }+y \left (\frac {3 \cos \left (x \right )^{3}}{8}+\left (\frac {3}{4}-\frac {x}{2}+\frac {\sin \left (x \right ) x}{8}-\frac {7 x^{2}}{4}\right ) \cos \left (x \right )^{2}+\left (\left (-\frac {1}{2} x^{2}+\frac {7}{2} x \right ) \sin \left (x \right )-\frac {3}{4}-3 x^{2}+\frac {x^{3}}{2}\right ) \cos \left (x \right )+x \left (\left (-3 x^{2}-\frac {3}{4}\right ) \sin \left (x \right )+x^{3}+\frac {11 x}{4}\right )\right )\right )\\ 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 \\ &= 16 \left (\left (\frac {\left (-3-\frac {x}{2}\right ) \cos \left (x \right )^{4}}{4}+\frac {\left (-3+\left (-\frac {x}{2}+1\right ) \sin \left (x \right )-\frac {87 x}{4}+7 x^{2}\right ) \cos \left (x \right )^{3}}{2}+\left (\left (5-7 x -\frac {27}{8} x^{2}\right ) \sin \left (x \right )+6 x^{2}-5 x +\frac {3}{2}+\frac {97 x^{3}}{8}\right ) \cos \left (x \right )^{2}+\left (\left (-3-25 x^{2}+\frac {3}{2} x +6 x^{3}\right ) \sin \left (x \right )+10 x^{3}+\frac {35 x}{2}-\frac {11 x^{2}}{2}-2 x^{4}\right ) \cos \left (x \right )+5 \left (\frac {5}{2} x^{2}+2 x^{4}\right ) \sin \left (x \right )-3 x -2 x^{5}-\frac {35 x^{3}}{2}\right ) y^{\prime }+\left (\frac {\cos \left (x \right )^{4} x}{16}+\frac {\left (1-\frac {11 x^{2}}{4}-\sin \left (x \right )+\frac {45 x}{2}\right ) \cos \left (x \right )^{3}}{4}+\frac {\left (\left (-5+\frac {11}{4} x +\frac {15}{4} x^{2}\right ) \sin \left (x \right )+5 x -\frac {15 x^{2}}{4}-\frac {99 x^{3}}{8}\right ) \cos \left (x \right )^{2}}{2}+\left (\frac {\left (3+25 x^{2}-\frac {15}{4} x^{3}\right ) \sin \left (x \right )}{2}+\frac {9 x^{2}}{8}-\frac {35 x}{4}-5 x^{3}+\frac {3 x^{4}}{4}\right ) \cos \left (x \right )+x \left (\frac {3}{2}+5 \left (-x^{3}-\frac {5}{4} x \right ) \sin \left (x \right )+x^{4}+\frac {35 x^{2}}{4}\right )\right ) y\right ) \sec \left (x \right )^{5} \end{align*}
And so on. Evaluating all the above at initial conditions \(x = 0\) and \(y \left (0\right ) = y \left (0\right )\) and \(y^{\prime }\left (0\right ) = y^{\prime }\left (0\right )\) gives
\begin{align*} F_0 &= 0\\ F_1 &= -2 y^{\prime }\left (0\right )+y \left (0\right )\\ F_2 &= 2 y^{\prime }\left (0\right )\\ F_3 &= 6 y^{\prime }\left (0\right )-3 y \left (0\right )\\ F_4 &= -12 y^{\prime }\left (0\right )+4 y \left (0\right ) \end{align*}
Substituting all the above in (7) and simplifying gives the solution as
\[
y = \left (1+\frac {1}{6} x^{3}-\frac {1}{40} x^{5}\right ) y \left (0\right )+\left (x -\frac {1}{3} x^{3}+\frac {1}{12} x^{4}+\frac {1}{20} x^{5}\right ) y^{\prime }\left (0\right )+O\left (x^{6}\right )
\]
Since the expansion point
\(x = 0\) is
an ordinary point, then this can also be solved using the standard power series method. 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*} \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right ) \cos \left (x \right )+2 \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right ) x -x \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right ) = 0\tag {1} \end{align*}
Hence the ODE in Eq (1) becomes
\[
\left (1-\frac {1}{2} x^{2}+\frac {1}{24} x^{4}-\frac {1}{720} x^{6}\right ) \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+2 \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right ) x -x \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right ) = 0
\]
Expanding the first term in (1) gives
\[
1\cdot \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )-\frac {x^{2}}{2}\cdot \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+\frac {x^{4}}{24}\cdot \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )-\frac {x^{6}}{720}\cdot \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+2 \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right ) x -x \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right ) = 0
\]
Which simplifies to
\begin{equation}
\tag{2} \moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {n \,x^{n +4} a_{n} \left (n -1\right )}{720}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}\frac {n \,x^{n +2} a_{n} \left (n -1\right )}{24}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {n a_{n} x^{n} \left (n -1\right )}{2}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 n a_{n} x^{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n} a_{n}\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 }}\left (-\frac {n \,x^{n +4} a_{n} \left (n -1\right )}{720}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {\left (n -4\right ) a_{n -4} \left (n -5\right ) x^{n}}{720}\right ) \\
\moverset {\infty }{\munderset {n =2}{\sum }}\frac {n \,x^{n +2} a_{n} \left (n -1\right )}{24} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {\left (n -2\right ) a_{n -2} \left (n -3\right ) x^{n}}{24} \\
\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 (1+n \right ) x^{n} \\
\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n}\right ) \\
\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} \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {\left (n -4\right ) a_{n -4} \left (n -5\right ) x^{n}}{720}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {\left (n -2\right ) a_{n -2} \left (n -3\right ) x^{n}}{24}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {n a_{n} x^{n} \left (n -1\right )}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +2\right ) a_{n +2} \left (1+n \right ) x^{n}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 n a_{n} x^{n}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n}\right ) = 0
\end{equation}
\(n=1\) gives
\[
6 a_{3}+2 a_{1}-a_{0}=0
\]
Which after substituting earlier equations, simplifies to
\[
a_{3} = \frac {a_{0}}{6}-\frac {a_{1}}{3}
\]
\(n=2\)
gives
\[
3 a_{2}+12 a_{4}-a_{1}=0
\]
Which after substituting earlier equations, simplifies to
\[
a_{4} = \frac {a_{1}}{12}
\]
\(n=3\) gives
\[
3 a_{3}+20 a_{5}-a_{2} = 0
\]
Which after substituting
earlier equations, simplifies to
\[
\frac {a_{0}}{2}-a_{1}+20 a_{5} = 0
\]
Or
\[
a_{5} = -\frac {a_{0}}{40}+\frac {a_{1}}{20}
\]
\(n=4\) gives
\[
\frac {a_{2}}{12}+2 a_{4}+30 a_{6}-a_{3} = 0
\]
Which after substituting earlier equations,
simplifies to
\[
\frac {a_{1}}{2}+30 a_{6}-\frac {a_{0}}{6} = 0
\]
Or
\[
a_{6} = \frac {a_{0}}{180}-\frac {a_{1}}{60}
\]
\(n=5\) gives
\[
\frac {a_{3}}{4}+42 a_{7}-a_{4} = 0
\]
Which after substituting earlier equations, simplifies to
\[
\frac {a_{0}}{24}-\frac {a_{1}}{6}+42 a_{7} = 0
\]
Or
\[
a_{7} = -\frac {a_{0}}{1008}+\frac {a_{1}}{252}
\]
For
\(6\le n\),
the recurrence equation is
\begin{equation}
\tag{4} -\frac {\left (n -4\right ) a_{n -4} \left (n -5\right )}{720}+\frac {\left (n -2\right ) a_{n -2} \left (n -3\right )}{24}-\frac {n a_{n} \left (n -1\right )}{2}+\left (n +2\right ) a_{n +2} \left (1+n \right )+2 n a_{n}-a_{n -1} = 0
\end{equation}
Solving for
\(a_{n +2}\), gives
\begin{align*}
\tag{5} a_{n +2}&= \frac {360 n^{2} a_{n}+n^{2} a_{n -4}-30 n^{2} a_{n -2}-1800 n a_{n}-9 n a_{n -4}+150 n a_{n -2}+20 a_{n -4}-180 a_{n -2}+720 a_{n -1}}{720 \left (n +2\right ) \left (1+n \right )} \\
&= \frac {\left (360 n^{2}-1800 n \right ) a_{n}}{720 \left (n +2\right ) \left (1+n \right )}+\frac {\left (n^{2}-9 n +20\right ) a_{n -4}}{720 \left (n +2\right ) \left (1+n \right )}+\frac {\left (-30 n^{2}+150 n -180\right ) a_{n -2}}{720 \left (n +2\right ) \left (1+n \right )}+\frac {a_{n -1}}{\left (n +2\right ) \left (1+n \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
\[
y = a_{0}+a_{1} x +\left (\frac {a_{0}}{6}-\frac {a_{1}}{3}\right ) x^{3}+\frac {a_{1} x^{4}}{12}+\left (-\frac {a_{0}}{40}+\frac {a_{1}}{20}\right ) x^{5}+\dots
\]
Collecting terms, the solution
becomes
\begin{equation}
\tag{3} y = \left (1+\frac {1}{6} x^{3}-\frac {1}{40} x^{5}\right ) a_{0}+\left (x -\frac {1}{3} x^{3}+\frac {1}{12} x^{4}+\frac {1}{20} x^{5}\right ) a_{1}+O\left (x^{6}\right )
\end{equation}
At
\(x = 0\) the solution above becomes
\[
y = \left (1+\frac {1}{6} x^{3}-\frac {1}{40} x^{5}\right ) y \left (0\right )+\left (x -\frac {1}{3} x^{3}+\frac {1}{12} x^{4}+\frac {1}{20} x^{5}\right ) y^{\prime }\left (0\right )+O\left (x^{6}\right )
\]
✓ Maple. Time used: 0.005 (sec). Leaf size: 44
Order:=6;
ode:=cos(x)*diff(diff(y(x),x),x)+2*diff(y(x),x)*x-y(x)*x = 0;
dsolve(ode,y(x),type='series',x=0);
\[
y = \left (1+\frac {1}{6} x^{3}-\frac {1}{40} x^{5}\right ) y \left (0\right )+\left (x -\frac {1}{3} x^{3}+\frac {1}{12} x^{4}+\frac {1}{20} x^{5}\right ) y^{\prime }\left (0\right )+O\left (x^{6}\right )
\]
Maple trace
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
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power \
@ Moebius
-> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x),\
dx)) * 2F1([a1, a2], [b1], f)
-> Trying changes of variables to rationalize or make the ODE simpler
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a pow\
er @ Moebius
-> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(\
x), dx)) * 2F1([a1, a2], [b1], f)
trying a symmetry of the form [xi=0, eta=F(x)]
trying 2nd order exact linear
trying symmetries linear in x and y(x)
trying to convert to a linear ODE with constant coefficients
-> trying with_periodic_functions in the coefficients
--- Trying Lie symmetry methods, 2nd order ---
-> Computing symmetries using: way = 5
[0, u]
<- successful computation of symmetries.
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a pow\
er @ Moebius
-> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(\
x), dx)) * 2F1([a1, a2], [b1], f)
trying a symmetry of the form [xi=0, eta=F(x)]
trying 2nd order exact linear
trying symmetries linear in x and y(x)
trying to convert to a linear ODE with constant coefficients
<- unable to find a useful change of variables
trying a symmetry of the form [xi=0, eta=F(x)]
trying differential order: 2; exact nonlinear
trying symmetries linear in x and y(x)
trying to convert to a linear ODE with constant coefficients
trying 2nd order, integrating factor of the form mu(x,y)
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a pow\
er @ Moebius
-> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(\
x), dx)) * 2F1([a1, a2], [b1], f)
-> Trying changes of variables to rationalize or make the ODE simpler
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a \
power @ Moebius
-> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int\
(r(x), dx)) * 2F1([a1, a2], [b1], f)
trying a symmetry of the form [xi=0, eta=F(x)]
trying 2nd order exact linear
trying symmetries linear in x and y(x)
trying to convert to a linear ODE with constant coefficients
-> trying with_periodic_functions in the coefficients
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a \
power @ Moebius
-> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int\
(r(x), dx)) * 2F1([a1, a2], [b1], f)
trying a symmetry of the form [xi=0, eta=F(x)]
trying 2nd order exact linear
trying symmetries linear in x and y(x)
trying to convert to a linear ODE with constant coefficients
-> trying with_periodic_functions in the coefficients
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a \
power @ Moebius
-> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int\
(r(x), dx)) * 2F1([a1, a2], [b1], f)
trying a symmetry of the form [xi=0, eta=F(x)]
trying 2nd order exact linear
trying symmetries linear in x and y(x)
trying to convert to a linear ODE with constant coefficients
<- unable to find a useful change of variables
trying a symmetry of the form [xi=0, eta=F(x)]
trying to convert to an ODE of Bessel type
-> trying reduction of order to Riccati
trying Riccati sub-methods:
trying Riccati_symmetries
-> trying a symmetry pattern of the form [F(x)*G(y), 0]
-> trying a symmetry pattern of the form [0, F(x)*G(y)]
-> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)]
-> trying with_periodic_functions in the coefficients
--- Trying Lie symmetry methods, 2nd order ---
-> Computing symmetries using: way = 5
[0, y]
<- successful computation of symmetries.
--- Trying Lie symmetry methods, 2nd order ---
-> Computing symmetries using: way = 3
[0, y]
<- successful computation of symmetries.
-> Computing symmetries using: way = 5
✓ Mathematica. Time used: 0.002 (sec). Leaf size: 49
ode=Cos[x]*D[y[x],{x,2}]+2*x*D[y[x],x]-x*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
\[
y(x)\to c_1 \left (-\frac {x^5}{40}+\frac {x^3}{6}+1\right )+c_2 \left (\frac {x^5}{20}+\frac {x^4}{12}-\frac {x^3}{3}+x\right )
\]
✓ Sympy. Time used: 0.408 (sec). Leaf size: 60
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(-x*y(x) + 2*x*Derivative(y(x), x) + cos(x)*Derivative(y(x), (x, 2)),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_ordinary",x0=0,n=6)
\[
y{\left (x \right )} = C_{2} \left (- \frac {x^{5}}{20 \cos ^{2}{\left (x \right )}} + \frac {x^{3}}{6 \cos {\left (x \right )}} + 1\right ) + C_{1} x \left (\frac {x^{4}}{10 \cos ^{2}{\left (x \right )}} + \frac {x^{3}}{12 \cos {\left (x \right )}} - \frac {x^{2}}{3 \cos {\left (x \right )}} + 1\right ) + O\left (x^{6}\right )
\]