Internal problem ID [1780]
Internal file name [OUTPUT/1781_Sunday_June_05_2022_02_31_05_AM_62074486/index.tex]
Book: Differential equations and their applications, 3rd ed., M. Braun
Section: Section 2.8, Series solutions. Page 195
Problem number: 15.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Ordinary point", "second order series method. Taylor series method"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime }+y^{\prime } t +{\mathrm e}^{t} y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 0] \end {align*}
With the expansion point for the power series method at \(t = 0\).
This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime \prime } + p(t)y^{\prime } + q(t) y &= F \end {align*}
Where here \begin {align*} p(t) &=t\\ q(t) &={\mathrm e}^{t}\\ F &=0 \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+y^{\prime } t +{\mathrm e}^{t} y = 0 \end {align*}
The domain of \(p(t)=t\) 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 } t -{\mathrm e}^{t} y\\ F_1 &= \frac {d F_0}{dt} \\ &= \frac {\partial F_{0}}{\partial t}+ \frac {\partial F_{0}}{\partial y} y^{\prime }+ \frac {\partial F_{0}}{\partial y^{\prime }} F_0 \\ &= \left (t^{2}-{\mathrm e}^{t}-1\right ) y^{\prime }+y \,{\mathrm e}^{t} \left (t -1\right )\\ F_2 &= \frac {d F_1}{dt} \\ &= \frac {\partial F_{1}}{\partial t}+ \frac {\partial F_{1}}{\partial y} y^{\prime }+ \frac {\partial F_{1}}{\partial y^{\prime }} F_1 \\ &= {\mathrm e}^{2 t} y+\left (\left (2 t -2\right ) {\mathrm e}^{t}-t^{3}+3 t \right ) y^{\prime }-y \,{\mathrm e}^{t} \left (t^{2}-t -1\right )\\ F_3 &= \frac {d F_2}{dt} \\ &= \frac {\partial F_{2}}{\partial t}+ \frac {\partial F_{2}}{\partial y} y^{\prime }+ \frac {\partial F_{2}}{\partial y^{\prime }} F_2 \\ &= \left (y^{\prime }+\left (-2 t +4\right ) y\right ) {\mathrm e}^{2 t}+\left (\left (-3 t^{2}+5 t +1\right ) {\mathrm e}^{t}+t^{4}-6 t^{2}+3\right ) y^{\prime }+y \,{\mathrm e}^{t} \left (t^{3}-t^{2}-4 t +2\right )\\ F_4 &= \frac {d F_3}{dt} \\ &= \frac {\partial F_{3}}{\partial t}+ \frac {\partial F_{3}}{\partial y} y^{\prime }+ \frac {\partial F_{3}}{\partial y^{\prime }} F_3 \\ &= \left (\left (-3 t +6\right ) y^{\prime }+3 y \left (t^{2}-3 t +\frac {5}{3}\right )\right ) {\mathrm e}^{2 t}-y \,{\mathrm e}^{3 t}+\left (\left (4 t^{3}-9 t^{2}-6 t +8\right ) {\mathrm e}^{t}-t^{5}+10 t^{3}-15 t \right ) y^{\prime }-y \,{\mathrm e}^{t} \left (t^{4}-t^{3}-8 t^{2}+6 t +5\right ) \end {align*}
And so on. Evaluating all the above at initial conditions \(t = 0\) and \(y \left (0\right ) = 1\) and \(y^{\prime }\left (0\right ) = 0\) gives \begin {align*} F_0 &= -1\\ F_1 &= -1\\ F_2 &= 2\\ F_3 &= 6\\ F_4 &= -1 \end {align*}
Substituting all the above in (7) and simplifying gives the solution as \[
y = 1-\frac {t^{2}}{2}-\frac {t^{3}}{6}+\frac {t^{4}}{12}+\frac {t^{5}}{20}-\frac {t^{6}}{720}+O\left (t^{6}\right )
\] \[
y = 1-\frac {t^{2}}{2}-\frac {t^{3}}{6}+\frac {t^{4}}{12}+\frac {t^{5}}{20}-\frac {t^{6}}{720}+O\left (t^{6}\right )
\] Since the expansion
point \(t = 0\) is an ordinary, we can also solve this using standard power series Let the solution be
represented as power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n} \] Then \begin {align*} y^{\prime } &= \moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1}\\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} t^{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} t^{n -2} = -\left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1}\right ) t -{\mathrm e}^{t} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right )\tag {1} \end {align*}
Expanding \({\mathrm e}^{t}\) as Taylor series around \(t=0\) and keeping only the first \(6\) terms gives \begin {align*} {\mathrm e}^{t} &= 1+t +\frac {1}{2} t^{2}+\frac {1}{6} t^{3}+\frac {1}{24} t^{4}+\frac {1}{120} t^{5}+\frac {1}{720} t^{6} + \dots \\ &= 1+t +\frac {1}{2} t^{2}+\frac {1}{6} t^{3}+\frac {1}{24} t^{4}+\frac {1}{120} t^{5}+\frac {1}{720} t^{6} \end {align*}
Hence the ODE in Eq (1) becomes \[
\left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} t^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1}\right ) t +\left (1+t +\frac {1}{2} t^{2}+\frac {1}{6} t^{3}+\frac {1}{24} t^{4}+\frac {1}{120} t^{5}+\frac {1}{720} t^{6}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right ) = 0
\] Expanding the third term in (1) gives \[
\left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} t^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1}\right ) t +\left (1\eslowast \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right )\right )+t \eslowast \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right )+\frac {t^{2}}{2}\eslowast \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right )+\frac {t^{3}}{6}\eslowast \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right )+\frac {t^{4}}{24}\eslowast \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right )+\frac {t^{5}}{120}\eslowast \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right )+\frac {t^{6}}{720}\eslowast \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right ) = 0
\] Which simplifies
to \begin{equation}
\tag{2} \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} t^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}n \,t^{n} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{1+n} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +2} a_{n}}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +3} a_{n}}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +4} a_{n}}{24}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +5} a_{n}}{120}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +6} a_{n}}{720}\right ) = 0
\end{equation} The next step is to make all powers of \(t\) be \(n\) in each summation term. Going over each
summation term above with power of \(t\) in it which is not already \(t^{n}\) and adjusting the power
and the corresponding index gives \begin{align*}
\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} t^{n -2} &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +2\right ) a_{n +2} \left (1+n \right ) t^{n} \\
\moverset {\infty }{\munderset {n =0}{\sum }}t^{1+n} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} t^{n} \\
\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +2} a_{n}}{2} &= \moverset {\infty }{\munderset {n =2}{\sum }}\frac {a_{n -2} t^{n}}{2} \\
\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +3} a_{n}}{6} &= \moverset {\infty }{\munderset {n =3}{\sum }}\frac {a_{n -3} t^{n}}{6} \\
\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +4} a_{n}}{24} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} t^{n}}{24} \\
\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +5} a_{n}}{120} &= \moverset {\infty }{\munderset {n =5}{\sum }}\frac {a_{n -5} t^{n}}{120} \\
\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +6} a_{n}}{720} &= \moverset {\infty }{\munderset {n =6}{\sum }}\frac {a_{n -6} t^{n}}{720} \\
\end{align*} Substituting all the above in Eq (2) gives the
following equation where now all powers of \(t\) 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 (1+n \right ) t^{n}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}n \,t^{n} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} t^{n}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}\frac {a_{n -2} t^{n}}{2}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}\frac {a_{n -3} t^{n}}{6}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} t^{n}}{24}\right )+\left (\moverset {\infty }{\munderset {n =5}{\sum }}\frac {a_{n -5} t^{n}}{120}\right )+\left (\moverset {\infty }{\munderset {n =6}{\sum }}\frac {a_{n -6} t^{n}}{720}\right ) = 0
\end{equation} \(n=0\) gives \[
2 a_{2}+a_{0}=0
\]
\[
a_{2} = -\frac {a_{0}}{2}
\] \(n=1\) 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 \[
12 a_{4}+3 a_{2}+a_{1}+\frac {a_{0}}{2}=0
\] Which
after substituting earlier equations, simplifies to \[
a_{4} = \frac {a_{0}}{12}-\frac {a_{1}}{12}
\] \(n=3\) gives \[
20 a_{5}+4 a_{3}+a_{2}+\frac {a_{1}}{2}+\frac {a_{0}}{6}=0
\] Which after substituting
earlier equations, simplifies to \[
a_{5} = \frac {a_{0}}{20}+\frac {a_{1}}{24}
\] \(n=4\) gives \[
30 a_{6}+5 a_{4}+a_{3}+\frac {a_{2}}{2}+\frac {a_{1}}{6}+\frac {a_{0}}{24}=0
\] Which after substituting earlier equations,
simplifies to \[
a_{6} = -\frac {a_{0}}{720}+\frac {7 a_{1}}{360}
\] \(n=5\) gives \[
42 a_{7}+6 a_{5}+a_{4}+\frac {a_{3}}{2}+\frac {a_{2}}{6}+\frac {a_{1}}{24}+\frac {a_{0}}{120}=0
\] Which after substituting earlier equations, simplifies to \[
a_{7} = -\frac {3 a_{0}}{560}-\frac {a_{1}}{1008}
\] For \(6\le n\), the
recurrence equation is \begin{equation}
\tag{4} \left (n +2\right ) a_{n +2} \left (1+n \right )+n a_{n}+a_{n}+a_{n -1}+\frac {a_{n -2}}{2}+\frac {a_{n -3}}{6}+\frac {a_{n -4}}{24}+\frac {a_{n -5}}{120}+\frac {a_{n -6}}{720} = 0
\end{equation} Solving for \(a_{n +2}\), gives \begin{align*}
\tag{5} a_{n +2}&= -\frac {720 n a_{n}+720 a_{n}+a_{n -6}+6 a_{n -5}+30 a_{n -4}+120 a_{n -3}+360 a_{n -2}+720 a_{n -1}}{720 \left (n +2\right ) \left (1+n \right )} \\
&= -\frac {\left (720 n +720\right ) a_{n}}{720 \left (n +2\right ) \left (1+n \right )}-\frac {a_{n -6}}{720 \left (n +2\right ) \left (1+n \right )}-\frac {a_{n -5}}{120 \left (n +2\right ) \left (1+n \right )}-\frac {a_{n -4}}{24 \left (n +2\right ) \left (1+n \right )}-\frac {a_{n -3}}{6 \left (n +2\right ) \left (1+n \right )}-\frac {a_{n -2}}{2 \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} t^{n}\\ &= a_{3} t^{3}+a_{2} t^{2}+a_{1} t +a_{0} + \dots \end {align*}
Substituting the values for \(a_{n}\) found above, the solution becomes \[
y = a_{0}+a_{1} t -\frac {a_{0} t^{2}}{2}+\left (-\frac {a_{0}}{6}-\frac {a_{1}}{3}\right ) t^{3}+\left (\frac {a_{0}}{12}-\frac {a_{1}}{12}\right ) t^{4}+\left (\frac {a_{0}}{20}+\frac {a_{1}}{24}\right ) t^{5}+\dots
\]
Collecting terms, the solution becomes \begin{equation}
\tag{3} y = \left (1-\frac {1}{2} t^{2}-\frac {1}{6} t^{3}+\frac {1}{12} t^{4}+\frac {1}{20} t^{5}\right ) a_{0}+\left (t -\frac {1}{3} t^{3}-\frac {1}{12} t^{4}+\frac {1}{24} t^{5}\right ) a_{1}+O\left (t^{6}\right )
\end{equation} At \(t = 0\) the solution above becomes \[
y = \left (1-\frac {1}{2} t^{2}-\frac {1}{6} t^{3}+\frac {1}{12} t^{4}+\frac {1}{20} t^{5}\right ) c_{1} +\left (t -\frac {1}{3} t^{3}-\frac {1}{12} t^{4}+\frac {1}{24} t^{5}\right ) c_{2} +O\left (t^{6}\right )
\] \[
y = 1-\frac {t^{2}}{2}-\frac {t^{3}}{6}+\frac {t^{4}}{12}+\frac {t^{5}}{20}+O\left (t^{6}\right )
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= 1-\frac {t^{2}}{2}-\frac {t^{3}}{6}+\frac {t^{4}}{12}+\frac {t^{5}}{20}-\frac {t^{6}}{720}+O\left (t^{6}\right ) \\
\tag{2} y &= 1-\frac {t^{2}}{2}-\frac {t^{3}}{6}+\frac {t^{4}}{12}+\frac {t^{5}}{20}+O\left (t^{6}\right ) \\
\end{align*} Verification of solutions
\[
y = 1-\frac {t^{2}}{2}-\frac {t^{3}}{6}+\frac {t^{4}}{12}+\frac {t^{5}}{20}-\frac {t^{6}}{720}+O\left (t^{6}\right )
\] Verified OK.
\[
y = 1-\frac {t^{2}}{2}-\frac {t^{3}}{6}+\frac {t^{4}}{12}+\frac {t^{5}}{20}+O\left (t^{6}\right )
\] Verified OK. Maple trace
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 18
\[
y \left (t \right ) = 1-\frac {1}{2} t^{2}-\frac {1}{6} t^{3}+\frac {1}{12} t^{4}+\frac {1}{20} t^{5}+\operatorname {O}\left (t^{6}\right )
\]
✓ Solution by Mathematica
Time used: 0.001 (sec). Leaf size: 33
\[
y(t)\to \frac {t^5}{20}+\frac {t^4}{12}-\frac {t^3}{6}-\frac {t^2}{2}+1
\]
`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 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 2nd order exact linear
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 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 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 Lie symmetry methods, 2nd order ---
`, `-> Computing symmetries using: way = 3`[0, y]
Order:=6;
dsolve([diff(y(t),t$2)+t*diff(y(t),t)+exp(t)*y(t)=0,y(0) = 1, D(y)(0) = 0],y(t),type='series',t=0);
AsymptoticDSolveValue[{y''[t]+t*y'[t]+Exp[t]*y[t]==0,{y[0]==1,y'[0]==0}},y[t],{t,0,5}]