13.8 problem 8

Internal problem ID [1249]
Internal file name [OUTPUT/1250_Sunday_June_05_2022_02_06_31_AM_53835349/index.tex]

Book: Elementary differential equations with boundary value problems. William F. Trench. Brooks/Cole 2001
Section: Chapter 7 Series Solutions of Linear Second Equations. 7.3 SERIES SOLUTIONS NEAR AN ORDINARY POINT II. Exercises 7.3. Page 338
Problem number: 8.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "exact linear second order ode", "second_order_integrable_as_is", "second order series method. Ordinary point", "second order series method. Taylor series method"

Maple gives the following as the ode type

[[_2nd_order, _exact, _linear, _homogeneous]]

\[ \boxed {\left (2 x^{2}-3 x +2\right ) y^{\prime \prime }-\left (4-6 x \right ) y^{\prime }+2 y=0} \] With initial conditions \begin {align*} [y \left (1\right ) = 1, y^{\prime }\left (1\right ) = -1] \end {align*}

With the expansion point for the power series method at \(x = 1\).

13.8.1 Existence and uniqueness analysis

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) &=\frac {6 x -4}{2 x^{2}-3 x +2}\\ q(x) &=\frac {2}{2 x^{2}-3 x +2}\\ F &=0 \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+\frac {\left (6 x -4\right ) y^{\prime }}{2 x^{2}-3 x +2}+\frac {2 y}{2 x^{2}-3 x +2} = 0 \end {align*}

The domain of \(p(x)=\frac {6 x -4}{2 x^{2}-3 x +2}\) is \[ \{-\infty

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 \[ t = x -1 \] The ode is converted to be in terms of the new independent variable \(t\). This results in \[ \left (2 \left (t +1\right )^{2}-3 t -1\right ) \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )+\left (6 t +2\right ) \left (\frac {d}{d t}y \left (t \right )\right )+2 y \left (t \right ) = 0 \] 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 &= -\frac {2 \left (3 \left (\frac {d}{d t}y \left (t \right )\right ) t +\frac {d}{d t}y \left (t \right )+y \left (t \right )\right )}{2 t^{2}+t +1}\\ 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 \\ &= \frac {\left (44 t^{2}+30 t -2\right ) \left (\frac {d}{d t}y \left (t \right )\right )+\left (20 t +6\right ) y \left (t \right )}{\left (2 t^{2}+t +1\right )^{2}}\\ 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 (-400 t^{3}-416 t^{2}+52 t +44\right ) \left (\frac {d}{d t}y \left (t \right )\right )-208 y \left (t \right ) \left (t^{2}+\frac {8}{13} t -\frac {3}{52}\right )}{\left (2 t^{2}+t +1\right )^{3}}\\ 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 (4384 t^{4}+6160 t^{3}-1096 t^{2}-1948 t -156\right ) \left (\frac {d}{d t}y \left (t \right )\right )+2464 \left (t^{3}+\frac {145}{154} t^{2}-\frac {51}{308} t -\frac {9}{88}\right ) y \left (t \right )}{\left (2 t^{2}+t +1\right )^{4}}\\ F_4 &= \frac {d F_3}{dt} \\ &= \frac {\partial F_{3}}{\partial t}+ \frac {\partial F_{3}}{\partial y} \frac {d}{d t}y \left (t \right )+ \frac {\partial F_{3}}{\partial \frac {d}{d t}y \left (t \right )} F_3 \\ &= \frac {\left (-56448 t^{5}-100224 t^{4}+22752 t^{3}+63232 t^{2}+10320 t -1264\right ) \left (\frac {d}{d t}y \left (t \right )\right )-33408 y \left (t \right ) \left (t^{4}+\frac {37}{29} t^{3}-\frac {37}{116} t^{2}-\frac {431}{1044} t -\frac {19}{696}\right )}{\left (2 t^{2}+t +1\right )^{5}} \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 &= 0\\ F_1 &= 8\\ F_2 &= -32\\ F_3 &= -96\\ F_4 &= 2176 \end {align*}

Substituting all the above in (7) and simplifying gives the solution as \[ y \left (t \right ) = 1-t +\frac {4 t^{3}}{3}-\frac {4 t^{4}}{3}-\frac {4 t^{5}}{5}+\frac {136 t^{6}}{45}+O\left (t^{6}\right ) \] \[ y \left (t \right ) = 1-t +\frac {4 t^{3}}{3}-\frac {4 t^{4}}{3}-\frac {4 t^{5}}{5}+\frac {136 t^{6}}{45}+O\left (t^{6}\right ) \] Since the expansion point \(t = 0\) is an ordinary, we can also solve this using standard power series The ode is normalized to be \[ \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) \left (2 t^{2}+t +1\right )+\left (6 t +2\right ) \left (\frac {d}{d t}y \left (t \right )\right )+2 y \left (t \right ) = 0 \] Let the solution be represented as power series of the form \[ y \left (t \right ) = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n} \] Then \begin {align*} \frac {d}{d t}y \left (t \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1}\\ \frac {d^{2}}{d t^{2}}y \left (t \right ) &= \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*} \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} t^{n -2}\right ) \left (2 t^{2}+t +1\right )+\left (6 t +2\right ) \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1}\right )+2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right ) = 0\tag {1} \end {align*}

Which simplifies to \begin{equation} \tag{2} \left (\moverset {\infty }{\munderset {n =2}{\sum }}2 t^{n} a_{n} n \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}n \,t^{n -1} a_{n} \left (n -1\right )\right )+\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 }}6 n a_{n} t^{n}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 n a_{n} t^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} t^{n}\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 \,t^{n -1} a_{n} \left (n -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (n +1\right ) a_{n +1} n \,t^{n} \\ \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 (n +1\right ) t^{n} \\ \moverset {\infty }{\munderset {n =1}{\sum }}2 n a_{n} t^{n -1} &= \moverset {\infty }{\munderset {n =0}{\sum }}2 \left (n +1\right ) a_{n +1} t^{n} \\ \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 =2}{\sum }}2 t^{n} a_{n} n \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\left (n +1\right ) a_{n +1} n \,t^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +2\right ) a_{n +2} \left (n +1\right ) t^{n}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}6 n a_{n} t^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 \left (n +1\right ) a_{n +1} t^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} t^{n}\right ) = 0 \end{equation} \(n=0\) gives \[ 2 a_{2}+2 a_{1}+2 a_{0}=0 \] \[ a_{2} = -a_{0}-a_{1} \] \(n=1\) gives \[ 6 a_{2}+6 a_{3}+8 a_{1}=0 \] Which after substituting earlier equations, simplifies to \[ a_{3} = a_{0}-\frac {a_{1}}{3} \] For \(2\le n\), the recurrence equation is \begin{equation} \tag{4} 2 n a_{n} \left (n -1\right )+\left (n +1\right ) a_{n +1} n +\left (n +2\right ) a_{n +2} \left (n +1\right )+6 n a_{n}+2 \left (n +1\right ) a_{n +1}+2 a_{n} = 0 \end{equation} Solving for \(a_{n +2}\), gives \begin{align*} \tag{5} a_{n +2}&= -\frac {2 n a_{n}+n a_{n +1}+2 a_{n}+2 a_{n +1}}{n +2} \\ &= -\frac {\left (2 n +2\right ) a_{n}}{n +2}-a_{n +1} \\ \end{align*} For \(n = 2\) the recurrence equation gives \[ 18 a_{2}+12 a_{3}+12 a_{4} = 0 \] Which after substituting the earlier terms found becomes \[ a_{4} = \frac {a_{0}}{2}+\frac {11 a_{1}}{6} \] For \(n = 3\) the recurrence equation gives \[ 32 a_{3}+20 a_{4}+20 a_{5} = 0 \] Which after substituting the earlier terms found becomes \[ a_{5} = -\frac {21 a_{0}}{10}-\frac {13 a_{1}}{10} \] For \(n = 4\) the recurrence equation gives \[ 50 a_{4}+30 a_{5}+30 a_{6} = 0 \] Which after substituting the earlier terms found becomes \[ a_{6} = \frac {19 a_{0}}{15}-\frac {79 a_{1}}{45} \] For \(n = 5\) the recurrence equation gives \[ 72 a_{5}+42 a_{6}+42 a_{7} = 0 \] Which after substituting the earlier terms found becomes \[ a_{7} = \frac {7 a_{0}}{3}+\frac {251 a_{1}}{63} \] And so on. Therefore the solution is \begin {align*} y \left (t \right ) &= \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 \[ y \left (t \right ) = a_{0}+a_{1} t +\left (-a_{0}-a_{1}\right ) t^{2}+\left (a_{0}-\frac {a_{1}}{3}\right ) t^{3}+\left (\frac {a_{0}}{2}+\frac {11 a_{1}}{6}\right ) t^{4}+\left (-\frac {21 a_{0}}{10}-\frac {13 a_{1}}{10}\right ) t^{5}+\dots \] Collecting terms, the solution becomes \begin{equation} \tag{3} y \left (t \right ) = \left (1-t^{2}+t^{3}+\frac {1}{2} t^{4}-\frac {21}{10} t^{5}\right ) a_{0}+\left (t -t^{2}-\frac {1}{3} t^{3}+\frac {11}{6} t^{4}-\frac {13}{10} t^{5}\right ) a_{1}+O\left (t^{6}\right ) \end{equation} At \(t = 0\) the solution above becomes \[ y \left (t \right ) = \left (1-t^{2}+t^{3}+\frac {1}{2} t^{4}-\frac {21}{10} t^{5}\right ) c_{1} +\left (t -t^{2}-\frac {1}{3} t^{3}+\frac {11}{6} t^{4}-\frac {13}{10} t^{5}\right ) c_{2} +O\left (t^{6}\right ) \] \[ y \left (t \right ) = 1+\frac {4 t^{3}}{3}-\frac {4 t^{4}}{3}-\frac {4 t^{5}}{5}-t +O\left (t^{6}\right ) \] Replacing \(t\) in the above with the original independent variable \(xs\)using \(t = x -1\) results in \[ y = 2-x +\frac {4 \left (x -1\right )^{3}}{3}-\frac {4 \left (x -1\right )^{4}}{3}-\frac {4 \left (x -1\right )^{5}}{5}+\frac {136 \left (x -1\right )^{6}}{45}+O\left (\left (x -1\right )^{6}\right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= 2-x +\frac {4 \left (x -1\right )^{3}}{3}-\frac {4 \left (x -1\right )^{4}}{3}-\frac {4 \left (x -1\right )^{5}}{5}+\frac {136 \left (x -1\right )^{6}}{45}+O\left (\left (x -1\right )^{6}\right ) \\ \end{align*}

Verification of solutions

\[ y = 2-x +\frac {4 \left (x -1\right )^{3}}{3}-\frac {4 \left (x -1\right )^{4}}{3}-\frac {4 \left (x -1\right )^{5}}{5}+\frac {136 \left (x -1\right )^{6}}{45}+O\left (\left (x -1\right )^{6}\right ) \] Verified OK.

13.8.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\left (2 x^{2}-3 x +2\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (6 x -4\right ) y^{\prime }+2 y=0, y \left (1\right )=1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=-1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {2 y}{2 x^{2}-3 x +2}-\frac {2 \left (3 x -2\right ) y^{\prime }}{2 x^{2}-3 x +2} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {2 \left (3 x -2\right ) y^{\prime }}{2 x^{2}-3 x +2}+\frac {2 y}{2 x^{2}-3 x +2}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {2 \left (3 x -2\right )}{2 x^{2}-3 x +2}, P_{3}\left (x \right )=\frac {2}{2 x^{2}-3 x +2}\right ] \\ {} & \circ & \left (\frac {\mathrm {I} \sqrt {7}}{4}+x -\frac {3}{4}\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-\frac {\mathrm {I} \sqrt {7}}{4}+\frac {3}{4} \\ {} & {} & \left (\left (\frac {\mathrm {I} \sqrt {7}}{4}+x -\frac {3}{4}\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-\frac {\mathrm {I} \sqrt {7}}{4}+\frac {3}{4}}}}=0 \\ {} & \circ & \left (\frac {\mathrm {I} \sqrt {7}}{4}+x -\frac {3}{4}\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-\frac {\mathrm {I} \sqrt {7}}{4}+\frac {3}{4} \\ {} & {} & \left (\left (\frac {\mathrm {I} \sqrt {7}}{4}+x -\frac {3}{4}\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-\frac {\mathrm {I} \sqrt {7}}{4}+\frac {3}{4}}}}=0 \\ {} & \circ & x =-\frac {\mathrm {I} \sqrt {7}}{4}+\frac {3}{4}\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=-\frac {\mathrm {I} \sqrt {7}}{4}+\frac {3}{4} \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (2 x^{2}-3 x +2\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (6 x -4\right ) y^{\prime }+2 y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -\frac {\mathrm {I} \sqrt {7}}{4}+\frac {3}{4}\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (2 u^{2}-\mathrm {I} u \sqrt {7}\right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (6 u -\frac {3 \,\mathrm {I} \sqrt {7}}{2}+\frac {1}{2}\right ) \left (\frac {d}{d u}y \left (u \right )\right )+2 y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -\frac {\mathrm {I} \sqrt {7}\, r \left (\mathrm {I} \sqrt {7}+7+14 r \right ) a_{0} u^{-1+r}}{14}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-\frac {\mathrm {I} \sqrt {7}\, \left (k +r +1\right ) \left (\mathrm {I} \sqrt {7}+14 k +21+14 r \right ) a_{k +1}}{14}+2 a_{k} \left (k +r +1\right )^{2}\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -\frac {\mathrm {I}}{14} \sqrt {7}\, r \left (\mathrm {I} \sqrt {7}+7+14 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -\frac {1}{2}-\frac {\mathrm {I} \sqrt {7}}{14}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -\left (k +r +1\right ) \left (\mathrm {I} \left (k +r +\frac {3}{2}\right ) a_{k +1} \sqrt {7}-\frac {a_{k +1}}{2}+\left (-2 k -2 r -2\right ) a_{k}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {4 a_{k} \left (k +r +1\right )}{-1+2 \,\mathrm {I} \sqrt {7}\, k +2 \,\mathrm {I} \sqrt {7}\, r +3 \,\mathrm {I} \sqrt {7}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {4 a_{k} \left (k +1\right )}{-1+2 \,\mathrm {I} \sqrt {7}\, k +3 \,\mathrm {I} \sqrt {7}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +1}=\frac {4 a_{k} \left (k +1\right )}{-1+2 \,\mathrm {I} \sqrt {7}\, k +3 \,\mathrm {I} \sqrt {7}}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =\frac {\mathrm {I} \sqrt {7}}{4}+x -\frac {3}{4} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (\frac {\mathrm {I} \sqrt {7}}{4}+x -\frac {3}{4}\right )^{k}, a_{k +1}=\frac {4 a_{k} \left (k +1\right )}{-1+2 \,\mathrm {I} \sqrt {7}\, k +3 \,\mathrm {I} \sqrt {7}}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{2}-\frac {\mathrm {I} \sqrt {7}}{14} \\ {} & {} & a_{k +1}=\frac {4 a_{k} \left (k +\frac {1}{2}-\frac {\mathrm {I} \sqrt {7}}{14}\right )}{-1+2 \,\mathrm {I} \sqrt {7}\, k +2 \,\mathrm {I} \sqrt {7}\, \left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {7}}{14}\right )+3 \,\mathrm {I} \sqrt {7}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {1}{2}-\frac {\mathrm {I} \sqrt {7}}{14} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -\frac {1}{2}-\frac {\mathrm {I} \sqrt {7}}{14}}, a_{k +1}=\frac {4 a_{k} \left (k +\frac {1}{2}-\frac {\mathrm {I} \sqrt {7}}{14}\right )}{-1+2 \,\mathrm {I} \sqrt {7}\, k +2 \,\mathrm {I} \sqrt {7}\, \left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {7}}{14}\right )+3 \,\mathrm {I} \sqrt {7}}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =\frac {\mathrm {I} \sqrt {7}}{4}+x -\frac {3}{4} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (\frac {\mathrm {I} \sqrt {7}}{4}+x -\frac {3}{4}\right )^{k -\frac {1}{2}-\frac {\mathrm {I} \sqrt {7}}{14}}, a_{k +1}=\frac {4 a_{k} \left (k +\frac {1}{2}-\frac {\mathrm {I} \sqrt {7}}{14}\right )}{-1+2 \,\mathrm {I} \sqrt {7}\, k +2 \,\mathrm {I} \sqrt {7}\, \left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {7}}{14}\right )+3 \,\mathrm {I} \sqrt {7}}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (\frac {\mathrm {I} \sqrt {7}}{4}+x -\frac {3}{4}\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (\frac {\mathrm {I} \sqrt {7}}{4}+x -\frac {3}{4}\right )^{k -\frac {1}{2}-\frac {\mathrm {I} \sqrt {7}}{14}}\right ), a_{k +1}=\frac {4 a_{k} \left (k +1\right )}{-1+2 \,\mathrm {I} \sqrt {7}\, k +3 \,\mathrm {I} \sqrt {7}}, b_{k +1}=\frac {4 b_{k} \left (k +\frac {1}{2}-\frac {\mathrm {I} \sqrt {7}}{14}\right )}{-1+2 \,\mathrm {I} \sqrt {7}\, k +2 \,\mathrm {I} \sqrt {7}\, \left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {7}}{14}\right )+3 \,\mathrm {I} \sqrt {7}}\right ] \end {array} \]

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
checking if the LODE is of Euler type 
trying a symmetry of the form [xi=0, eta=F(x)] 
   One independent solution has integrals. Trying a hypergeometric solution free of integrals... 
   -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
   <- hyper3 successful: received ODE is equivalent to the 2F1 ODE 
   -> Trying to convert hypergeometric functions to elementary form... 
   <- elementary form is not straightforward to achieve - returning hypergeometric solution free of uncomputed integrals 
<- linear_1 successful`
 

✓ Solution by Maple

Time used: 0.0 (sec). Leaf size: 18

Order:=6; 
dsolve([(2-3*x+2*x^2)*diff(y(x),x$2)-(4-6*x)*diff(y(x),x)+2*y(x)=0,y(1) = 1, D(y)(1) = -1],y(x),type='series',x=1);
 

\[ y \left (x \right ) = 1-\left (x -1\right )+\frac {4}{3} \left (x -1\right )^{3}-\frac {4}{3} \left (x -1\right )^{4}-\frac {4}{5} \left (x -1\right )^{5}+\operatorname {O}\left (\left (x -1\right )^{6}\right ) \]

✓ Solution by Mathematica

Time used: 0.001 (sec). Leaf size: 35

AsymptoticDSolveValue[{(2-3*x+2*x^2)*y''[x]-(4-6*x)*y'[x]+2*y[x]==0,{y[1]==1,y'[1]==-1}},y[x],{x,1,5}]
 

\[ y(x)\to -\frac {4}{5} (x-1)^5-\frac {4}{3} (x-1)^4+\frac {4}{3} (x-1)^3-x+2 \]