Internal problem ID [5576]
Internal file name [OUTPUT/4824_Sunday_June_05_2022_03_06_57_PM_24876484/index.tex]
Book: A FIRST COURSE IN DIFFERENTIAL EQUATIONS with Modeling Applications.
Dennis G. Zill. 9th edition. Brooks/Cole. CA, USA.
Section: Chapter 6. SERIES SOLUTIONS OF LINEAR EQUATIONS. Exercises. 6.2 page
239
Problem number: 21.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference not integer"
Maple gives the following as the ode type
[_Laguerre]
\[ \boxed {2 x y^{\prime \prime }-\left (2 x +3\right ) y^{\prime }+y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ 2 x y^{\prime \prime }+\left (-2 x -3\right ) y^{\prime }+y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}
Where \begin {align*} p(x) &= -\frac {2 x +3}{2 x}\\ q(x) &= \frac {1}{2 x}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 2 x y^{\prime \prime }+\left (-2 x -3\right ) y^{\prime }+y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} 2 x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (-2 x -3\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r -1} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )-3 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ 2 x^{-1+r} a_{0} r \left (-1+r \right )-3 r a_{0} x^{-1+r} = 0 \] Or \[ \left (2 x^{-1+r} r \left (-1+r \right )-3 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (-5+2 r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 2 r^{2}-5 r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {5}{2}}\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (-5+2 r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {5}{2}}, 0\right ]\).
Since \(r_1 - r_2 = {\frac {5}{2}}\) is not an integer, then we can construct two linearly independent solutions \begin {align*} y_{1}\left (x \right ) &= x^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= x^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {5}{2}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n} \end {align*}
We start by finding \(y_{1}\left (x \right )\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 2 a_{n} \left (n +r \right ) \left (n +r -1\right )-2 a_{n -1} \left (n +r -1\right )-3 a_{n} \left (n +r \right )+a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -1} \left (2 n +2 r -3\right )}{2 n^{2}+4 n r +2 r^{2}-5 n -5 r}\tag {4} \] Which for the root \(r = {\frac {5}{2}}\) becomes \[ a_{n} = \frac {2 a_{n -1} \left (n +1\right )}{n \left (2 n +5\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {5}{2}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ a_{1}=\frac {-1+2 r}{2 r^{2}-r -3} \] Which for the root \(r = {\frac {5}{2}}\) becomes \[ a_{1}={\frac {4}{7}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1+2 r}{2 r^{2}-r -3}\) | \(\frac {4}{7}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {1+2 r}{2 r^{3}+3 r^{2}-5 r -6} \] Which for the root \(r = {\frac {5}{2}}\) becomes \[ a_{2}={\frac {4}{21}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1+2 r}{2 r^{2}-r -3}\) | \(\frac {4}{7}\) |
| \(a_{2}\) | \(\frac {1+2 r}{2 r^{3}+3 r^{2}-5 r -6}\) | \(\frac {4}{21}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {3+2 r}{2 r^{4}+9 r^{3}+4 r^{2}-21 r -18} \] Which for the root \(r = {\frac {5}{2}}\) becomes \[ a_{3}={\frac {32}{693}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1+2 r}{2 r^{2}-r -3}\) | \(\frac {4}{7}\) |
| \(a_{2}\) | \(\frac {1+2 r}{2 r^{3}+3 r^{2}-5 r -6}\) | \(\frac {4}{21}\) |
| \(a_{3}\) | \(\frac {3+2 r}{2 r^{4}+9 r^{3}+4 r^{2}-21 r -18}\) | \(\frac {32}{693}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {5+2 r}{2 r^{5}+17 r^{4}+40 r^{3}-5 r^{2}-102 r -72} \] Which for the root \(r = {\frac {5}{2}}\) becomes \[ a_{4}={\frac {80}{9009}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1+2 r}{2 r^{2}-r -3}\) | \(\frac {4}{7}\) |
| \(a_{2}\) | \(\frac {1+2 r}{2 r^{3}+3 r^{2}-5 r -6}\) | \(\frac {4}{21}\) |
| \(a_{3}\) | \(\frac {3+2 r}{2 r^{4}+9 r^{3}+4 r^{2}-21 r -18}\) | \(\frac {32}{693}\) |
| \(a_{4}\) | \(\frac {5+2 r}{2 r^{5}+17 r^{4}+40 r^{3}-5 r^{2}-102 r -72}\) | \(\frac {80}{9009}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {7+2 r}{\left (r +5\right ) \left (2 r^{5}+17 r^{4}+40 r^{3}-5 r^{2}-102 r -72\right )} \] Which for the root \(r = {\frac {5}{2}}\) becomes \[ a_{5}={\frac {64}{45045}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1+2 r}{2 r^{2}-r -3}\) | \(\frac {4}{7}\) |
| \(a_{2}\) | \(\frac {1+2 r}{2 r^{3}+3 r^{2}-5 r -6}\) | \(\frac {4}{21}\) |
| \(a_{3}\) | \(\frac {3+2 r}{2 r^{4}+9 r^{3}+4 r^{2}-21 r -18}\) | \(\frac {32}{693}\) |
| \(a_{4}\) | \(\frac {5+2 r}{2 r^{5}+17 r^{4}+40 r^{3}-5 r^{2}-102 r -72}\) | \(\frac {80}{9009}\) |
| \(a_{5}\) | \(\frac {7+2 r}{\left (r +5\right ) \left (2 r^{5}+17 r^{4}+40 r^{3}-5 r^{2}-102 r -72\right )}\) | \(\frac {64}{45045}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {5}{2}} \left (a_{0}+a_{1} x +a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}+a_{6} x^{6}\dots \right ) \\ &= x^{\frac {5}{2}} \left (1+\frac {4 x}{7}+\frac {4 x^{2}}{21}+\frac {32 x^{3}}{693}+\frac {80 x^{4}}{9009}+\frac {64 x^{5}}{45045}+O\left (x^{6}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Eq (2B) derived above is now used to find all \(b_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 2 b_{n} \left (n +r \right ) \left (n +r -1\right )-2 b_{n -1} \left (n +r -1\right )-3 \left (n +r \right ) b_{n}+b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {b_{n -1} \left (2 n +2 r -3\right )}{2 n^{2}+4 n r +2 r^{2}-5 n -5 r}\tag {4} \] Which for the root \(r = 0\) becomes \[ b_{n} = \frac {b_{n -1} \left (2 n -3\right )}{n \left (2 n -5\right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ b_{1}=\frac {-1+2 r}{2 r^{2}-r -3} \] Which for the root \(r = 0\) becomes \[ b_{1}={\frac {1}{3}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-1+2 r}{2 r^{2}-r -3}\) | \(\frac {1}{3}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {1+2 r}{2 r^{3}+3 r^{2}-5 r -6} \] Which for the root \(r = 0\) becomes \[ b_{2}=-{\frac {1}{6}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-1+2 r}{2 r^{2}-r -3}\) | \(\frac {1}{3}\) |
| \(b_{2}\) | \(\frac {1+2 r}{2 r^{3}+3 r^{2}-5 r -6}\) | \(-{\frac {1}{6}}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {3+2 r}{2 r^{4}+9 r^{3}+4 r^{2}-21 r -18} \] Which for the root \(r = 0\) becomes \[ b_{3}=-{\frac {1}{6}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-1+2 r}{2 r^{2}-r -3}\) | \(\frac {1}{3}\) |
| \(b_{2}\) | \(\frac {1+2 r}{2 r^{3}+3 r^{2}-5 r -6}\) | \(-{\frac {1}{6}}\) |
| \(b_{3}\) | \(\frac {3+2 r}{2 r^{4}+9 r^{3}+4 r^{2}-21 r -18}\) | \(-{\frac {1}{6}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {5+2 r}{2 r^{5}+17 r^{4}+40 r^{3}-5 r^{2}-102 r -72} \] Which for the root \(r = 0\) becomes \[ b_{4}=-{\frac {5}{72}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-1+2 r}{2 r^{2}-r -3}\) | \(\frac {1}{3}\) |
| \(b_{2}\) | \(\frac {1+2 r}{2 r^{3}+3 r^{2}-5 r -6}\) | \(-{\frac {1}{6}}\) |
| \(b_{3}\) | \(\frac {3+2 r}{2 r^{4}+9 r^{3}+4 r^{2}-21 r -18}\) | \(-{\frac {1}{6}}\) |
| \(b_{4}\) | \(\frac {5+2 r}{2 r^{5}+17 r^{4}+40 r^{3}-5 r^{2}-102 r -72}\) | \(-{\frac {5}{72}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {7+2 r}{\left (r +5\right ) \left (2 r^{5}+17 r^{4}+40 r^{3}-5 r^{2}-102 r -72\right )} \] Which for the root \(r = 0\) becomes \[ b_{5}=-{\frac {7}{360}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-1+2 r}{2 r^{2}-r -3}\) | \(\frac {1}{3}\) |
| \(b_{2}\) | \(\frac {1+2 r}{2 r^{3}+3 r^{2}-5 r -6}\) | \(-{\frac {1}{6}}\) |
| \(b_{3}\) | \(\frac {3+2 r}{2 r^{4}+9 r^{3}+4 r^{2}-21 r -18}\) | \(-{\frac {1}{6}}\) |
| \(b_{4}\) | \(\frac {5+2 r}{2 r^{5}+17 r^{4}+40 r^{3}-5 r^{2}-102 r -72}\) | \(-{\frac {5}{72}}\) |
| \(b_{5}\) | \(\frac {7+2 r}{\left (r +5\right ) \left (2 r^{5}+17 r^{4}+40 r^{3}-5 r^{2}-102 r -72\right )}\) | \(-{\frac {7}{360}}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}\dots \\ &= 1+\frac {x}{3}-\frac {x^{2}}{6}-\frac {x^{3}}{6}-\frac {5 x^{4}}{72}-\frac {7 x^{5}}{360}+O\left (x^{6}\right ) \end {align*}
Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} x^{\frac {5}{2}} \left (1+\frac {4 x}{7}+\frac {4 x^{2}}{21}+\frac {32 x^{3}}{693}+\frac {80 x^{4}}{9009}+\frac {64 x^{5}}{45045}+O\left (x^{6}\right )\right ) + c_{2} \left (1+\frac {x}{3}-\frac {x^{2}}{6}-\frac {x^{3}}{6}-\frac {5 x^{4}}{72}-\frac {7 x^{5}}{360}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {5}{2}} \left (1+\frac {4 x}{7}+\frac {4 x^{2}}{21}+\frac {32 x^{3}}{693}+\frac {80 x^{4}}{9009}+\frac {64 x^{5}}{45045}+O\left (x^{6}\right )\right )+c_{2} \left (1+\frac {x}{3}-\frac {x^{2}}{6}-\frac {x^{3}}{6}-\frac {5 x^{4}}{72}-\frac {7 x^{5}}{360}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {5}{2}} \left (1+\frac {4 x}{7}+\frac {4 x^{2}}{21}+\frac {32 x^{3}}{693}+\frac {80 x^{4}}{9009}+\frac {64 x^{5}}{45045}+O\left (x^{6}\right )\right )+c_{2} \left (1+\frac {x}{3}-\frac {x^{2}}{6}-\frac {x^{3}}{6}-\frac {5 x^{4}}{72}-\frac {7 x^{5}}{360}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {5}{2}} \left (1+\frac {4 x}{7}+\frac {4 x^{2}}{21}+\frac {32 x^{3}}{693}+\frac {80 x^{4}}{9009}+\frac {64 x^{5}}{45045}+O\left (x^{6}\right )\right )+c_{2} \left (1+\frac {x}{3}-\frac {x^{2}}{6}-\frac {x^{3}}{6}-\frac {5 x^{4}}{72}-\frac {7 x^{5}}{360}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x y^{\prime \prime }+\left (-2 x -3\right ) y^{\prime }+y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {y}{2 x}+\frac {\left (2 x +3\right ) y^{\prime }}{2 x} \\ \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} \\ {} & {} & y^{\prime \prime }-\frac {\left (2 x +3\right ) y^{\prime }}{2 x}+\frac {y}{2 x}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=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 x +3}{2 x}, P_{3}\left (x \right )=\frac {1}{2 x}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=-\frac {3}{2} \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=0 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 2 x y^{\prime \prime }+\left (-2 x -3\right ) y^{\prime }+y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (-5+2 r \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (2 k -3+2 r \right )-a_{k} \left (2 k +2 r -1\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-5+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {5}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 2 \left (k -\frac {3}{2}+r \right ) \left (k +1+r \right ) a_{k +1}-2 \left (k -\frac {1}{2}+r \right ) a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {\left (2 k +2 r -1\right ) a_{k}}{\left (2 k -3+2 r \right ) \left (k +1+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {\left (2 k -1\right ) a_{k}}{\left (2 k -3\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +1}=\frac {\left (2 k -1\right ) a_{k}}{\left (2 k -3\right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {5}{2} \\ {} & {} & a_{k +1}=\frac {\left (2 k +4\right ) a_{k}}{\left (2 k +2\right ) \left (k +\frac {7}{2}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {5}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {5}{2}}, a_{k +1}=\frac {\left (2 k +4\right ) a_{k}}{\left (2 k +2\right ) \left (k +\frac {7}{2}\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {5}{2}}\right ), a_{k +1}=\frac {\left (2 k -1\right ) a_{k}}{\left (2 k -3\right ) \left (k +1\right )}, b_{k +1}=\frac {\left (2 k +4\right ) b_{k}}{\left (2 k +2\right ) \left (k +\frac {7}{2}\right )}\right ] \end {array} \]
Maple trace Kovacic algorithm successful
`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)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Reducible group (found an exponential solution) Group is reducible, not completely reducible Solution has integrals. Trying a special function solution free of integrals... -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre <- Kummer successful <- special function solution successful Solution using Kummer functions still has integrals. Trying a hypergeometric solution. -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE -> Trying to convert hypergeometric functions to elementary form... <- elementary form for at least one hypergeometric solution is achieved - returning with no uncomputed integrals <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 44
Order:=6; dsolve(2*x*diff(y(x),x$2)-(3+2*x)*diff(y(x),x)+y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{\frac {5}{2}} \left (1+\frac {4}{7} x +\frac {4}{21} x^{2}+\frac {32}{693} x^{3}+\frac {80}{9009} x^{4}+\frac {64}{45045} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (1+\frac {1}{3} x -\frac {1}{6} x^{2}-\frac {1}{6} x^{3}-\frac {5}{72} x^{4}-\frac {7}{360} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 85
AsymptoticDSolveValue[2*x*y''[x]-(3+2*x)*y'[x]+y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \left (-\frac {7 x^5}{360}-\frac {5 x^4}{72}-\frac {x^3}{6}-\frac {x^2}{6}+\frac {x}{3}+1\right )+c_1 \left (\frac {64 x^5}{45045}+\frac {80 x^4}{9009}+\frac {32 x^3}{693}+\frac {4 x^2}{21}+\frac {4 x}{7}+1\right ) x^{5/2} \]