Internal problem ID [12073]
Internal file name [OUTPUT/10726_Monday_September_11_2023_12_50_06_AM_57487830/index.tex]
Book: AN INTRODUCTION TO ORDINARY DIFFERENTIAL EQUATIONS by JAMES C.
ROBINSON. Cambridge University Press 2004
Section: Chapter 20, Series solutions of second order linear equations. Exercises page
195
Problem number: 20.2 (iii).
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
[[_Emden, _Fowler], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]
\[ \boxed {2 x y^{\prime \prime }+y^{\prime }-2 y=0} \] 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. \[ 2 x y^{\prime \prime }+y^{\prime }-2 y = 0 \] 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 {1}{2 x}\\ q(x) &= -\frac {1}{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 \[ 2 x y^{\prime \prime }+y^{\prime }-2 y = 0 \] 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 (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )-2 \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 )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 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 a_{n} x^{n +r}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} x^{n +r -1}\right ) \\ \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 )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+\left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ 2 x^{-1+r} a_{0} r \left (-1+r \right )+r a_{0} x^{-1+r} = 0 \] Or \[ \left (2 x^{-1+r} r \left (-1+r \right )+r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (-1+2 r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 2 r^{2}-r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{2}}\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (-1+2 r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{2}}, 0\right ]\).
Since \(r_1 - r_2 = {\frac {1}{2}}\) 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 {1}{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 )+a_{n} \left (n +r \right )-2 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {2 a_{n -1}}{2 n^{2}+4 n r +2 r^{2}-n -r}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{n} = \frac {2 a_{n -1}}{2 n^{2}+n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{2}}\) 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 \[ a_{1}=\frac {2}{2 r^{2}+3 r +1} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{1}={\frac {2}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2}{2 r^{2}+3 r +1}\) | \(\frac {2}{3}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {4}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{2}={\frac {2}{15}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2}{2 r^{2}+3 r +1}\) | \(\frac {2}{3}\) |
| \(a_{2}\) | \(\frac {4}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {2}{15}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {8}{8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{3}={\frac {4}{315}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2}{2 r^{2}+3 r +1}\) | \(\frac {2}{3}\) |
| \(a_{2}\) | \(\frac {4}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {2}{15}\) |
| \(a_{3}\) | \(\frac {8}{8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90}\) | \(\frac {4}{315}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {16}{16 r^{8}+288 r^{7}+2184 r^{6}+9072 r^{5}+22449 r^{4}+33642 r^{3}+29531 r^{2}+13698 r +2520} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{4}={\frac {2}{2835}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2}{2 r^{2}+3 r +1}\) | \(\frac {2}{3}\) |
| \(a_{2}\) | \(\frac {4}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {2}{15}\) |
| \(a_{3}\) | \(\frac {8}{8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90}\) | \(\frac {4}{315}\) |
| \(a_{4}\) | \(\frac {16}{16 r^{8}+288 r^{7}+2184 r^{6}+9072 r^{5}+22449 r^{4}+33642 r^{3}+29531 r^{2}+13698 r +2520}\) | \(\frac {2}{2835}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {32}{32 r^{10}+880 r^{9}+10560 r^{8}+72600 r^{7}+315546 r^{6}+902055 r^{5}+1708465 r^{4}+2102375 r^{3}+1594197 r^{2}+664290 r +113400} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{5}={\frac {4}{155925}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2}{2 r^{2}+3 r +1}\) | \(\frac {2}{3}\) |
| \(a_{2}\) | \(\frac {4}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {2}{15}\) |
| \(a_{3}\) | \(\frac {8}{8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90}\) | \(\frac {4}{315}\) |
| \(a_{4}\) | \(\frac {16}{16 r^{8}+288 r^{7}+2184 r^{6}+9072 r^{5}+22449 r^{4}+33642 r^{3}+29531 r^{2}+13698 r +2520}\) | \(\frac {2}{2835}\) |
| \(a_{5}\) | \(\frac {32}{32 r^{10}+880 r^{9}+10560 r^{8}+72600 r^{7}+315546 r^{6}+902055 r^{5}+1708465 r^{4}+2102375 r^{3}+1594197 r^{2}+664290 r +113400}\) | \(\frac {4}{155925}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= \sqrt {x} \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 ) \\ &= \sqrt {x}\, \left (1+\frac {2 x}{3}+\frac {2 x^{2}}{15}+\frac {4 x^{3}}{315}+\frac {2 x^{4}}{2835}+\frac {4 x^{5}}{155925}+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 )+\left (n +r \right ) b_{n}-2 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {2 b_{n -1}}{2 n^{2}+4 n r +2 r^{2}-n -r}\tag {4} \] Which for the root \(r = 0\) becomes \[ b_{n} = \frac {2 b_{n -1}}{n \left (2 n -1\right )}\tag {5} \] 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 \[ b_{1}=\frac {2}{2 r^{2}+3 r +1} \] Which for the root \(r = 0\) becomes \[ b_{1}=2 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2}{2 r^{2}+3 r +1}\) | \(2\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {4}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6} \] Which for the root \(r = 0\) becomes \[ b_{2}={\frac {2}{3}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2}{2 r^{2}+3 r +1}\) | \(2\) |
| \(b_{2}\) | \(\frac {4}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {2}{3}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {8}{8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90} \] Which for the root \(r = 0\) becomes \[ b_{3}={\frac {4}{45}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2}{2 r^{2}+3 r +1}\) | \(2\) |
| \(b_{2}\) | \(\frac {4}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {2}{3}\) |
| \(b_{3}\) | \(\frac {8}{8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90}\) | \(\frac {4}{45}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {16}{16 r^{8}+288 r^{7}+2184 r^{6}+9072 r^{5}+22449 r^{4}+33642 r^{3}+29531 r^{2}+13698 r +2520} \] Which for the root \(r = 0\) becomes \[ b_{4}={\frac {2}{315}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2}{2 r^{2}+3 r +1}\) | \(2\) |
| \(b_{2}\) | \(\frac {4}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {2}{3}\) |
| \(b_{3}\) | \(\frac {8}{8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90}\) | \(\frac {4}{45}\) |
| \(b_{4}\) | \(\frac {16}{16 r^{8}+288 r^{7}+2184 r^{6}+9072 r^{5}+22449 r^{4}+33642 r^{3}+29531 r^{2}+13698 r +2520}\) | \(\frac {2}{315}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {32}{32 r^{10}+880 r^{9}+10560 r^{8}+72600 r^{7}+315546 r^{6}+902055 r^{5}+1708465 r^{4}+2102375 r^{3}+1594197 r^{2}+664290 r +113400} \] Which for the root \(r = 0\) becomes \[ b_{5}={\frac {4}{14175}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2}{2 r^{2}+3 r +1}\) | \(2\) |
| \(b_{2}\) | \(\frac {4}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {2}{3}\) |
| \(b_{3}\) | \(\frac {8}{8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90}\) | \(\frac {4}{45}\) |
| \(b_{4}\) | \(\frac {16}{16 r^{8}+288 r^{7}+2184 r^{6}+9072 r^{5}+22449 r^{4}+33642 r^{3}+29531 r^{2}+13698 r +2520}\) | \(\frac {2}{315}\) |
| \(b_{5}\) | \(\frac {32}{32 r^{10}+880 r^{9}+10560 r^{8}+72600 r^{7}+315546 r^{6}+902055 r^{5}+1708465 r^{4}+2102375 r^{3}+1594197 r^{2}+664290 r +113400}\) | \(\frac {4}{14175}\) |
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+2 x +\frac {2 x^{2}}{3}+\frac {4 x^{3}}{45}+\frac {2 x^{4}}{315}+\frac {4 x^{5}}{14175}+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} \sqrt {x}\, \left (1+\frac {2 x}{3}+\frac {2 x^{2}}{15}+\frac {4 x^{3}}{315}+\frac {2 x^{4}}{2835}+\frac {4 x^{5}}{155925}+O\left (x^{6}\right )\right ) + c_{2} \left (1+2 x +\frac {2 x^{2}}{3}+\frac {4 x^{3}}{45}+\frac {2 x^{4}}{315}+\frac {4 x^{5}}{14175}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \sqrt {x}\, \left (1+\frac {2 x}{3}+\frac {2 x^{2}}{15}+\frac {4 x^{3}}{315}+\frac {2 x^{4}}{2835}+\frac {4 x^{5}}{155925}+O\left (x^{6}\right )\right )+c_{2} \left (1+2 x +\frac {2 x^{2}}{3}+\frac {4 x^{3}}{45}+\frac {2 x^{4}}{315}+\frac {4 x^{5}}{14175}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \sqrt {x}\, \left (1+\frac {2 x}{3}+\frac {2 x^{2}}{15}+\frac {4 x^{3}}{315}+\frac {2 x^{4}}{2835}+\frac {4 x^{5}}{155925}+O\left (x^{6}\right )\right )+c_{2} \left (1+2 x +\frac {2 x^{2}}{3}+\frac {4 x^{3}}{45}+\frac {2 x^{4}}{315}+\frac {4 x^{5}}{14175}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \sqrt {x}\, \left (1+\frac {2 x}{3}+\frac {2 x^{2}}{15}+\frac {4 x^{3}}{315}+\frac {2 x^{4}}{2835}+\frac {4 x^{5}}{155925}+O\left (x^{6}\right )\right )+c_{2} \left (1+2 x +\frac {2 x^{2}}{3}+\frac {4 x^{3}}{45}+\frac {2 x^{4}}{315}+\frac {4 x^{5}}{14175}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x \left (\frac {d}{d x}y^{\prime }\right )+y^{\prime }-2 y=0 \\ \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 {y^{\prime }}{2 x}+\frac {y}{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} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {y^{\prime }}{2 x}-\frac {y}{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 {1}{2 x}, P_{3}\left (x \right )=-\frac {1}{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 {1}{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 \left (\frac {d}{d x}y^{\prime }\right )+y^{\prime }-2 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} y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime }\right )=\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 \left (\frac {d}{d x}y^{\prime }\right )=\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 (-1+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 +1+2 r \right )-2 a_{k}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-1+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 2 \left (k +\frac {1}{2}+r \right ) \left (k +1+r \right ) a_{k +1}-2 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {2 a_{k}}{\left (2 k +1+2 r \right ) \left (k +1+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {2 a_{k}}{\left (2 k +1\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 {2 a_{k}}{\left (2 k +1\right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +1}=\frac {2 a_{k}}{\left (2 k +2\right ) \left (k +\frac {3}{2}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}}, a_{k +1}=\frac {2 a_{k}}{\left (2 k +2\right ) \left (k +\frac {3}{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 {1}{2}}\right ), a_{k +1}=\frac {2 a_{k}}{\left (2 k +1\right ) \left (k +1\right )}, b_{k +1}=\frac {2 b_{k}}{\left (2 k +2\right ) \left (k +\frac {3}{2}\right )}\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)] <- linear_1 successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 44
Order:=6; dsolve(2*x*diff(y(x),x$2)+diff(y(x),x)-2*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} \sqrt {x}\, \left (1+\frac {2}{3} x +\frac {2}{15} x^{2}+\frac {4}{315} x^{3}+\frac {2}{2835} x^{4}+\frac {4}{155925} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (1+2 x +\frac {2}{3} x^{2}+\frac {4}{45} x^{3}+\frac {2}{315} x^{4}+\frac {4}{14175} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 83
AsymptoticDSolveValue[2*x*y''[x]+y'[x]-2*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \sqrt {x} \left (\frac {4 x^5}{155925}+\frac {2 x^4}{2835}+\frac {4 x^3}{315}+\frac {2 x^2}{15}+\frac {2 x}{3}+1\right )+c_2 \left (\frac {4 x^5}{14175}+\frac {2 x^4}{315}+\frac {4 x^3}{45}+\frac {2 x^2}{3}+2 x+1\right ) \]