Internal problem ID [15482]
Internal file name [OUTPUT/15483_Friday_May_10_2024_05_47_26_PM_97534733/index.tex]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 18.2. Expanding a solution in generalized power
series. Bessels equation. Exercises page 177
Problem number: 739.
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)]`]]
\[ \boxed {4 x y^{\prime \prime }+2 y^{\prime }+y=0} \] 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. \[ 4 x y^{\prime \prime }+2 y^{\prime }+y = 0 \] 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 {1}{2 x}\\ q(x) &= \frac {1}{4 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 \[ 4 x y^{\prime \prime }+2 y^{\prime }+y = 0 \] 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} 4 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) x +2 \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 }}4 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 \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 }}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 }}4 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 \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 \[ 4 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+2 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ 4 x^{-1+r} a_{0} r \left (-1+r \right )+2 r a_{0} x^{-1+r} = 0 \] Or \[ \left (4 x^{-1+r} r \left (-1+r \right )+2 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (4 r^{2}-2 r \right ) x^{-1+r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 4 r^{2}-2 r = 0 \] 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 \[ \left (4 r^{2}-2 r \right ) x^{-1+r} = 0 \] 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} 4 a_{n} \left (n +r \right ) \left (n +r -1\right )+2 a_{n} \left (n +r \right )+a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1}}{2 \left (2 n^{2}+4 n r +2 r^{2}-n -r \right )}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{n} = -\frac {a_{n -1}}{4 n^{2}+2 n}\tag {5} \] 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 \[ a_{1}=-\frac {1}{4 r^{2}+6 r +2} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{1}=-{\frac {1}{6}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{4 r^{2}+6 r +2}\) | \(-{\frac {1}{6}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {1}{16 r^{4}+80 r^{3}+140 r^{2}+100 r +24} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{2}={\frac {1}{120}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{4 r^{2}+6 r +2}\) | \(-{\frac {1}{6}}\) |
| \(a_{2}\) | \(\frac {1}{16 r^{4}+80 r^{3}+140 r^{2}+100 r +24}\) | \(\frac {1}{120}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {1}{64 r^{6}+672 r^{5}+2800 r^{4}+5880 r^{3}+6496 r^{2}+3528 r +720} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{3}=-{\frac {1}{5040}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{4 r^{2}+6 r +2}\) | \(-{\frac {1}{6}}\) |
| \(a_{2}\) | \(\frac {1}{16 r^{4}+80 r^{3}+140 r^{2}+100 r +24}\) | \(\frac {1}{120}\) |
| \(a_{3}\) | \(-\frac {1}{64 r^{6}+672 r^{5}+2800 r^{4}+5880 r^{3}+6496 r^{2}+3528 r +720}\) | \(-{\frac {1}{5040}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {1}{16 \left (8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90\right ) \left (2 r^{2}+15 r +28\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{4}={\frac {1}{362880}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{4 r^{2}+6 r +2}\) | \(-{\frac {1}{6}}\) |
| \(a_{2}\) | \(\frac {1}{16 r^{4}+80 r^{3}+140 r^{2}+100 r +24}\) | \(\frac {1}{120}\) |
| \(a_{3}\) | \(-\frac {1}{64 r^{6}+672 r^{5}+2800 r^{4}+5880 r^{3}+6496 r^{2}+3528 r +720}\) | \(-{\frac {1}{5040}}\) |
| \(a_{4}\) | \(\frac {1}{16 \left (8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90\right ) \left (2 r^{2}+15 r +28\right )}\) | \(\frac {1}{362880}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {1}{32 \left (8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90\right ) \left (2 r^{2}+15 r +28\right ) \left (2 r^{2}+19 r +45\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{5}=-{\frac {1}{39916800}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{4 r^{2}+6 r +2}\) | \(-{\frac {1}{6}}\) |
| \(a_{2}\) | \(\frac {1}{16 r^{4}+80 r^{3}+140 r^{2}+100 r +24}\) | \(\frac {1}{120}\) |
| \(a_{3}\) | \(-\frac {1}{64 r^{6}+672 r^{5}+2800 r^{4}+5880 r^{3}+6496 r^{2}+3528 r +720}\) | \(-{\frac {1}{5040}}\) |
| \(a_{4}\) | \(\frac {1}{16 \left (8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90\right ) \left (2 r^{2}+15 r +28\right )}\) | \(\frac {1}{362880}\) |
| \(a_{5}\) | \(-\frac {1}{32 \left (8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90\right ) \left (2 r^{2}+15 r +28\right ) \left (2 r^{2}+19 r +45\right )}\) | \(-{\frac {1}{39916800}}\) |
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 {x}{6}+\frac {x^{2}}{120}-\frac {x^{3}}{5040}+\frac {x^{4}}{362880}-\frac {x^{5}}{39916800}+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} 4 b_{n} \left (n +r \right ) \left (n +r -1\right )+2 \left (n +r \right ) b_{n}+b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {b_{n -1}}{2 \left (2 n^{2}+4 n r +2 r^{2}-n -r \right )}\tag {4} \] Which for the root \(r = 0\) becomes \[ b_{n} = -\frac {b_{n -1}}{4 n^{2}-2 n}\tag {5} \] 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 \[ b_{1}=-\frac {1}{4 r^{2}+6 r +2} \] Which for the root \(r = 0\) becomes \[ b_{1}=-{\frac {1}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{4 r^{2}+6 r +2}\) | \(-{\frac {1}{2}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {1}{16 r^{4}+80 r^{3}+140 r^{2}+100 r +24} \] Which for the root \(r = 0\) becomes \[ b_{2}={\frac {1}{24}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{4 r^{2}+6 r +2}\) | \(-{\frac {1}{2}}\) |
| \(b_{2}\) | \(\frac {1}{16 r^{4}+80 r^{3}+140 r^{2}+100 r +24}\) | \(\frac {1}{24}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {1}{64 r^{6}+672 r^{5}+2800 r^{4}+5880 r^{3}+6496 r^{2}+3528 r +720} \] Which for the root \(r = 0\) becomes \[ b_{3}=-{\frac {1}{720}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{4 r^{2}+6 r +2}\) | \(-{\frac {1}{2}}\) |
| \(b_{2}\) | \(\frac {1}{16 r^{4}+80 r^{3}+140 r^{2}+100 r +24}\) | \(\frac {1}{24}\) |
| \(b_{3}\) | \(-\frac {1}{64 r^{6}+672 r^{5}+2800 r^{4}+5880 r^{3}+6496 r^{2}+3528 r +720}\) | \(-{\frac {1}{720}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {1}{16 \left (8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90\right ) \left (2 r^{2}+15 r +28\right )} \] Which for the root \(r = 0\) becomes \[ b_{4}={\frac {1}{40320}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{4 r^{2}+6 r +2}\) | \(-{\frac {1}{2}}\) |
| \(b_{2}\) | \(\frac {1}{16 r^{4}+80 r^{3}+140 r^{2}+100 r +24}\) | \(\frac {1}{24}\) |
| \(b_{3}\) | \(-\frac {1}{64 r^{6}+672 r^{5}+2800 r^{4}+5880 r^{3}+6496 r^{2}+3528 r +720}\) | \(-{\frac {1}{720}}\) |
| \(b_{4}\) | \(\frac {1}{16 \left (8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90\right ) \left (2 r^{2}+15 r +28\right )}\) | \(\frac {1}{40320}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {1}{32 \left (8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90\right ) \left (2 r^{2}+15 r +28\right ) \left (2 r^{2}+19 r +45\right )} \] Which for the root \(r = 0\) becomes \[ b_{5}=-{\frac {1}{3628800}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{4 r^{2}+6 r +2}\) | \(-{\frac {1}{2}}\) |
| \(b_{2}\) | \(\frac {1}{16 r^{4}+80 r^{3}+140 r^{2}+100 r +24}\) | \(\frac {1}{24}\) |
| \(b_{3}\) | \(-\frac {1}{64 r^{6}+672 r^{5}+2800 r^{4}+5880 r^{3}+6496 r^{2}+3528 r +720}\) | \(-{\frac {1}{720}}\) |
| \(b_{4}\) | \(\frac {1}{16 \left (8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90\right ) \left (2 r^{2}+15 r +28\right )}\) | \(\frac {1}{40320}\) |
| \(b_{5}\) | \(-\frac {1}{32 \left (8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90\right ) \left (2 r^{2}+15 r +28\right ) \left (2 r^{2}+19 r +45\right )}\) | \(-{\frac {1}{3628800}}\) |
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}{2}+\frac {x^{2}}{24}-\frac {x^{3}}{720}+\frac {x^{4}}{40320}-\frac {x^{5}}{3628800}+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 {x}{6}+\frac {x^{2}}{120}-\frac {x^{3}}{5040}+\frac {x^{4}}{362880}-\frac {x^{5}}{39916800}+O\left (x^{6}\right )\right ) + c_{2} \left (1-\frac {x}{2}+\frac {x^{2}}{24}-\frac {x^{3}}{720}+\frac {x^{4}}{40320}-\frac {x^{5}}{3628800}+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 {x}{6}+\frac {x^{2}}{120}-\frac {x^{3}}{5040}+\frac {x^{4}}{362880}-\frac {x^{5}}{39916800}+O\left (x^{6}\right )\right )+c_{2} \left (1-\frac {x}{2}+\frac {x^{2}}{24}-\frac {x^{3}}{720}+\frac {x^{4}}{40320}-\frac {x^{5}}{3628800}+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 {x}{6}+\frac {x^{2}}{120}-\frac {x^{3}}{5040}+\frac {x^{4}}{362880}-\frac {x^{5}}{39916800}+O\left (x^{6}\right )\right )+c_{2} \left (1-\frac {x}{2}+\frac {x^{2}}{24}-\frac {x^{3}}{720}+\frac {x^{4}}{40320}-\frac {x^{5}}{3628800}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \sqrt {x}\, \left (1-\frac {x}{6}+\frac {x^{2}}{120}-\frac {x^{3}}{5040}+\frac {x^{4}}{362880}-\frac {x^{5}}{39916800}+O\left (x^{6}\right )\right )+c_{2} \left (1-\frac {x}{2}+\frac {x^{2}}{24}-\frac {x^{3}}{720}+\frac {x^{4}}{40320}-\frac {x^{5}}{3628800}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 y^{\prime \prime } x +2 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}{4 x}-\frac {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 {y^{\prime }}{2 x}+\frac {y}{4 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}{4 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} \\ {} & {} & 4 y^{\prime \prime } x +2 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} 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 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} \\ {} & {} & 2 a_{0} r \left (-1+2 r \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (2 a_{k +1} \left (k +1+r \right ) \left (2 k +1+2 r \right )+a_{k}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 2 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} \\ {} & {} & 4 \left (k +1+r \right ) \left (k +\frac {1}{2}+r \right ) a_{k +1}+a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k}}{2 \left (k +1+r \right ) \left (2 k +1+2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {a_{k}}{2 \left (k +1\right ) \left (2 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 {a_{k}}{2 \left (k +1\right ) \left (2 k +1\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +1}=-\frac {a_{k}}{2 \left (k +\frac {3}{2}\right ) \left (2 k +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 {a_{k}}{2 \left (k +\frac {3}{2}\right ) \left (2 k +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 {a_{k}}{2 \left (k +1\right ) \left (2 k +1\right )}, b_{k +1}=-\frac {b_{k}}{2 \left (k +\frac {3}{2}\right ) \left (2 k +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: 0.016 (sec). Leaf size: 44
Order:=6; dsolve(4*x*diff(y(x),x$2)+2*diff(y(x),x)+y(x)=0,y(x),type='series',x=0);
\[ y = c_{1} \sqrt {x}\, \left (1-\frac {1}{6} x +\frac {1}{120} x^{2}-\frac {1}{5040} x^{3}+\frac {1}{362880} x^{4}-\frac {1}{39916800} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (1-\frac {1}{2} x +\frac {1}{24} x^{2}-\frac {1}{720} x^{3}+\frac {1}{40320} x^{4}-\frac {1}{3628800} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 85
AsymptoticDSolveValue[4*x*y''[x]+2*y'[x]+y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \sqrt {x} \left (-\frac {x^5}{39916800}+\frac {x^4}{362880}-\frac {x^3}{5040}+\frac {x^2}{120}-\frac {x}{6}+1\right )+c_2 \left (-\frac {x^5}{3628800}+\frac {x^4}{40320}-\frac {x^3}{720}+\frac {x^2}{24}-\frac {x}{2}+1\right ) \]