Internal problem ID [5478]
Internal file name [OUTPUT/4726_Sunday_June_05_2022_03_04_10_PM_91628320/index.tex]
Book: A treatise on Differential Equations by A. R. Forsyth. 6th edition. 1929. Macmillan Co. ltd.
New York, reprinted 1956
Section: Chapter VI. Note I. Integration of linear equations in series by the method of Frobenius.
page 243
Problem number: Ex. 8(i), page 258.
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
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} y^{\prime \prime }+4 \left (x +a \right ) y=0} \] 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. \[ x^{2} y^{\prime \prime }+\left (4 a +4 x \right ) y = 0 \] 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) &= 0\\ q(x) &= \frac {4 a +4 x}{x^{2}}\\ \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 \[ x^{2} y^{\prime \prime }+\left (4 a +4 x \right ) y = 0 \] 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} x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (4 a +4 x \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 }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 a \,x^{n +r} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n +r} a_{n}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} x^{n +r} \\ \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 \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 a \,x^{n +r} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+4 a \,x^{n +r} a_{n} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+4 a \,x^{r} a_{0} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+4 a \,x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}+4 a -r \right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}+4 a -r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= \frac {1}{2}+\frac {\sqrt {1-16 a}}{2}\\ r_2 &= \frac {1}{2}-\frac {\sqrt {1-16 a}}{2} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}+4 a -r \right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [\frac {1}{2}+\frac {\sqrt {1-16 a}}{2}, \frac {1}{2}-\frac {\sqrt {1-16 a}}{2}\right ]\).
Assuming the roots differ by non-integer Since \(r_1 - r_2 = \sqrt {1-16 a}\) 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}+\frac {\sqrt {1-16 a}}{2}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {1}{2}-\frac {\sqrt {1-16 a}}{2}} \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} a_{n} \left (n +r \right ) \left (n +r -1\right )+4 a a_{n}+4 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {4 a_{n -1}}{n^{2}+2 n r +r^{2}+4 a -n -r}\tag {4} \] Which for the root \(r = \frac {1}{2}+\frac {\sqrt {1-16 a}}{2}\) becomes \[ a_{n} = -\frac {4 a_{n -1}}{n \left (\sqrt {1-16 a}+n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = \frac {1}{2}+\frac {\sqrt {1-16 a}}{2}\) 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 \[ a_{1}=-\frac {4}{r^{2}+4 a +r} \] Which for the root \(r = \frac {1}{2}+\frac {\sqrt {1-16 a}}{2}\) becomes \[ a_{1}=-\frac {4}{1+\sqrt {1-16 a}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {4}{r^{2}+4 a +r}\) | \(-\frac {4}{1+\sqrt {1-16 a}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {16}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right )} \] Which for the root \(r = \frac {1}{2}+\frac {\sqrt {1-16 a}}{2}\) becomes \[ a_{2}=\frac {8}{\left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right )} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {4}{r^{2}+4 a +r}\) | \(-\frac {4}{1+\sqrt {1-16 a}}\) |
| \(a_{2}\) | \(\frac {16}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right )}\) | \(\frac {8}{\left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right )}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {64}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right ) \left (r^{2}+4 a +5 r +6\right )} \] Which for the root \(r = \frac {1}{2}+\frac {\sqrt {1-16 a}}{2}\) becomes \[ a_{3}=-\frac {32}{3 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right )} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {4}{r^{2}+4 a +r}\) | \(-\frac {4}{1+\sqrt {1-16 a}}\) |
| \(a_{2}\) | \(\frac {16}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right )}\) | \(\frac {8}{\left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right )}\) |
| \(a_{3}\) | \(-\frac {64}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right ) \left (r^{2}+4 a +5 r +6\right )}\) | \(-\frac {32}{3 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right )}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {256}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right ) \left (r^{2}+4 a +5 r +6\right ) \left (r^{2}+4 a +7 r +12\right )} \] Which for the root \(r = \frac {1}{2}+\frac {\sqrt {1-16 a}}{2}\) becomes \[ a_{4}=\frac {32}{3 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right ) \left (4+\sqrt {1-16 a}\right )} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {4}{r^{2}+4 a +r}\) | \(-\frac {4}{1+\sqrt {1-16 a}}\) |
| \(a_{2}\) | \(\frac {16}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right )}\) | \(\frac {8}{\left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right )}\) |
| \(a_{3}\) | \(-\frac {64}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right ) \left (r^{2}+4 a +5 r +6\right )}\) | \(-\frac {32}{3 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right )}\) |
| \(a_{4}\) | \(\frac {256}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right ) \left (r^{2}+4 a +5 r +6\right ) \left (r^{2}+4 a +7 r +12\right )}\) | \(\frac {32}{3 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right ) \left (4+\sqrt {1-16 a}\right )}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {1024}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right ) \left (r^{2}+4 a +5 r +6\right ) \left (r^{2}+4 a +7 r +12\right ) \left (r^{2}+4 a +9 r +20\right )} \] Which for the root \(r = \frac {1}{2}+\frac {\sqrt {1-16 a}}{2}\) becomes \[ a_{5}=-\frac {128}{15 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right ) \left (4+\sqrt {1-16 a}\right ) \left (5+\sqrt {1-16 a}\right )} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {4}{r^{2}+4 a +r}\) | \(-\frac {4}{1+\sqrt {1-16 a}}\) |
| \(a_{2}\) | \(\frac {16}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right )}\) | \(\frac {8}{\left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right )}\) |
| \(a_{3}\) | \(-\frac {64}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right ) \left (r^{2}+4 a +5 r +6\right )}\) | \(-\frac {32}{3 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right )}\) |
| \(a_{4}\) | \(\frac {256}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right ) \left (r^{2}+4 a +5 r +6\right ) \left (r^{2}+4 a +7 r +12\right )}\) | \(\frac {32}{3 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right ) \left (4+\sqrt {1-16 a}\right )}\) |
| \(a_{5}\) | \(-\frac {1024}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right ) \left (r^{2}+4 a +5 r +6\right ) \left (r^{2}+4 a +7 r +12\right ) \left (r^{2}+4 a +9 r +20\right )}\) | \(-\frac {128}{15 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right ) \left (4+\sqrt {1-16 a}\right ) \left (5+\sqrt {1-16 a}\right )}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {1}{2}+\frac {\sqrt {1-16 a}}{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 {1}{2}+\frac {\sqrt {1-16 a}}{2}} \left (1-\frac {4 x}{1+\sqrt {1-16 a}}+\frac {8 x^{2}}{\left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right )}-\frac {32 x^{3}}{3 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right )}+\frac {32 x^{4}}{3 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right ) \left (4+\sqrt {1-16 a}\right )}-\frac {128 x^{5}}{15 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right ) \left (4+\sqrt {1-16 a}\right ) \left (5+\sqrt {1-16 a}\right )}+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} b_{n} \left (n +r \right ) \left (n +r -1\right )+4 a b_{n}+4 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {4 b_{n -1}}{n^{2}+2 n r +r^{2}+4 a -n -r}\tag {4} \] Which for the root \(r = \frac {1}{2}-\frac {\sqrt {1-16 a}}{2}\) becomes \[ b_{n} = \frac {4 b_{n -1}}{n \left (\sqrt {1-16 a}-n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = \frac {1}{2}-\frac {\sqrt {1-16 a}}{2}\) 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 \[ b_{1}=-\frac {4}{r^{2}+4 a +r} \] Which for the root \(r = \frac {1}{2}-\frac {\sqrt {1-16 a}}{2}\) becomes \[ b_{1}=\frac {4}{-1+\sqrt {1-16 a}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {4}{r^{2}+4 a +r}\) | \(\frac {4}{-1+\sqrt {1-16 a}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {16}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right )} \] Which for the root \(r = \frac {1}{2}-\frac {\sqrt {1-16 a}}{2}\) becomes \[ b_{2}=\frac {8}{\left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right )} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {4}{r^{2}+4 a +r}\) | \(\frac {4}{-1+\sqrt {1-16 a}}\) |
| \(b_{2}\) | \(\frac {16}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right )}\) | \(\frac {8}{\left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right )}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {64}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right ) \left (r^{2}+4 a +5 r +6\right )} \] Which for the root \(r = \frac {1}{2}-\frac {\sqrt {1-16 a}}{2}\) becomes \[ b_{3}=\frac {32}{3 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right )} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {4}{r^{2}+4 a +r}\) | \(\frac {4}{-1+\sqrt {1-16 a}}\) |
| \(b_{2}\) | \(\frac {16}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right )}\) | \(\frac {8}{\left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right )}\) |
| \(b_{3}\) | \(-\frac {64}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right ) \left (r^{2}+4 a +5 r +6\right )}\) | \(\frac {32}{3 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right )}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {256}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right ) \left (r^{2}+4 a +5 r +6\right ) \left (r^{2}+4 a +7 r +12\right )} \] Which for the root \(r = \frac {1}{2}-\frac {\sqrt {1-16 a}}{2}\) becomes \[ b_{4}=\frac {32}{3 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right ) \left (-4+\sqrt {1-16 a}\right )} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {4}{r^{2}+4 a +r}\) | \(\frac {4}{-1+\sqrt {1-16 a}}\) |
| \(b_{2}\) | \(\frac {16}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right )}\) | \(\frac {8}{\left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right )}\) |
| \(b_{3}\) | \(-\frac {64}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right ) \left (r^{2}+4 a +5 r +6\right )}\) | \(\frac {32}{3 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right )}\) |
| \(b_{4}\) | \(\frac {256}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right ) \left (r^{2}+4 a +5 r +6\right ) \left (r^{2}+4 a +7 r +12\right )}\) | \(\frac {32}{3 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right ) \left (-4+\sqrt {1-16 a}\right )}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {1024}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right ) \left (r^{2}+4 a +5 r +6\right ) \left (r^{2}+4 a +7 r +12\right ) \left (r^{2}+4 a +9 r +20\right )} \] Which for the root \(r = \frac {1}{2}-\frac {\sqrt {1-16 a}}{2}\) becomes \[ b_{5}=\frac {128}{15 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right ) \left (-4+\sqrt {1-16 a}\right ) \left (-5+\sqrt {1-16 a}\right )} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {4}{r^{2}+4 a +r}\) | \(\frac {4}{-1+\sqrt {1-16 a}}\) |
| \(b_{2}\) | \(\frac {16}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right )}\) | \(\frac {8}{\left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right )}\) |
| \(b_{3}\) | \(-\frac {64}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right ) \left (r^{2}+4 a +5 r +6\right )}\) | \(\frac {32}{3 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right )}\) |
| \(b_{4}\) | \(\frac {256}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right ) \left (r^{2}+4 a +5 r +6\right ) \left (r^{2}+4 a +7 r +12\right )}\) | \(\frac {32}{3 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right ) \left (-4+\sqrt {1-16 a}\right )}\) |
| \(b_{5}\) | \(-\frac {1024}{\left (r^{2}+4 a +r \right ) \left (r^{2}+4 a +3 r +2\right ) \left (r^{2}+4 a +5 r +6\right ) \left (r^{2}+4 a +7 r +12\right ) \left (r^{2}+4 a +9 r +20\right )}\) | \(\frac {128}{15 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right ) \left (-4+\sqrt {1-16 a}\right ) \left (-5+\sqrt {1-16 a}\right )}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{\frac {1}{2}+\frac {\sqrt {1-16 a}}{2}} \left (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 \right ) \\ &= x^{\frac {1}{2}-\frac {\sqrt {1-16 a}}{2}} \left (1+\frac {4 x}{-1+\sqrt {1-16 a}}+\frac {8 x^{2}}{\left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right )}+\frac {32 x^{3}}{3 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right )}+\frac {32 x^{4}}{3 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right ) \left (-4+\sqrt {1-16 a}\right )}+\frac {128 x^{5}}{15 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right ) \left (-4+\sqrt {1-16 a}\right ) \left (-5+\sqrt {1-16 a}\right )}+O\left (x^{6}\right )\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 {1}{2}+\frac {\sqrt {1-16 a}}{2}} \left (1-\frac {4 x}{1+\sqrt {1-16 a}}+\frac {8 x^{2}}{\left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right )}-\frac {32 x^{3}}{3 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right )}+\frac {32 x^{4}}{3 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right ) \left (4+\sqrt {1-16 a}\right )}-\frac {128 x^{5}}{15 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right ) \left (4+\sqrt {1-16 a}\right ) \left (5+\sqrt {1-16 a}\right )}+O\left (x^{6}\right )\right ) + c_{2} x^{\frac {1}{2}-\frac {\sqrt {1-16 a}}{2}} \left (1+\frac {4 x}{-1+\sqrt {1-16 a}}+\frac {8 x^{2}}{\left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right )}+\frac {32 x^{3}}{3 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right )}+\frac {32 x^{4}}{3 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right ) \left (-4+\sqrt {1-16 a}\right )}+\frac {128 x^{5}}{15 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right ) \left (-4+\sqrt {1-16 a}\right ) \left (-5+\sqrt {1-16 a}\right )}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {1}{2}+\frac {\sqrt {1-16 a}}{2}} \left (1-\frac {4 x}{1+\sqrt {1-16 a}}+\frac {8 x^{2}}{\left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right )}-\frac {32 x^{3}}{3 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right )}+\frac {32 x^{4}}{3 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right ) \left (4+\sqrt {1-16 a}\right )}-\frac {128 x^{5}}{15 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right ) \left (4+\sqrt {1-16 a}\right ) \left (5+\sqrt {1-16 a}\right )}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{2}-\frac {\sqrt {1-16 a}}{2}} \left (1+\frac {4 x}{-1+\sqrt {1-16 a}}+\frac {8 x^{2}}{\left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right )}+\frac {32 x^{3}}{3 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right )}+\frac {32 x^{4}}{3 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right ) \left (-4+\sqrt {1-16 a}\right )}+\frac {128 x^{5}}{15 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right ) \left (-4+\sqrt {1-16 a}\right ) \left (-5+\sqrt {1-16 a}\right )}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {1}{2}+\frac {\sqrt {1-16 a}}{2}} \left (1-\frac {4 x}{1+\sqrt {1-16 a}}+\frac {8 x^{2}}{\left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right )}-\frac {32 x^{3}}{3 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right )}+\frac {32 x^{4}}{3 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right ) \left (4+\sqrt {1-16 a}\right )}-\frac {128 x^{5}}{15 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right ) \left (4+\sqrt {1-16 a}\right ) \left (5+\sqrt {1-16 a}\right )}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{2}-\frac {\sqrt {1-16 a}}{2}} \left (1+\frac {4 x}{-1+\sqrt {1-16 a}}+\frac {8 x^{2}}{\left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right )}+\frac {32 x^{3}}{3 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right )}+\frac {32 x^{4}}{3 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right ) \left (-4+\sqrt {1-16 a}\right )}+\frac {128 x^{5}}{15 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right ) \left (-4+\sqrt {1-16 a}\right ) \left (-5+\sqrt {1-16 a}\right )}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {1}{2}+\frac {\sqrt {1-16 a}}{2}} \left (1-\frac {4 x}{1+\sqrt {1-16 a}}+\frac {8 x^{2}}{\left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right )}-\frac {32 x^{3}}{3 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right )}+\frac {32 x^{4}}{3 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right ) \left (4+\sqrt {1-16 a}\right )}-\frac {128 x^{5}}{15 \left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right ) \left (4+\sqrt {1-16 a}\right ) \left (5+\sqrt {1-16 a}\right )}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{2}-\frac {\sqrt {1-16 a}}{2}} \left (1+\frac {4 x}{-1+\sqrt {1-16 a}}+\frac {8 x^{2}}{\left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right )}+\frac {32 x^{3}}{3 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right )}+\frac {32 x^{4}}{3 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right ) \left (-4+\sqrt {1-16 a}\right )}+\frac {128 x^{5}}{15 \left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right ) \left (-4+\sqrt {1-16 a}\right ) \left (-5+\sqrt {1-16 a}\right )}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (4 a +4 x \right ) 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 {4 \left (x +a \right ) y}{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 {4 \left (x +a \right ) y}{x^{2}}=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 )=0, P_{3}\left (x \right )=\frac {4 \left (x +a \right )}{x^{2}}\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}}}=0 \\ {} & \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}}}=4 a \\ {} & \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} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (4 a +4 x \right ) 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\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\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} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (r^{2}+4 a -r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (k^{2}+2 k r +r^{2}+4 a -k -r \right )+4 a_{k -1}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r^{2}+4 a -r =0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\frac {1}{2}-\frac {\sqrt {1-16 a}}{2}, \frac {1}{2}+\frac {\sqrt {1-16 a}}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k^{2}+\left (2 r -1\right ) k +r^{2}+4 a -r \right ) a_{k}+4 a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (\left (k +1\right )^{2}+\left (2 r -1\right ) \left (k +1\right )+r^{2}+4 a -r \right ) a_{k +1}+4 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {4 a_{k}}{k^{2}+2 k r +r^{2}+4 a +k +r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2}-\frac {\sqrt {1-16 a}}{2} \\ {} & {} & a_{k +1}=-\frac {4 a_{k}}{k^{2}+2 k \left (\frac {1}{2}-\frac {\sqrt {1-16 a}}{2}\right )+\left (\frac {1}{2}-\frac {\sqrt {1-16 a}}{2}\right )^{2}+4 a +k +\frac {1}{2}-\frac {\sqrt {1-16 a}}{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2}-\frac {\sqrt {1-16 a}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}-\frac {\sqrt {1-16 a}}{2}}, a_{k +1}=-\frac {4 a_{k}}{k^{2}+2 k \left (\frac {1}{2}-\frac {\sqrt {1-16 a}}{2}\right )+\left (\frac {1}{2}-\frac {\sqrt {1-16 a}}{2}\right )^{2}+4 a +k +\frac {1}{2}-\frac {\sqrt {1-16 a}}{2}}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2}+\frac {\sqrt {1-16 a}}{2} \\ {} & {} & a_{k +1}=-\frac {4 a_{k}}{k^{2}+2 k \left (\frac {1}{2}+\frac {\sqrt {1-16 a}}{2}\right )+\left (\frac {1}{2}+\frac {\sqrt {1-16 a}}{2}\right )^{2}+4 a +k +\frac {1}{2}+\frac {\sqrt {1-16 a}}{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2}+\frac {\sqrt {1-16 a}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}+\frac {\sqrt {1-16 a}}{2}}, a_{k +1}=-\frac {4 a_{k}}{k^{2}+2 k \left (\frac {1}{2}+\frac {\sqrt {1-16 a}}{2}\right )+\left (\frac {1}{2}+\frac {\sqrt {1-16 a}}{2}\right )^{2}+4 a +k +\frac {1}{2}+\frac {\sqrt {1-16 a}}{2}}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{2}-\frac {\sqrt {1-16 a}}{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k +\frac {1}{2}+\frac {\sqrt {1-16 a}}{2}}\right ), b_{k +1}=-\frac {4 b_{k}}{k^{2}+2 k \left (\frac {1}{2}-\frac {\sqrt {1-16 a}}{2}\right )+\left (\frac {1}{2}-\frac {\sqrt {1-16 a}}{2}\right )^{2}+4 a +k +\frac {1}{2}-\frac {\sqrt {1-16 a}}{2}}, c_{k +1}=-\frac {4 c_{k}}{k^{2}+2 k \left (\frac {1}{2}+\frac {\sqrt {1-16 a}}{2}\right )+\left (\frac {1}{2}+\frac {\sqrt {1-16 a}}{2}\right )^{2}+4 a +k +\frac {1}{2}+\frac {\sqrt {1-16 a}}{2}}\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)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel <- Bessel successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 947
Order:=6; dsolve(x^2*diff(y(x),x$2)+4*(x+a)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \sqrt {x}\, \left (c_{1} x^{-\frac {\sqrt {1-16 a}}{2}} \left (1+4 \frac {1}{-1+\sqrt {1-16 a}} x +8 \frac {1}{\left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right )} x^{2}+\frac {32}{3} \frac {1}{\left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right )} x^{3}+\frac {32}{3} \frac {1}{\left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right ) \left (-4+\sqrt {1-16 a}\right )} x^{4}+\frac {128}{15} \frac {1}{\left (-1+\sqrt {1-16 a}\right ) \left (-2+\sqrt {1-16 a}\right ) \left (-3+\sqrt {1-16 a}\right ) \left (-4+\sqrt {1-16 a}\right ) \left (-5+\sqrt {1-16 a}\right )} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} x^{\frac {\sqrt {1-16 a}}{2}} \left (1-4 \frac {1}{1+\sqrt {1-16 a}} x +8 \frac {1}{\left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right )} x^{2}-\frac {32}{3} \frac {1}{\left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right )} x^{3}+\frac {32}{3} \frac {1}{\left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right ) \left (4+\sqrt {1-16 a}\right )} x^{4}-\frac {128}{15} \frac {1}{\left (1+\sqrt {1-16 a}\right ) \left (2+\sqrt {1-16 a}\right ) \left (3+\sqrt {1-16 a}\right ) \left (4+\sqrt {1-16 a}\right ) \left (5+\sqrt {1-16 a}\right )} x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 1356
AsymptoticDSolveValue[x^2*y''[x]+4*(x+a)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to \left (-\frac {1024 x^5}{\left (\left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+1\right ) \left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+2\right )+4 a\right ) \left (\left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+2\right ) \left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+3\right )+4 a\right ) \left (\left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+3\right ) \left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+4\right )+4 a\right ) \left (\left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+4\right ) \left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+5\right )+4 a\right ) \left (\frac {1}{2} \left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+1\right ) \left (1-\sqrt {1-16 a}\right )+4 a\right )}+\frac {256 x^4}{\left (\left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+1\right ) \left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+2\right )+4 a\right ) \left (\left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+2\right ) \left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+3\right )+4 a\right ) \left (\left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+3\right ) \left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+4\right )+4 a\right ) \left (\frac {1}{2} \left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+1\right ) \left (1-\sqrt {1-16 a}\right )+4 a\right )}-\frac {64 x^3}{\left (\left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+1\right ) \left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+2\right )+4 a\right ) \left (\left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+2\right ) \left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+3\right )+4 a\right ) \left (\frac {1}{2} \left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+1\right ) \left (1-\sqrt {1-16 a}\right )+4 a\right )}+\frac {16 x^2}{\left (\left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+1\right ) \left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+2\right )+4 a\right ) \left (\frac {1}{2} \left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+1\right ) \left (1-\sqrt {1-16 a}\right )+4 a\right )}-\frac {4 x}{\frac {1}{2} \left (\frac {1}{2} \left (1-\sqrt {1-16 a}\right )+1\right ) \left (1-\sqrt {1-16 a}\right )+4 a}+1\right ) c_2 x^{\frac {1}{2} \left (1-\sqrt {1-16 a}\right )}+\left (-\frac {1024 x^5}{\left (\left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+1\right ) \left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+2\right )+4 a\right ) \left (\left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+2\right ) \left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+3\right )+4 a\right ) \left (\left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+3\right ) \left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+4\right )+4 a\right ) \left (\left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+4\right ) \left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+5\right )+4 a\right ) \left (\frac {1}{2} \left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+1\right ) \left (\sqrt {1-16 a}+1\right )+4 a\right )}+\frac {256 x^4}{\left (\left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+1\right ) \left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+2\right )+4 a\right ) \left (\left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+2\right ) \left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+3\right )+4 a\right ) \left (\left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+3\right ) \left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+4\right )+4 a\right ) \left (\frac {1}{2} \left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+1\right ) \left (\sqrt {1-16 a}+1\right )+4 a\right )}-\frac {64 x^3}{\left (\left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+1\right ) \left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+2\right )+4 a\right ) \left (\left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+2\right ) \left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+3\right )+4 a\right ) \left (\frac {1}{2} \left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+1\right ) \left (\sqrt {1-16 a}+1\right )+4 a\right )}+\frac {16 x^2}{\left (\left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+1\right ) \left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+2\right )+4 a\right ) \left (\frac {1}{2} \left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+1\right ) \left (\sqrt {1-16 a}+1\right )+4 a\right )}-\frac {4 x}{\frac {1}{2} \left (\frac {1}{2} \left (\sqrt {1-16 a}+1\right )+1\right ) \left (\sqrt {1-16 a}+1\right )+4 a}+1\right ) c_1 x^{\frac {1}{2} \left (\sqrt {1-16 a}+1\right )} \]