4.56 problem 53

Internal problem ID [7925]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 53
Date solved : Monday, October 21, 2024 at 04:35:03 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

\begin{align*} x^{2} y^{\prime \prime }+x y^{\prime }+\left (x^{2}-\frac {1}{4}\right ) y&=0 \end{align*}

Using series expansion around \(x=0\)

The type of the expansion point is first determined. This is done on the homogeneous part of the ODE.

\[ x^{2} y^{\prime \prime }+x y^{\prime }+\left (x^{2}-\frac {1}{4}\right ) y = 0 \]

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) &= \frac {1}{x}\\ q(x) &= \frac {4 x^{2}-1}{4 x^{2}}\\ \end{align*}

Table 102: Table \(p(x),q(x)\) singularites.

\(p(x)=\frac {1}{x}\)
singularity	type
\(x = 0\)	\(\text {``regular''}\)

\(q(x)=\frac {4 x^{2}-1}{4 x^{2}}\)
singularity	type
\(x = 0\)	\(\text {``regular''}\)

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

\[ x^{2} y^{\prime \prime }+x y^{\prime }+\left (x^{2}-\frac {1}{4}\right ) y = 0 \]

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 )+x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (x^{2}-\frac {1}{4}\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 }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {a_{n} x^{n +r}}{4}\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 }}x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} 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 }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {a_{n} x^{n +r}}{4}\right ) = 0 \end{equation}

The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives

\[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right )-\frac {a_{n} x^{n +r}}{4} = 0 \]

When \(n = 0\) the above becomes

\[ x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r -\frac {a_{0} x^{r}}{4} = 0 \]

Or

\[ \left (x^{r} r \left (-1+r \right )+x^{r} r -\frac {x^{r}}{4}\right ) a_{0} = 0 \]

Since \(a_{0}\neq 0\) then the above simplifies to

\[ \frac {\left (4 r^{2}-1\right ) x^{r}}{4} = 0 \]

Since the above is true for all \(x\) then the indicial equation becomes

\[ r^{2}-\frac {1}{4} = 0 \]

Solving for \(r\) gives the roots of the indicial equation as

\begin{align*} r_1 &= {\frac {1}{2}}\\ r_2 &= -{\frac {1}{2}} \end{align*}

Since \(a_{0}\neq 0\) then the indicial equation becomes

\[ \frac {\left (4 r^{2}-1\right ) x^{r}}{4} = 0 \]

Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{2}}, -{\frac {1}{2}}\right ]\).

Since \(r_1 - r_2 = 1\) is 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 ) &= C y_{1}\left (x \right ) \ln \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 ) &= \sqrt {x}\, \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\frac {\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}}{\sqrt {x}} \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 ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {1}{2}}\right ) \end{align*}

Where \(C\) above can be zero. We start by finding \(y_{1}\). 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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives

\[ a_{1} = 0 \]

For \(2\le n\) the recursive equation is

\begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n} \left (n +r \right )+a_{n -2}-\frac {a_{n}}{4} = 0 \end{equation}

Solving for \(a_{n}\) from recursive equation (4) gives

\[ a_{n} = -\frac {4 a_{n -2}}{4 n^{2}+8 n r +4 r^{2}-1}\tag {4} \]

Which for the root \(r = {\frac {1}{2}}\) becomes

\[ a_{n} = -\frac {a_{n -2}}{n \left (n +1\right )}\tag {5} \]

At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{2}}\) and after as more terms are found using the above recursive equation.

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(0\)	\(0\)

For \(n = 2\), using the above recursive equation gives

\[ a_{2}=-\frac {4}{4 r^{2}+16 r +15} \]

Which for the root \(r = {\frac {1}{2}}\) becomes

\[ a_{2}=-{\frac {1}{6}} \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(0\)	\(0\)
\(a_{2}\)	\(-\frac {4}{4 r^{2}+16 r +15}\)	\(-{\frac {1}{6}}\)

For \(n = 3\), using the above recursive equation gives

\[ a_{3}=0 \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(0\)	\(0\)
\(a_{2}\)	\(-\frac {4}{4 r^{2}+16 r +15}\)	\(-{\frac {1}{6}}\)
\(a_{3}\)	\(0\)	\(0\)

For \(n = 4\), using the above recursive equation gives

\[ a_{4}=\frac {16}{\left (4 r^{2}+16 r +15\right ) \left (4 r^{2}+32 r +63\right )} \]

Which for the root \(r = {\frac {1}{2}}\) becomes

\[ a_{4}={\frac {1}{120}} \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(0\)	\(0\)
\(a_{2}\)	\(-\frac {4}{4 r^{2}+16 r +15}\)	\(-{\frac {1}{6}}\)
\(a_{3}\)	\(0\)	\(0\)
\(a_{4}\)	\(\frac {16}{\left (4 r^{2}+16 r +15\right ) \left (4 r^{2}+32 r +63\right )}\)	\(\frac {1}{120}\)

For \(n = 5\), using the above recursive equation gives

\[ a_{5}=0 \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(0\)	\(0\)
\(a_{2}\)	\(-\frac {4}{4 r^{2}+16 r +15}\)	\(-{\frac {1}{6}}\)
\(a_{3}\)	\(0\)	\(0\)
\(a_{4}\)	\(\frac {16}{\left (4 r^{2}+16 r +15\right ) \left (4 r^{2}+32 r +63\right )}\)	\(\frac {1}{120}\)
\(a_{5}\)	\(0\)	\(0\)

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^{2}}{6}+\frac {x^{4}}{120}+O\left (x^{6}\right )\right ) \end{align*}

Now the second solution \(y_{2}\left (x \right )\) is found. Let

\[ r_{1}-r_{2} = N \]

Where \(N\) is positive integer which is the difference between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=1\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{1}\left (r \right )\). If this limit exists, then \(C = 0\), else we need to keep the log term and \(C \neq 0\). The above table shows that

\begin{align*} a_N &= a_{1} \\ &= 0 \end{align*}

Therefore

\begin{align*} \lim _{r\rightarrow r_{2}}0&= \lim _{r\rightarrow -{\frac {1}{2}}}0\\ &= 0 \end{align*}

The limit is \(0\). Since the limit exists then the log term is not needed and we can set \(C = 0\). Therefore the second solution has the form

\begin{align*} y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r}\\ &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {1}{2}} \end{align*}

Eq (3) derived above is 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 \(b_{0} = 1\). Substituting \(n = 1\) in Eq(3) gives

\[ b_{1} = 0 \]

For \(2\le n\) the recursive equation is

\begin{equation} \tag{4} b_{n} \left (n +r \right ) \left (n +r -1\right )+b_{n} \left (n +r \right )+b_{n -2}-\frac {b_{n}}{4} = 0 \end{equation}

Which for for the root \(r = -{\frac {1}{2}}\) becomes

\begin{equation} \tag{4A} b_{n} \left (n -\frac {1}{2}\right ) \left (n -\frac {3}{2}\right )+b_{n} \left (n -\frac {1}{2}\right )+b_{n -2}-\frac {b_{n}}{4} = 0 \end{equation}

Solving for \(b_{n}\) from the recursive equation (4) gives

\[ b_{n} = -\frac {4 b_{n -2}}{4 n^{2}+8 n r +4 r^{2}-1}\tag {5} \]

Which for the root \(r = -{\frac {1}{2}}\) becomes

\[ b_{n} = -\frac {4 b_{n -2}}{4 n^{2}-4 n}\tag {6} \]

At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -{\frac {1}{2}}\) and after as more terms are found using the above recursive equation.

\(n\)	\(b_{n ,r}\)	\(b_{n}\)
\(b_{0}\)	\(1\)	\(1\)
\(b_{1}\)	\(0\)	\(0\)

For \(n = 2\), using the above recursive equation gives

\[ b_{2}=-\frac {4}{4 r^{2}+16 r +15} \]

Which for the root \(r = -{\frac {1}{2}}\) becomes

\[ b_{2}=-{\frac {1}{2}} \]

And the table now becomes

\(n\)	\(b_{n ,r}\)	\(b_{n}\)
\(b_{0}\)	\(1\)	\(1\)
\(b_{1}\)	\(0\)	\(0\)
\(b_{2}\)	\(-\frac {4}{4 r^{2}+16 r +15}\)	\(-{\frac {1}{2}}\)

For \(n = 3\), using the above recursive equation gives

\[ b_{3}=0 \]

And the table now becomes

\(n\)	\(b_{n ,r}\)	\(b_{n}\)
\(b_{0}\)	\(1\)	\(1\)
\(b_{1}\)	\(0\)	\(0\)
\(b_{2}\)	\(-\frac {4}{4 r^{2}+16 r +15}\)	\(-{\frac {1}{2}}\)
\(b_{3}\)	\(0\)	\(0\)

For \(n = 4\), using the above recursive equation gives

\[ b_{4}=\frac {16}{\left (4 r^{2}+16 r +15\right ) \left (4 r^{2}+32 r +63\right )} \]

Which for the root \(r = -{\frac {1}{2}}\) becomes

\[ b_{4}={\frac {1}{24}} \]

And the table now becomes

\(n\)	\(b_{n ,r}\)	\(b_{n}\)
\(b_{0}\)	\(1\)	\(1\)
\(b_{1}\)	\(0\)	\(0\)
\(b_{2}\)	\(-\frac {4}{4 r^{2}+16 r +15}\)	\(-{\frac {1}{2}}\)
\(b_{3}\)	\(0\)	\(0\)
\(b_{4}\)	\(\frac {16}{\left (4 r^{2}+16 r +15\right ) \left (4 r^{2}+32 r +63\right )}\)	\(\frac {1}{24}\)

For \(n = 5\), using the above recursive equation gives

\[ b_{5}=0 \]

And the table now becomes

\(n\)	\(b_{n ,r}\)	\(b_{n}\)
\(b_{0}\)	\(1\)	\(1\)
\(b_{1}\)	\(0\)	\(0\)
\(b_{2}\)	\(-\frac {4}{4 r^{2}+16 r +15}\)	\(-{\frac {1}{2}}\)
\(b_{3}\)	\(0\)	\(0\)
\(b_{4}\)	\(\frac {16}{\left (4 r^{2}+16 r +15\right ) \left (4 r^{2}+32 r +63\right )}\)	\(\frac {1}{24}\)
\(b_{5}\)	\(0\)	\(0\)

Using the above table, then the solution \(y_{2}\left (x \right )\) is

\begin{align*} y_{2}\left (x \right )&= \sqrt {x} \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 ) \\ &= \frac {1-\frac {x^{2}}{2}+\frac {x^{4}}{24}+O\left (x^{6}\right )}{\sqrt {x}} \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^{2}}{6}+\frac {x^{4}}{120}+O\left (x^{6}\right )\right ) + \frac {c_2 \left (1-\frac {x^{2}}{2}+\frac {x^{4}}{24}+O\left (x^{6}\right )\right )}{\sqrt {x}} \\ \end{align*}

Hence the final solution is

\begin{align*} y &= y_h \\ &= c_1 \sqrt {x}\, \left (1-\frac {x^{2}}{6}+\frac {x^{4}}{120}+O\left (x^{6}\right )\right )+\frac {c_2 \left (1-\frac {x^{2}}{2}+\frac {x^{4}}{24}+O\left (x^{6}\right )\right )}{\sqrt {x}} \\ \end{align*}

4.56.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )+x y^{\prime }+\left (x^{2}-\frac {1}{4}\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 {\left (4 x^{2}-1\right ) y}{4 x^{2}}-\frac {y^{\prime }}{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 }}{x}+\frac {\left (4 x^{2}-1\right ) y}{4 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 )=\frac {1}{x}, P_{3}\left (x \right )=\frac {4 x^{2}-1}{4 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}}}=1 \\ {} & \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}}}=-\frac {1}{4} \\ {} & \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 x^{2} \left (\frac {d}{d x}y^{\prime }\right )+4 x y^{\prime }+\left (4 x^{2}-1\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..2 \\ {} & {} & 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 \cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) 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 (1+2 r \right ) \left (-1+2 r \right ) x^{r}+a_{1} \left (3+2 r \right ) \left (1+2 r \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (2 k +2 r +1\right ) \left (2 k +2 r -1\right )+4 a_{k -2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (1+2 r \right ) \left (-1+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {1}{2}, \frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (3+2 r \right ) \left (1+2 r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (4 k^{2}+8 k r +4 r^{2}-1\right )+4 a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (4 \left (k +2\right )^{2}+8 \left (k +2\right ) r +4 r^{2}-1\right )+4 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {4 a_{k}}{4 k^{2}+8 k r +4 r^{2}+16 k +16 r +15} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{2} \\ {} & {} & a_{k +2}=-\frac {4 a_{k}}{4 k^{2}+12 k +8} \\ \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 +2}=-\frac {4 a_{k}}{4 k^{2}+12 k +8}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +2}=-\frac {4 a_{k}}{4 k^{2}+20 k +24} \\ \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 +2}=-\frac {4 a_{k}}{4 k^{2}+20 k +24}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{2}}\right ), a_{k +2}=-\frac {4 a_{k}}{4 k^{2}+12 k +8}, a_{1}=0, b_{k +2}=-\frac {4 b_{k}}{4 k^{2}+20 k +24}, b_{1}=0\right ] \end {array} \]

4.56.2 Maple trace

Methods for second order ODEs:
 

4.56.3 Maple dsolve solution

Solving time : 0.016 (sec)
Leaf size : 34

dsolve(x^2*diff(diff(y(x),x),x)+x*diff(y(x),x)+(x^2-1/4)*y(x) = 0,y(x), 
       series,x=0)
 

\[ y = \frac {c_1 x \left (1-\frac {1}{6} x^{2}+\frac {1}{120} x^{4}+\operatorname {O}\left (x^{6}\right )\right )+c_2 \left (1-\frac {1}{2} x^{2}+\frac {1}{24} x^{4}+\operatorname {O}\left (x^{6}\right )\right )}{\sqrt {x}} \]

4.56.4 Mathematica DSolve solution

Solving time : 0.013 (sec)
Leaf size : 58

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

\[ y(x)\to c_1 \left (\frac {x^{7/2}}{24}-\frac {x^{3/2}}{2}+\frac {1}{\sqrt {x}}\right )+c_2 \left (\frac {x^{9/2}}{120}-\frac {x^{5/2}}{6}+\sqrt {x}\right ) \]