2.4.36 problem 33
Internal
problem
ID
[8351]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
4.0
Problem
number
:
33
Date
solved
:
Sunday, November 10, 2024 at 03:39:10 AM
CAS
classification
:
[[_Emden, _Fowler], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]
Solve
\begin{align*} 4 x y^{\prime \prime }+2 y^{\prime }+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.
\[ 4 x y^{\prime \prime }+2 y^{\prime }+y = 0 \]
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*}
Table 2.83: Table \(p(x),q(x)\) singularites.
| |
| \(p(x)=\frac {1}{2 x}\) |
| |
|
singularity | type |
| |
| \(x = 0\) | \(\text {``regular''}\) |
| |
| |
| \(q(x)=\frac {1}{4 x}\) |
| |
|
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
\[ 4 x y^{\prime \prime }+2 y^{\prime }+y = 0 \]
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 x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+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*}
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right ) x +2 \frac {d}{d x}y \left (x \right )+y \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=-\frac {y \left (x \right )}{4 x}-\frac {\frac {d}{d x}y \left (x \right )}{2 x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (x \right )\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^{2}}{d x^{2}}y \left (x \right )+\frac {\frac {d}{d x}y \left (x \right )}{2 x}+\frac {y \left (x \right )}{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 \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right ) x +2 \frac {d}{d x}y \left (x \right )+y \left (x \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=\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} \frac {d}{d x}y \left (x \right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\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 \\ {} & {} & \frac {d}{d x}y \left (x \right )=\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^{2}}{d x^{2}}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\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^{2}}{d x^{2}}y \left (x \right )\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} \\ {} & {} & 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 \left (x \right )=\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 \left (x \right )=\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 (x \right )=\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`
Maple dsolve solution
Solving time : 0.020 (sec)
Leaf size : 44
dsolve(4*x*diff(diff(y(x),x),x)+2*diff(y(x),x)+y(x) = 0,y(x),
series,x=0)
\[
y = c_{1} \sqrt {x}\, \left (1-\frac {1}{6} x +\frac {1}{120} x^{2}-\frac {1}{5040} x^{3}+\frac {1}{362880} x^{4}-\frac {1}{39916800} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (1-\frac {1}{2} x +\frac {1}{24} x^{2}-\frac {1}{720} x^{3}+\frac {1}{40320} x^{4}-\frac {1}{3628800} x^{5}+\operatorname {O}\left (x^{6}\right )\right )
\]
Mathematica DSolve solution
Solving time : 0.004 (sec)
Leaf size : 85
AsymptoticDSolveValue[{4*x*D[y[x],{x,2}] +2*D[y[x],x]+y[x] == 0,{}},
y[x],{x,0,5}]
\[
y(x)\to c_1 \sqrt {x} \left (-\frac {x^5}{39916800}+\frac {x^4}{362880}-\frac {x^3}{5040}+\frac {x^2}{120}-\frac {x}{6}+1\right )+c_2 \left (-\frac {x^5}{3628800}+\frac {x^4}{40320}-\frac {x^3}{720}+\frac {x^2}{24}-\frac {x}{2}+1\right )
\]