2.4.41 problem 38
Internal
problem
ID
[8356]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
4.0
Problem
number
:
38
Date
solved
:
Sunday, November 10, 2024 at 03:39:16 AM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
Solve
\begin{align*} 2 x^{2} \left (2+x \right ) y^{\prime \prime }+5 x^{2} y^{\prime }+\left (1+x \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.
\[ \left (2 x^{3}+4 x^{2}\right ) y^{\prime \prime }+5 x^{2} y^{\prime }+\left (1+x \right ) y = 0 \]
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) &= \frac {5}{2 \left (2+x \right )}\\ q(x) &= \frac {1+x}{2 x^{2} \left (2+x \right )}\\ \end{align*}
Table 2.88: Table \(p(x),q(x)\) singularites.
| |
| \(p(x)=\frac {5}{2 \left (2+x \right )}\) |
| |
|
singularity | type |
| |
| \(x = -2\) | \(\text {``regular''}\) |
| |
| |
| \(q(x)=\frac {1+x}{2 x^{2} \left (2+x \right )}\) |
| |
|
singularity | type |
| |
| \(x = -2\) | \(\text {``regular''}\) |
| |
| \(x = 0\) |
\(\text {``regular''}\) |
| |
Combining everything together gives the following summary of singularities for the ode
as
Regular singular points : \([-2, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to
be
\[ 2 x^{2} \left (2+x \right ) y^{\prime \prime }+5 x^{2} y^{\prime }+\left (1+x \right ) y = 0 \]
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} 2 x^{2} \left (2+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+5 x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (1+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 }}2 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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 }}2 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\
\moverset {\infty }{\munderset {n =0}{\sum }}5 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}5 a_{n -1} \left (n +r -1\right ) x^{n +r} \\
\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}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 =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}5 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r}\right ) = 0
\end{equation}
The indicial
equation is obtained from \(n = 0\). From Eq (2B) this gives
\[ 4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n} x^{n +r} = 0 \]
When \(n = 0\) the above becomes
\[ 4 x^{r} a_{0} r \left (-1+r \right )+a_{0} x^{r} = 0 \]
Or
\[ \left (4 x^{r} r \left (-1+r \right )+x^{r}\right ) a_{0} = 0 \]
Since \(a_{0}\neq 0\) then the above simplifies to
\[ \left (2 r -1\right )^{2} x^{r} = 0 \]
Since the above is true for all \(x\) then the
indicial equation becomes
\[ \left (2 r -1\right )^{2} = 0 \]
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
\[ \left (2 r -1\right )^{2} x^{r} = 0 \]
Solving for \(r\) gives the roots of the indicial equation
as \(\left [{\frac {1}{2}}, {\frac {1}{2}}\right ]\).
Since the root of the indicial equation is repeated, then we can construct two linearly
independent solutions. The first solution has the form
\begin{align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\tag {1A} \end{align*}
Now the second solution \(y_{2}\) is found using
\begin{align*} y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right )\tag {1B} \end{align*}
Then the general solution will be
\[ y = c_1 y_{1}\left (x \right )+c_2 y_{2}\left (x \right ) \]
In Eq (1B) the sum starts from 1 and not zero. In Eq
(1A), \(a_{0}\) is never zero, and is arbitrary and is typically taken as \(a_{0} = 1\), and \(\{c_1, c_2\}\) are two arbitray
constants of integration which can be found from initial conditions. Using the value of the
indicial root found earlier, \(r = {\frac {1}{2}}\), Eqs (1A,1B) become
\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 ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +\frac {1}{2}}\right ) \end{align*}
We start by finding the first solution \(y_{1}\left (x \right )\). Eq (2B) derived above is now used to find all \(a_{n}\)
coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\)
is arbitrary and taken as \(a_{0} = 1\). For \(1\le n\) the recursive equation is
\begin{equation}
\tag{3} 2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+4 a_{n} \left (n +r \right ) \left (n +r -1\right )+5 a_{n -1} \left (n +r -1\right )+a_{n}+a_{n -1} = 0
\end{equation}
Solving for \(a_{n}\) from recursive equation
(4) gives
\[ a_{n} = -\frac {\left (n +r \right ) a_{n -1}}{-1+2 n +2 r}\tag {4} \]
Which for the root \(r = {\frac {1}{2}}\) becomes
\[ a_{n} = -\frac {\left (2 n +1\right ) a_{n -1}}{4 n}\tag {5} \]
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
\[ a_{1}=\frac {-1-r}{1+2 r} \]
Which for the root \(r = {\frac {1}{2}}\) becomes
\[ a_{1}=-{\frac {3}{4}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(\frac {-1-r}{1+2 r}\) | \(-{\frac {3}{4}}\) |
| | |
For \(n = 2\), using the above recursive equation gives
\[ a_{2}=\frac {r^{2}+3 r +2}{4 r^{2}+8 r +3} \]
Which for the root \(r = {\frac {1}{2}}\) becomes
\[ a_{2}={\frac {15}{32}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(\frac {-1-r}{1+2 r}\) | \(-{\frac {3}{4}}\) |
| | |
| \(a_{2}\) | \(\frac {r^{2}+3 r +2}{4 r^{2}+8 r +3}\) | \(\frac {15}{32}\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ a_{3}=\frac {-r^{3}-6 r^{2}-11 r -6}{8 r^{3}+36 r^{2}+46 r +15} \]
Which for the root \(r = {\frac {1}{2}}\) becomes
\[ a_{3}=-{\frac {35}{128}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {-1-r}{1+2 r}\) | \(-{\frac {3}{4}}\) |
| | |
| \(a_{2}\) | \(\frac {r^{2}+3 r +2}{4 r^{2}+8 r +3}\) | \(\frac {15}{32}\) |
| | |
| \(a_{3}\) |
\(\frac {-r^{3}-6 r^{2}-11 r -6}{8 r^{3}+36 r^{2}+46 r +15}\) |
\(-{\frac {35}{128}}\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ a_{4}=\frac {r^{4}+10 r^{3}+35 r^{2}+50 r +24}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105} \]
Which for the root \(r = {\frac {1}{2}}\) becomes
\[ a_{4}={\frac {315}{2048}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {-1-r}{1+2 r}\) | \(-{\frac {3}{4}}\) |
| | |
| \(a_{2}\) | \(\frac {r^{2}+3 r +2}{4 r^{2}+8 r +3}\) | \(\frac {15}{32}\) |
| | |
| \(a_{3}\) | \(\frac {-r^{3}-6 r^{2}-11 r -6}{8 r^{3}+36 r^{2}+46 r +15}\) | \(-{\frac {35}{128}}\) |
| | |
| \(a_{4}\) |
\(\frac {r^{4}+10 r^{3}+35 r^{2}+50 r +24}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) |
\(\frac {315}{2048}\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ a_{5}=\frac {-r^{5}-15 r^{4}-85 r^{3}-225 r^{2}-274 r -120}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945} \]
Which for the root \(r = {\frac {1}{2}}\) becomes
\[ a_{5}=-{\frac {693}{8192}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {-1-r}{1+2 r}\) |
\(-{\frac {3}{4}}\) |
| | |
| \(a_{2}\) |
\(\frac {r^{2}+3 r +2}{4 r^{2}+8 r +3}\) | \(\frac {15}{32}\) |
| | |
| \(a_{3}\) | \(\frac {-r^{3}-6 r^{2}-11 r -6}{8 r^{3}+36 r^{2}+46 r +15}\) | \(-{\frac {35}{128}}\) |
| | |
| \(a_{4}\) |
\(\frac {r^{4}+10 r^{3}+35 r^{2}+50 r +24}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) |
\(\frac {315}{2048}\) |
| | |
| \(a_{5}\) |
\(\frac {-r^{5}-15 r^{4}-85 r^{3}-225 r^{2}-274 r -120}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945}\) |
\(-{\frac {693}{8192}}\) |
| | |
Using the above table, then the first 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 {3 x}{4}+\frac {15 x^{2}}{32}-\frac {35 x^{3}}{128}+\frac {315 x^{4}}{2048}-\frac {693 x^{5}}{8192}+O\left (x^{6}\right )\right ) \\
\end{align*}
Now the second solution is found. The
second solution is given by
\[ y_{2}\left (x \right ) = y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right ) \]
Where \(b_{n}\) is found using
\[ b_{n} = \frac {d}{d r}a_{n ,r} \]
And the above is then evaluated at \(r = {\frac {1}{2}}\). The
above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| | | | |
| \(n\) |
\(b_{n ,r}\) |
\(a_{n}\) |
\(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) |
\(b_{n}\left (r =\frac {1}{2}\right )\) |
| | | | |
| \(b_{0}\) |
\(1\) |
\(1\) |
N/A since \(b_{n}\) starts from 1 |
N/A |
| | | | |
| \(b_{1}\) | \(\frac {-1-r}{1+2 r}\) | \(-{\frac {3}{4}}\) | \(\frac {1}{\left (1+2 r \right )^{2}}\) | \(\frac {1}{4}\) |
| | | | |
| \(b_{2}\) | \(\frac {r^{2}+3 r +2}{4 r^{2}+8 r +3}\) | \(\frac {15}{32}\) | \(\frac {-4 r^{2}-10 r -7}{\left (4 r^{2}+8 r +3\right )^{2}}\) | \(-{\frac {13}{64}}\) |
| | | | |
| \(b_{3}\) | \(\frac {-r^{3}-6 r^{2}-11 r -6}{8 r^{3}+36 r^{2}+46 r +15}\) | \(-{\frac {35}{128}}\) | \(\frac {12 r^{4}+84 r^{3}+219 r^{2}+252 r +111}{\left (8 r^{3}+36 r^{2}+46 r +15\right )^{2}}\) | \(\frac {101}{768}\) |
| | | | |
| \(b_{4}\) |
\(\frac {r^{4}+10 r^{3}+35 r^{2}+50 r +24}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) |
\(\frac {315}{2048}\) |
\(\frac {-32 r^{6}-432 r^{5}-2384 r^{4}-6876 r^{3}-10946 r^{2}-9162 r -3198}{\left (16 r^{4}+128 r^{3}+344 r^{2}+352 r +105\right )^{2}}\) |
\(-{\frac {641}{8192}}\) |
| | | | |
| \(b_{5}\) |
\(\frac {-r^{5}-15 r^{4}-85 r^{3}-225 r^{2}-274 r -120}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945}\) |
\(-{\frac {693}{8192}}\) |
\(\frac {80 r^{8}+1760 r^{7}+16600 r^{6}+87560 r^{5}+282265 r^{4}+569360 r^{3}+702575 r^{2}+486750 r +146430}{\left (32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945\right )^{2}}\) |
\(\frac {7303}{163840}\) |
| | | | |
The above table gives all values of \(b_{n}\) needed. Hence the second solution is
\begin{align*}
y_{2}\left (x \right )&=y_{1}\left (x \right ) \ln \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 \\
&= \sqrt {x}\, \left (1-\frac {3 x}{4}+\frac {15 x^{2}}{32}-\frac {35 x^{3}}{128}+\frac {315 x^{4}}{2048}-\frac {693 x^{5}}{8192}+O\left (x^{6}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (\frac {x}{4}-\frac {13 x^{2}}{64}+\frac {101 x^{3}}{768}-\frac {641 x^{4}}{8192}+\frac {7303 x^{5}}{163840}+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 \sqrt {x}\, \left (1-\frac {3 x}{4}+\frac {15 x^{2}}{32}-\frac {35 x^{3}}{128}+\frac {315 x^{4}}{2048}-\frac {693 x^{5}}{8192}+O\left (x^{6}\right )\right ) + c_2 \left (\sqrt {x}\, \left (1-\frac {3 x}{4}+\frac {15 x^{2}}{32}-\frac {35 x^{3}}{128}+\frac {315 x^{4}}{2048}-\frac {693 x^{5}}{8192}+O\left (x^{6}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (\frac {x}{4}-\frac {13 x^{2}}{64}+\frac {101 x^{3}}{768}-\frac {641 x^{4}}{8192}+\frac {7303 x^{5}}{163840}+O\left (x^{6}\right )\right )\right ) \\
\end{align*}
Hence the final solution is
\begin{align*}
y &= y_h \\
&= c_1 \sqrt {x}\, \left (1-\frac {3 x}{4}+\frac {15 x^{2}}{32}-\frac {35 x^{3}}{128}+\frac {315 x^{4}}{2048}-\frac {693 x^{5}}{8192}+O\left (x^{6}\right )\right )+c_2 \left (\sqrt {x}\, \left (1-\frac {3 x}{4}+\frac {15 x^{2}}{32}-\frac {35 x^{3}}{128}+\frac {315 x^{4}}{2048}-\frac {693 x^{5}}{8192}+O\left (x^{6}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (\frac {x}{4}-\frac {13 x^{2}}{64}+\frac {101 x^{3}}{768}-\frac {641 x^{4}}{8192}+\frac {7303 x^{5}}{163840}+O\left (x^{6}\right )\right )\right ) \\
\end{align*}
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x^{2} \left (x +2\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+5 x^{2} \left (\frac {d}{d x}y \left (x \right )\right )+\left (x +1\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 {\left (x +1\right ) y \left (x \right )}{2 \left (x +2\right ) x^{2}}-\frac {5 \left (\frac {d}{d x}y \left (x \right )\right )}{2 \left (x +2\right )} \\ \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 {5 \left (\frac {d}{d x}y \left (x \right )\right )}{2 \left (x +2\right )}+\frac {\left (x +1\right ) y \left (x \right )}{2 \left (x +2\right ) x^{2}}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {5}{2 \left (x +2\right )}, P_{3}\left (x \right )=\frac {x +1}{2 \left (x +2\right ) x^{2}}\right ] \\ {} & \circ & \left (x +2\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-2 \\ {} & {} & \left (\left (x +2\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-2}}}=\frac {5}{2} \\ {} & \circ & \left (x +2\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-2 \\ {} & {} & \left (\left (x +2\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-2}}}=0 \\ {} & \circ & x =-2\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=-2 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 2 x^{2} \left (x +2\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+5 x^{2} \left (\frac {d}{d x}y \left (x \right )\right )+\left (x +1\right ) y \left (x \right )=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -2\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (2 u^{3}-8 u^{2}+8 u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (5 u^{2}-20 u +20\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (u -1\right ) y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 4 a_{0} r \left (3+2 r \right ) u^{-1+r}+\left (4 a_{1} \left (1+r \right ) \left (5+2 r \right )-a_{0} \left (8 r^{2}+12 r +1\right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (4 a_{k +1} \left (k +r +1\right ) \left (2 k +5+2 r \right )-a_{k} \left (8 k^{2}+16 k r +8 r^{2}+12 k +12 r +1\right )+a_{k -1} \left (k +r \right ) \left (2 k -1+2 r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 4 r \left (3+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -\frac {3}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 4 a_{1} \left (1+r \right ) \left (5+2 r \right )-a_{0} \left (8 r^{2}+12 r +1\right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 2 \left (-4 a_{k}+a_{k -1}+4 a_{k +1}\right ) k^{2}+\left (4 \left (-4 a_{k}+a_{k -1}+4 a_{k +1}\right ) r -12 a_{k}-a_{k -1}+28 a_{k +1}\right ) k +2 \left (-4 a_{k}+a_{k -1}+4 a_{k +1}\right ) r^{2}+\left (-12 a_{k}-a_{k -1}+28 a_{k +1}\right ) r -a_{k}+20 a_{k +1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 2 \left (-4 a_{k +1}+a_{k}+4 a_{k +2}\right ) \left (k +1\right )^{2}+\left (4 \left (-4 a_{k +1}+a_{k}+4 a_{k +2}\right ) r -12 a_{k +1}-a_{k}+28 a_{k +2}\right ) \left (k +1\right )+2 \left (-4 a_{k +1}+a_{k}+4 a_{k +2}\right ) r^{2}+\left (-12 a_{k +1}-a_{k}+28 a_{k +2}\right ) r -a_{k +1}+20 a_{k +2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}+4 k r a_{k}-16 k r a_{k +1}+2 r^{2} a_{k}-8 r^{2} a_{k +1}+3 k a_{k}-28 k a_{k +1}+3 r a_{k}-28 r a_{k +1}+a_{k}-21 a_{k +1}}{4 \left (2 k^{2}+4 k r +2 r^{2}+11 k +11 r +14\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}+3 k a_{k}-28 k a_{k +1}+a_{k}-21 a_{k +1}}{4 \left (2 k^{2}+11 k +14\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}+3 k a_{k}-28 k a_{k +1}+a_{k}-21 a_{k +1}}{4 \left (2 k^{2}+11 k +14\right )}, 20 a_{1}-a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +2 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +2\right )^{k}, a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}+3 k a_{k}-28 k a_{k +1}+a_{k}-21 a_{k +1}}{4 \left (2 k^{2}+11 k +14\right )}, 20 a_{1}-a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {3}{2} \\ {} & {} & a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}-3 k a_{k}-4 k a_{k +1}+a_{k}+3 a_{k +1}}{4 \left (2 k^{2}+5 k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {3}{2} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -\frac {3}{2}}, a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}-3 k a_{k}-4 k a_{k +1}+a_{k}+3 a_{k +1}}{4 \left (2 k^{2}+5 k +2\right )}, -4 a_{1}-a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +2 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +2\right )^{k -\frac {3}{2}}, a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}-3 k a_{k}-4 k a_{k +1}+a_{k}+3 a_{k +1}}{4 \left (2 k^{2}+5 k +2\right )}, -4 a_{1}-a_{0}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y \left (x \right )=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +2\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +2\right )^{k -\frac {3}{2}}\right ), a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}+3 k a_{k}-28 k a_{k +1}+a_{k}-21 a_{k +1}}{4 \left (2 k^{2}+11 k +14\right )}, 20 a_{1}-a_{0}=0, b_{k +2}=-\frac {2 k^{2} b_{k}-8 k^{2} b_{k +1}-3 k b_{k}-4 k b_{k +1}+b_{k}+3 b_{k +1}}{4 \left (2 k^{2}+5 k +2\right )}, -4 b_{1}-b_{0}=0\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
A Liouvillian solution exists
Reducible group (found an exponential solution)
Group is reducible, not completely reducible
<- Kovacics algorithm successful`
Maple dsolve solution
Solving time : 0.021 (sec)
Leaf size : 48
dsolve(2*x^2*(x+2)*diff(diff(y(x),x),x)+5*diff(y(x),x)*x^2+y(x)*(x+1) = 0,y(x),
series,x=0)
\[
y = \sqrt {x}\, \left (\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1-\frac {3}{4} x +\frac {15}{32} x^{2}-\frac {35}{128} x^{3}+\frac {315}{2048} x^{4}-\frac {693}{8192} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (\frac {1}{4} x -\frac {13}{64} x^{2}+\frac {101}{768} x^{3}-\frac {641}{8192} x^{4}+\frac {7303}{163840} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) c_{2} \right )
\]
Mathematica DSolve solution
Solving time : 0.022 (sec)
Leaf size : 134
AsymptoticDSolveValue[{2*x^2*(2+x)*D[y[x],{x,2}] +5*x^2*D[y[x],x]+(1+x)*y[x] == 0,{}},
y[x],{x,0,5}]
\[
y(x)\to c_1 \sqrt {x} \left (-\frac {693 x^5}{8192}+\frac {315 x^4}{2048}-\frac {35 x^3}{128}+\frac {15 x^2}{32}-\frac {3 x}{4}+1\right )+c_2 \left (\sqrt {x} \left (\frac {7303 x^5}{163840}-\frac {641 x^4}{8192}+\frac {101 x^3}{768}-\frac {13 x^2}{64}+\frac {x}{4}\right )+\sqrt {x} \left (-\frac {693 x^5}{8192}+\frac {315 x^4}{2048}-\frac {35 x^3}{128}+\frac {15 x^2}{32}-\frac {3 x}{4}+1\right ) \log (x)\right )
\]