2.4.30 problem 26
Internal
problem
ID
[8345]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
4.0
Problem
number
:
26
Date
solved
:
Sunday, November 10, 2024 at 03:38:57 AM
CAS
classification
:
[[_2nd_order, _exact, _linear, _homogeneous]]
Solve
\begin{align*} x^{2} \left (3+x \right ) y^{\prime \prime }+5 x \left (1+x \right ) y^{\prime }-\left (1-4 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.
\[ \left (x^{3}+3 x^{2}\right ) y^{\prime \prime }+\left (5 x^{2}+5 x \right ) y^{\prime }+\left (4 x -1\right ) y = 0 \]
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+5 x}{x \left (3+x \right )}\\ q(x) &= \frac {4 x -1}{x^{2} \left (3+x \right )}\\ \end{align*}
Table 2.79: Table \(p(x),q(x)\) singularites.
| |
| \(p(x)=\frac {5+5 x}{x \left (3+x \right )}\) |
| |
|
singularity | type |
| |
| \(x = -3\) | \(\text {``regular''}\) |
| |
| \(x = 0\) |
\(\text {``regular''}\) |
| |
| |
| \(q(x)=\frac {4 x -1}{x^{2} \left (3+x \right )}\) |
| |
|
singularity | type |
| |
| \(x = -3\) | \(\text {``regular''}\) |
| |
| \(x = 0\) |
\(\text {``regular''}\) |
| |
Combining everything together gives the following summary of singularities for the ode
as
Regular singular points : \([-3, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to
be
\[ x^{2} \left (3+x \right ) y^{\prime \prime }+\left (5 x^{2}+5 x \right ) y^{\prime }+\left (4 x -1\right ) y = 0 \]
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 (3+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 )+\left (5 x^{2}+5 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (4 x -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 }}x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 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 }}5 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n +r}\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^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}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 }}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 =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 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 }}5 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n +r}\right ) = 0
\end{equation}
The indicial
equation is obtained from \(n = 0\). From Eq (2B) this gives
\[ 3 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+5 x^{n +r} a_{n} \left (n +r \right )-a_{n} x^{n +r} = 0 \]
When \(n = 0\) the above becomes
\[ 3 x^{r} a_{0} r \left (-1+r \right )+5 x^{r} a_{0} r -a_{0} x^{r} = 0 \]
Or
\[ \left (3 x^{r} r \left (-1+r \right )+5 x^{r} r -x^{r}\right ) a_{0} = 0 \]
Since \(a_{0}\neq 0\) then the above simplifies to
\[ \left (3 r^{2}+2 r -1\right ) x^{r} = 0 \]
Since the above is true for all \(x\) then the
indicial equation becomes
\[ 3 r^{2}+2 r -1 = 0 \]
Solving for \(r\) gives the roots of the indicial equation as
\begin{align*} r_1 &= {\frac {1}{3}}\\ r_2 &= -1 \end{align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes
\[ \left (3 r^{2}+2 r -1\right ) x^{r} = 0 \]
Solving for \(r\) gives the roots of the indicial equation
as \(\left [{\frac {1}{3}}, -1\right ]\).
Since \(r_1 - r_2 = {\frac {4}{3}}\) 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}{3}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -1} \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 -1} \left (n +r -1\right ) \left (n +r -2\right )+3 a_{n} \left (n +r \right ) \left (n +r -1\right )+5 a_{n -1} \left (n +r -1\right )+5 a_{n} \left (n +r \right )+4 a_{n -1}-a_{n} = 0
\end{equation}
Solving for \(a_{n}\) from recursive equation (4) gives
\[ a_{n} = -\frac {\left (1+n +r \right ) a_{n -1}}{3 n +3 r -1}\tag {4} \]
Which for the
root \(r = {\frac {1}{3}}\) becomes
\[ a_{n} = -\frac {\left (4+3 n \right ) a_{n -1}}{9 n}\tag {5} \]
At this point, it is a good idea to keep track of \(a_{n}\) in a table both before
substituting \(r = {\frac {1}{3}}\) 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
\[ a_{1}=\frac {-2-r}{2+3 r} \]
Which for the root \(r = {\frac {1}{3}}\) becomes
\[ a_{1}=-{\frac {7}{9}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(\frac {-2-r}{2+3 r}\) | \(-{\frac {7}{9}}\) |
| | |
For \(n = 2\), using the above recursive equation gives
\[ a_{2}=\frac {r^{2}+5 r +6}{9 r^{2}+21 r +10} \]
Which for the root \(r = {\frac {1}{3}}\) becomes
\[ a_{2}={\frac {35}{81}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(\frac {-2-r}{2+3 r}\) | \(-{\frac {7}{9}}\) |
| | |
| \(a_{2}\) | \(\frac {r^{2}+5 r +6}{9 r^{2}+21 r +10}\) | \(\frac {35}{81}\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ a_{3}=\frac {-r^{3}-9 r^{2}-26 r -24}{27 r^{3}+135 r^{2}+198 r +80} \]
Which for the root \(r = {\frac {1}{3}}\) becomes
\[ a_{3}=-{\frac {455}{2187}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {-2-r}{2+3 r}\) | \(-{\frac {7}{9}}\) |
| | |
| \(a_{2}\) | \(\frac {r^{2}+5 r +6}{9 r^{2}+21 r +10}\) | \(\frac {35}{81}\) |
| | |
| \(a_{3}\) |
\(\frac {-r^{3}-9 r^{2}-26 r -24}{27 r^{3}+135 r^{2}+198 r +80}\) |
\(-{\frac {455}{2187}}\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ a_{4}=\frac {r^{4}+14 r^{3}+71 r^{2}+154 r +120}{81 r^{4}+702 r^{3}+2079 r^{2}+2418 r +880} \]
Which for the root \(r = {\frac {1}{3}}\) becomes
\[ a_{4}={\frac {1820}{19683}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {-2-r}{2+3 r}\) | \(-{\frac {7}{9}}\) |
| | |
| \(a_{2}\) | \(\frac {r^{2}+5 r +6}{9 r^{2}+21 r +10}\) | \(\frac {35}{81}\) |
| | |
| \(a_{3}\) | \(\frac {-r^{3}-9 r^{2}-26 r -24}{27 r^{3}+135 r^{2}+198 r +80}\) | \(-{\frac {455}{2187}}\) |
| | |
| \(a_{4}\) |
\(\frac {r^{4}+14 r^{3}+71 r^{2}+154 r +120}{81 r^{4}+702 r^{3}+2079 r^{2}+2418 r +880}\) |
\(\frac {1820}{19683}\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ a_{5}=\frac {-r^{5}-20 r^{4}-155 r^{3}-580 r^{2}-1044 r -720}{243 r^{5}+3240 r^{4}+16065 r^{3}+36360 r^{2}+36492 r +12320} \]
Which for the root \(r = {\frac {1}{3}}\) becomes
\[ a_{5}=-{\frac {6916}{177147}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {-2-r}{2+3 r}\) |
\(-{\frac {7}{9}}\) |
| | |
| \(a_{2}\) |
\(\frac {r^{2}+5 r +6}{9 r^{2}+21 r +10}\) | \(\frac {35}{81}\) |
| | |
| \(a_{3}\) | \(\frac {-r^{3}-9 r^{2}-26 r -24}{27 r^{3}+135 r^{2}+198 r +80}\) | \(-{\frac {455}{2187}}\) |
| | |
| \(a_{4}\) |
\(\frac {r^{4}+14 r^{3}+71 r^{2}+154 r +120}{81 r^{4}+702 r^{3}+2079 r^{2}+2418 r +880}\) |
\(\frac {1820}{19683}\) |
| | |
| \(a_{5}\) |
\(\frac {-r^{5}-20 r^{4}-155 r^{3}-580 r^{2}-1044 r -720}{243 r^{5}+3240 r^{4}+16065 r^{3}+36360 r^{2}+36492 r +12320}\) |
\(-{\frac {6916}{177147}}\) |
| | |
Using the above table, then the solution \(y_{1}\left (x \right )\) is
\begin{align*} y_{1}\left (x \right )&= x^{{1}/{3}} \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^{{1}/{3}} \left (1-\frac {7 x}{9}+\frac {35 x^{2}}{81}-\frac {455 x^{3}}{2187}+\frac {1820 x^{4}}{19683}-\frac {6916 x^{5}}{177147}+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 -1} \left (n +r -1\right ) \left (n +r -2\right )+3 b_{n} \left (n +r \right ) \left (n +r -1\right )+5 b_{n -1} \left (n +r -1\right )+5 b_{n} \left (n +r \right )+4 b_{n -1}-b_{n} = 0
\end{equation}
Solving for \(b_{n}\) from recursive equation (4) gives
\[ b_{n} = -\frac {\left (1+n +r \right ) b_{n -1}}{3 n +3 r -1}\tag {4} \]
Which for the root \(r = -1\) becomes
\[ b_{n} = -\frac {n b_{n -1}}{3 n -4}\tag {5} \]
At this point, it is a good idea to keep track of \(b_{n}\) in a table both
before substituting \(r = -1\) 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
\[ b_{1}=\frac {-2-r}{2+3 r} \]
Which for the root \(r = -1\) becomes
\[ b_{1}=1 \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) | \(1\) |
| | |
| \(b_{1}\) | \(\frac {-2-r}{2+3 r}\) | \(1\) |
| | |
For \(n = 2\), using the above recursive equation gives
\[ b_{2}=\frac {r^{2}+5 r +6}{9 r^{2}+21 r +10} \]
Which for the root \(r = -1\) becomes
\[ b_{2}=-1 \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) | \(1\) |
| | |
| \(b_{1}\) | \(\frac {-2-r}{2+3 r}\) | \(1\) |
| | |
| \(b_{2}\) | \(\frac {r^{2}+5 r +6}{9 r^{2}+21 r +10}\) | \(-1\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ b_{3}=\frac {-r^{3}-9 r^{2}-26 r -24}{27 r^{3}+135 r^{2}+198 r +80} \]
Which for the root \(r = -1\) becomes
\[ b_{3}={\frac {3}{5}} \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) |
\(1\) |
| | |
| \(b_{1}\) |
\(\frac {-2-r}{2+3 r}\) | \(1\) |
| | |
| \(b_{2}\) | \(\frac {r^{2}+5 r +6}{9 r^{2}+21 r +10}\) | \(-1\) |
| | |
| \(b_{3}\) |
\(\frac {-r^{3}-9 r^{2}-26 r -24}{27 r^{3}+135 r^{2}+198 r +80}\) |
\(\frac {3}{5}\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ b_{4}=\frac {r^{4}+14 r^{3}+71 r^{2}+154 r +120}{81 r^{4}+702 r^{3}+2079 r^{2}+2418 r +880} \]
Which for the root \(r = -1\) becomes
\[ b_{4}=-{\frac {3}{10}} \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) |
\(1\) |
| | |
| \(b_{1}\) |
\(\frac {-2-r}{2+3 r}\) | \(1\) |
| | |
| \(b_{2}\) | \(\frac {r^{2}+5 r +6}{9 r^{2}+21 r +10}\) | \(-1\) |
| | |
| \(b_{3}\) | \(\frac {-r^{3}-9 r^{2}-26 r -24}{27 r^{3}+135 r^{2}+198 r +80}\) | \(\frac {3}{5}\) |
| | |
| \(b_{4}\) |
\(\frac {r^{4}+14 r^{3}+71 r^{2}+154 r +120}{81 r^{4}+702 r^{3}+2079 r^{2}+2418 r +880}\) |
\(-{\frac {3}{10}}\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ b_{5}=\frac {-r^{5}-20 r^{4}-155 r^{3}-580 r^{2}-1044 r -720}{243 r^{5}+3240 r^{4}+16065 r^{3}+36360 r^{2}+36492 r +12320} \]
Which for the root \(r = -1\) becomes
\[ b_{5}={\frac {3}{22}} \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) |
\(1\) |
| | |
| \(b_{1}\) |
\(\frac {-2-r}{2+3 r}\) |
\(1\) |
| | |
| \(b_{2}\) |
\(\frac {r^{2}+5 r +6}{9 r^{2}+21 r +10}\) | \(-1\) |
| | |
| \(b_{3}\) | \(\frac {-r^{3}-9 r^{2}-26 r -24}{27 r^{3}+135 r^{2}+198 r +80}\) | \(\frac {3}{5}\) |
| | |
| \(b_{4}\) |
\(\frac {r^{4}+14 r^{3}+71 r^{2}+154 r +120}{81 r^{4}+702 r^{3}+2079 r^{2}+2418 r +880}\) |
\(-{\frac {3}{10}}\) |
| | |
| \(b_{5}\) |
\(\frac {-r^{5}-20 r^{4}-155 r^{3}-580 r^{2}-1044 r -720}{243 r^{5}+3240 r^{4}+16065 r^{3}+36360 r^{2}+36492 r +12320}\) |
\(\frac {3}{22}\) |
| | |
Using the above table, then the solution \(y_{2}\left (x \right )\) is
\begin{align*} y_{2}\left (x \right )&= x^{{1}/{3}} \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+x -x^{2}+\frac {3 x^{3}}{5}-\frac {3 x^{4}}{10}+\frac {3 x^{5}}{22}+O\left (x^{6}\right )}{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 \,x^{{1}/{3}} \left (1-\frac {7 x}{9}+\frac {35 x^{2}}{81}-\frac {455 x^{3}}{2187}+\frac {1820 x^{4}}{19683}-\frac {6916 x^{5}}{177147}+O\left (x^{6}\right )\right ) + \frac {c_2 \left (1+x -x^{2}+\frac {3 x^{3}}{5}-\frac {3 x^{4}}{10}+\frac {3 x^{5}}{22}+O\left (x^{6}\right )\right )}{x} \\
\end{align*}
Hence the final solution is
\begin{align*}
y &= y_h \\
&= c_1 \,x^{{1}/{3}} \left (1-\frac {7 x}{9}+\frac {35 x^{2}}{81}-\frac {455 x^{3}}{2187}+\frac {1820 x^{4}}{19683}-\frac {6916 x^{5}}{177147}+O\left (x^{6}\right )\right )+\frac {c_2 \left (1+x -x^{2}+\frac {3 x^{3}}{5}-\frac {3 x^{4}}{10}+\frac {3 x^{5}}{22}+O\left (x^{6}\right )\right )}{x} \\
\end{align*}
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (x +3\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+5 x \left (x +1\right ) \left (\frac {d}{d x}y \left (x \right )\right )-\left (1-4 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 {\left (4 x -1\right ) y \left (x \right )}{x^{2} \left (x +3\right )}-\frac {5 \left (x +1\right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x \left (x +3\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 (x +1\right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x \left (x +3\right )}+\frac {\left (4 x -1\right ) y \left (x \right )}{x^{2} \left (x +3\right )}=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 \left (x +1\right )}{x \left (x +3\right )}, P_{3}\left (x \right )=\frac {4 x -1}{x^{2} \left (x +3\right )}\right ] \\ {} & \circ & \left (x +3\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-3 \\ {} & {} & \left (\left (x +3\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-3}}}=\frac {10}{3} \\ {} & \circ & \left (x +3\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-3 \\ {} & {} & \left (\left (x +3\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-3}}}=0 \\ {} & \circ & x =-3\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}=-3 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{2} \left (x +3\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+5 x \left (x +1\right ) \left (\frac {d}{d x}y \left (x \right )\right )+\left (4 x -1\right ) y \left (x \right )=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -3\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{3}-6 u^{2}+9 u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (5 u^{2}-25 u +30\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (4 u -13\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} \\ {} & {} & 3 a_{0} r \left (7+3 r \right ) u^{-1+r}+\left (3 a_{1} \left (1+r \right ) \left (10+3 r \right )-a_{0} \left (13+6 r \right ) \left (1+r \right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (3 a_{k +1} \left (k +r +1\right ) \left (3 k +10+3 r \right )-a_{k} \left (6 k +6 r +13\right ) \left (k +r +1\right )+a_{k -1} \left (k +r +1\right )^{2}\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 3 r \left (7+3 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -\frac {7}{3}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 3 a_{1} \left (1+r \right ) \left (10+3 r \right )-a_{0} \left (13+6 r \right ) \left (1+r \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -6 \left (\left (a_{k}-\frac {a_{k -1}}{6}-\frac {3 a_{k +1}}{2}\right ) k +\left (a_{k}-\frac {a_{k -1}}{6}-\frac {3 a_{k +1}}{2}\right ) r +\frac {13 a_{k}}{6}-\frac {a_{k -1}}{6}-5 a_{k +1}\right ) \left (k +r +1\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & -6 \left (\left (a_{k +1}-\frac {a_{k}}{6}-\frac {3 a_{k +2}}{2}\right ) \left (k +1\right )+\left (a_{k +1}-\frac {a_{k}}{6}-\frac {3 a_{k +2}}{2}\right ) r +\frac {13 a_{k +1}}{6}-\frac {a_{k}}{6}-5 a_{k +2}\right ) \left (k +r +2\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k a_{k}-6 k a_{k +1}+r a_{k}-6 r a_{k +1}+2 a_{k}-19 a_{k +1}}{3 \left (3 k +13+3 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {k a_{k}-6 k a_{k +1}+2 a_{k}-19 a_{k +1}}{3 \left (3 k +13\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 {k a_{k}-6 k a_{k +1}+2 a_{k}-19 a_{k +1}}{3 \left (3 k +13\right )}, 30 a_{1}-13 a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +3 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +3\right )^{k}, a_{k +2}=-\frac {k a_{k}-6 k a_{k +1}+2 a_{k}-19 a_{k +1}}{3 \left (3 k +13\right )}, 30 a_{1}-13 a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {7}{3} \\ {} & {} & a_{k +2}=-\frac {k a_{k}-6 k a_{k +1}-\frac {1}{3} a_{k}-5 a_{k +1}}{3 \left (3 k +6\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {7}{3} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -\frac {7}{3}}, a_{k +2}=-\frac {k a_{k}-6 k a_{k +1}-\frac {1}{3} a_{k}-5 a_{k +1}}{3 \left (3 k +6\right )}, -12 a_{1}-\frac {4 a_{0}}{3}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +3 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +3\right )^{k -\frac {7}{3}}, a_{k +2}=-\frac {k a_{k}-6 k a_{k +1}-\frac {1}{3} a_{k}-5 a_{k +1}}{3 \left (3 k +6\right )}, -12 a_{1}-\frac {4 a_{0}}{3}=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 +3\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +3\right )^{k -\frac {7}{3}}\right ), a_{k +2}=-\frac {k a_{k}-6 k a_{k +1}+2 a_{k}-19 a_{k +1}}{3 \left (3 k +13\right )}, 30 a_{1}-13 a_{0}=0, b_{k +2}=-\frac {k b_{k}-6 k b_{k +1}-\frac {1}{3} b_{k}-5 b_{k +1}}{3 \left (3 k +6\right )}, -12 b_{1}-\frac {4 b_{0}}{3}=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)]
One independent solution has integrals. Trying a hypergeometric solution free of integrals...
-> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius
<- hyper3 successful: received ODE is equivalent to the 2F1 ODE
-> Trying to convert hypergeometric functions to elementary form...
<- elementary form for at least one hypergeometric solution is achieved - returning with no uncomputed integrals
<- linear_1 successful`
Maple dsolve solution
Solving time : 0.023 (sec)
Leaf size : 48
dsolve(x^2*(x+3)*diff(diff(y(x),x),x)+5*x*(x+1)*diff(y(x),x)-(1-4*x)*y(x) = 0,y(x),
series,x=0)
\[
y = \frac {c_{2} x^{{4}/{3}} \left (1-\frac {7}{9} x +\frac {35}{81} x^{2}-\frac {455}{2187} x^{3}+\frac {1820}{19683} x^{4}-\frac {6916}{177147} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{1} \left (1+x -x^{2}+\frac {3}{5} x^{3}-\frac {3}{10} x^{4}+\frac {3}{22} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{x}
\]
Mathematica DSolve solution
Solving time : 0.018 (sec)
Leaf size : 82
AsymptoticDSolveValue[{x^2*(3+x)*D[y[x],{x,2}] + 5*x*(1+x)*D[y[x],x] - (1-4*x)*y[x] == 0,{}},
y[x],{x,0,5}]
\[
y(x)\to c_1 \sqrt [3]{x} \left (-\frac {6916 x^5}{177147}+\frac {1820 x^4}{19683}-\frac {455 x^3}{2187}+\frac {35 x^2}{81}-\frac {7 x}{9}+1\right )+\frac {c_2 \left (\frac {3 x^5}{22}-\frac {3 x^4}{10}+\frac {3 x^3}{5}-x^2+x+1\right )}{x}
\]