4.61 problem 58
Internal
problem
ID
[7930]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
4.0
Problem
number
:
58
Date
solved
:
Monday, October 21, 2024 at 04:35:10 PM
CAS
classification
:
[[_Emden, _Fowler]]
Solve
\begin{align*} x^{2} y^{\prime \prime }-x 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.
\[ x^{2} y^{\prime \prime }-x y = 0 \]
The following is summary of singularities for the above ode. Writing the ode as
\begin{align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end{align*}
Where
\begin{align*} p(x) &= 0\\ q(x) &= -\frac {1}{x}\\ \end{align*}
Table 107: Table \(p(x),q(x)\) singularites.
| |
| \(p(x)=0\) |
| |
|
singularity | type |
| |
| |
| \(q(x)=-\frac {1}{x}\) |
| |
|
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
\[ x^{2} y^{\prime \prime }-x y = 0 \]
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 }}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 )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-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 }}\left (-x^{1+n +r} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n +r}\right ) \\
\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 )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n +r}\right ) = 0
\end{equation}
The indicial
equation is obtained from \(n = 0\). From Eq (2B) this gives
\[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) = 0 \]
When \(n = 0\) the above becomes
\[ x^{r} a_{0} r \left (-1+r \right ) = 0 \]
Or
\[ x^{r} a_{0} r \left (-1+r \right ) = 0 \]
Since \(a_{0}\neq 0\) then the above simplifies to
\[ x^{r} r \left (-1+r \right ) = 0 \]
Since the above is true for all \(x\) then the
indicial equation becomes
\[ r \left (-1+r \right ) = 0 \]
Solving for \(r\) gives the roots of the indicial equation as
\begin{align*} r_1 &= 1\\ r_2 &= 0 \end{align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes
\[ x^{r} r \left (-1+r \right ) = 0 \]
Solving for \(r\) gives the roots of the indicial equation
as \([1, 0]\).
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 ) &= 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 )+\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^{1+n}\\ 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}\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 \(a_{0} = 1\). For \(1\le n\) the recursive equation is
\begin{equation}
\tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )-a_{n -1} = 0
\end{equation}
Solving for \(a_{n}\) from
recursive equation (4) gives
\[ a_{n} = \frac {a_{n -1}}{\left (n +r \right ) \left (n +r -1\right )}\tag {4} \]
Which for the root \(r = 1\) becomes
\[ a_{n} = \frac {a_{n -1}}{\left (1+n \right ) n}\tag {5} \]
At this point, it is a good idea to
keep track of \(a_{n}\) in a table both before substituting \(r = 1\) 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
\[ a_{1}=\frac {1}{\left (1+r \right ) r} \]
Which for the root \(r = 1\) becomes
\[ a_{1}={\frac {1}{2}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(\frac {1}{\left (1+r \right ) r}\) | \(\frac {1}{2}\) |
| | |
For \(n = 2\), using the above recursive equation gives
\[ a_{2}=\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )} \]
Which for the root \(r = 1\) becomes
\[ a_{2}={\frac {1}{12}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(\frac {1}{\left (1+r \right ) r}\) | \(\frac {1}{2}\) |
| | |
| \(a_{2}\) | \(\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )}\) | \(\frac {1}{12}\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ a_{3}=\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )} \]
Which for the root \(r = 1\) becomes
\[ a_{3}={\frac {1}{144}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {1}{\left (1+r \right ) r}\) | \(\frac {1}{2}\) |
| | |
| \(a_{2}\) | \(\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )}\) | \(\frac {1}{12}\) |
| | |
| \(a_{3}\) |
\(\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )}\) |
\(\frac {1}{144}\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ a_{4}=\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )} \]
Which for the root \(r = 1\) becomes
\[ a_{4}={\frac {1}{2880}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {1}{\left (1+r \right ) r}\) | \(\frac {1}{2}\) |
| | |
| \(a_{2}\) | \(\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )}\) | \(\frac {1}{12}\) |
| | |
| \(a_{3}\) | \(\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )}\) | \(\frac {1}{144}\) |
| | |
| \(a_{4}\) |
\(\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )}\) |
\(\frac {1}{2880}\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ a_{5}=\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )} \]
Which for the root \(r = 1\) becomes
\[ a_{5}={\frac {1}{86400}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {1}{\left (1+r \right ) r}\) |
\(\frac {1}{2}\) |
| | |
| \(a_{2}\) |
\(\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )}\) | \(\frac {1}{12}\) |
| | |
| \(a_{3}\) | \(\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )}\) | \(\frac {1}{144}\) |
| | |
| \(a_{4}\) |
\(\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )}\) |
\(\frac {1}{2880}\) |
| | |
| \(a_{5}\) |
\(\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )}\) |
\(\frac {1}{86400}\) |
| | |
Using the above table, then the solution \(y_{1}\left (x \right )\) is
\begin{align*} y_{1}\left (x \right )&= 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 ) \\ &= x \left (1+\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{144}+\frac {x^{4}}{2880}+\frac {x^{5}}{86400}+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} \\ &= \frac {1}{\left (1+r \right ) r} \end{align*}
Therefore
\begin{align*} \lim _{r\rightarrow r_{2}}\frac {1}{\left (1+r \right ) r}&= \lim _{r\rightarrow 0}\frac {1}{\left (1+r \right ) r}\\ &= \textit {undefined} \end{align*}
Since the limit does not exist then the log term is needed. Therefore the second solution has
the form
\[ y_{2}\left (x \right ) = C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) \]
Therefore
\begin{align*}
\frac {d}{d x}y_{2}\left (x \right ) &= C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) \\
&= C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) \\
\frac {d^{2}}{d x^{2}}y_{2}\left (x \right ) &= C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right ) \\
&= C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n +r_{2}} b_{n} \left (n +r_{2}\right ) \left (-1+n +r_{2}\right )\right ) \\
\end{align*}
Substituting these back into the given ode \(x^{2} y^{\prime \prime }-x y = 0\) gives
\[
x^{2} \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )-x \left (C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )\right ) = 0
\]
Which can be written
as
\begin{equation}
\tag{7} \left (\left (y_{1}^{\prime \prime }\left (x \right ) x^{2}-y_{1}\left (x \right ) x \right ) \ln \left (x \right )+x^{2} \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )\right ) C +x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )-x \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0
\end{equation}
But since \(y_{1}\left (x \right )\) is a solution to the ode, then
\[ y_{1}^{\prime \prime }\left (x \right ) x^{2}-y_{1}\left (x \right ) x = 0 \]
Eq (7) simplifes to
\begin{equation}
\tag{8} x^{2} \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right ) C +x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )-x \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0
\end{equation}
Substituting \(y_{1} = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\) into
the above gives
\begin{equation}
\tag{9} \left (2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right ) x -\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C +\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n +r_{2}} b_{n} \left (n +r_{2}\right ) \left (-1+n +r_{2}\right )\right ) x^{2}-x \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0
\end{equation}
Since \(r_{1} = 1\) and \(r_{2} = 0\) then the above becomes
\begin{equation}
\tag{10} \left (2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n} a_{n} \left (1+n \right )\right ) x -\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{1+n}\right )\right ) C +\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n} b_{n} n \left (n -1\right )\right ) x^{2}-x \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) = 0
\end{equation}
Which simplifies to
\begin{equation}
\tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{1+n} a_{n} \left (1+n \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{1+n} C \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n} b_{n}\right ) = 0
\end{equation}
The
next step is to make all powers of \(x\) be \(n\) in each summation term. Going over each
summation term above with power of \(x\) in it which is not already \(x^{n}\) and adjusting the
power and the corresponding index gives
\begin{align*}
\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{1+n} a_{n} \left (1+n \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 C a_{n -1} n \,x^{n} \\
\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{1+n} C \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-C a_{n -1} x^{n}\right ) \\
\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n} b_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-b_{n -1} x^{n}\right ) \\
\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
\(n\).
\begin{equation}
\tag{2B} \left (\moverset {\infty }{\munderset {n =1}{\sum }}2 C a_{n -1} n \,x^{n}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-C a_{n -1} x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-b_{n -1} x^{n}\right ) = 0
\end{equation}
For \(n=0\) in Eq. (2B), we choose arbitray value for \(b_{0}\) as \(b_{0} = 1\). For \(n=N\), where \(N=1\) which is the
difference between the two roots, we are free to choose \(b_{1} = 0\). Hence for \(n=1\), Eq (2B) gives
\[ C -1 = 0 \]
Which is solved for \(C\). Solving for \(C\) gives
\[ C=1 \]
For \(n=2\), Eq (2B) gives
\[ 3 C a_{1}-b_{1}+2 b_{2} = 0 \]
Which when replacing
the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\),
gives
\[ 2 b_{2}+\frac {3}{2} = 0 \]
Solving the above for \(b_{2}\) gives
\[ b_{2}=-{\frac {3}{4}} \]
For \(n=3\), Eq (2B) gives
\[ 5 C a_{2}-b_{2}+6 b_{3} = 0 \]
Which when replacing
the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\),
gives
\[ 6 b_{3}+\frac {7}{6} = 0 \]
Solving the above for \(b_{3}\) gives
\[ b_{3}=-{\frac {7}{36}} \]
For \(n=4\), Eq (2B) gives
\[ 7 C a_{3}-b_{3}+12 b_{4} = 0 \]
Which when replacing
the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\),
gives
\[ 12 b_{4}+\frac {35}{144} = 0 \]
Solving the above for \(b_{4}\) gives
\[ b_{4}=-{\frac {35}{1728}} \]
For \(n=5\), Eq (2B) gives
\[ 9 C a_{4}-b_{4}+20 b_{5} = 0 \]
Which when replacing the
above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives
\[ 20 b_{5}+\frac {101}{4320} = 0 \]
Solving the above for \(b_{5}\) gives
\[ b_{5}=-{\frac {101}{86400}} \]
Now that we found all \(b_{n}\) and \(C\), we can calculate the
second solution from
\[ y_{2}\left (x \right ) = C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) \]
Using the above value found for \(C=1\) and all \(b_{n}\), then the second
solution becomes
\[
y_{2}\left (x \right )= 1\eslowast \left (x \left (1+\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{144}+\frac {x^{4}}{2880}+\frac {x^{5}}{86400}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}-\frac {7 x^{3}}{36}-\frac {35 x^{4}}{1728}-\frac {101 x^{5}}{86400}+O\left (x^{6}\right )
\]
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 \left (1+\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{144}+\frac {x^{4}}{2880}+\frac {x^{5}}{86400}+O\left (x^{6}\right )\right ) + c_2 \left (1\eslowast \left (x \left (1+\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{144}+\frac {x^{4}}{2880}+\frac {x^{5}}{86400}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}-\frac {7 x^{3}}{36}-\frac {35 x^{4}}{1728}-\frac {101 x^{5}}{86400}+O\left (x^{6}\right )\right ) \\
\end{align*}
Hence the final solution is
\begin{align*}
y &= y_h \\
&= c_1 x \left (1+\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{144}+\frac {x^{4}}{2880}+\frac {x^{5}}{86400}+O\left (x^{6}\right )\right )+c_2 \left (x \left (1+\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{144}+\frac {x^{4}}{2880}+\frac {x^{5}}{86400}+O\left (x^{6}\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}-\frac {7 x^{3}}{36}-\frac {35 x^{4}}{1728}-\frac {101 x^{5}}{86400}+O\left (x^{6}\right )\right ) \\
\end{align*}
4.61.1 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=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 {y}{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}{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 )=0, P_{3}\left (x \right )=-\frac {1}{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}}}=0 \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=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} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) x -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 \cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \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 -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime }\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} \\ {} & {} & a_{0} r \left (-1+r \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +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} \\ {} & {} & r \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1+r \right ) \left (k +r \right )-a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k}}{\left (k +1+r \right ) \left (k +r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k}}{\left (k +1\right ) k} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +1}=\frac {a_{k}}{\left (k +1\right ) k}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +1}=\frac {a_{k}}{\left (k +2\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}, a_{k +1}=\frac {a_{k}}{\left (k +2\right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +1}\right ), a_{k +1}=\frac {a_{k}}{\left (k +1\right ) k}, b_{k +1}=\frac {b_{k}}{\left (k +2\right ) \left (k +1\right )}\right ] \end {array} \]
4.61.2 Maple trace
Methods for second order ODEs:
4.61.3 Maple dsolve solution
Solving time : 0.016 (sec)
Leaf size : 58
dsolve(x^2*diff(diff(y(x),x),x)-x*y(x) = 0,y(x),
series,x=0)
\[
y = c_1 x \left (1+\frac {1}{2} x +\frac {1}{12} x^{2}+\frac {1}{144} x^{3}+\frac {1}{2880} x^{4}+\frac {1}{86400} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_2 \left (\ln \left (x \right ) \left (x +\frac {1}{2} x^{2}+\frac {1}{12} x^{3}+\frac {1}{144} x^{4}+\frac {1}{2880} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (1-\frac {3}{4} x^{2}-\frac {7}{36} x^{3}-\frac {35}{1728} x^{4}-\frac {101}{86400} x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right )
\]
4.61.4 Mathematica DSolve solution
Solving time : 0.018 (sec)
Leaf size : 85
AsymptoticDSolveValue[{x^2*D[y[x],{x,2}]-x*y[x]==0,{}},
y[x],{x,0,5}]
\[
y(x)\to c_1 \left (\frac {1}{144} x \left (x^3+12 x^2+72 x+144\right ) \log (x)+\frac {-47 x^4-480 x^3-2160 x^2-1728 x+1728}{1728}\right )+c_2 \left (\frac {x^5}{2880}+\frac {x^4}{144}+\frac {x^3}{12}+\frac {x^2}{2}+x\right )
\]