2.4.20 Problem 20
Internal
problem
ID
[10183]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
4.0
Problem
number
:
20
Date
solved
:
Thursday, November 27, 2025 at 10:24:31 AM
CAS
classification
:
[[_2nd_order, _linear, _nonhomogeneous]]
\begin{align*}
2 x^{2} y^{\prime \prime }+2 y^{\prime } x -x y&=1 \\
\end{align*}
Series expansion around
\(x=0\).
\begin{align*}
2 x^{2} y^{\prime \prime }+2 y^{\prime } x -x y&=1 \\
\end{align*}
Series expansion around
\(x=0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the
ODE.
\[ 2 x^{2} y^{\prime \prime }+2 y^{\prime } x -x y = 0 \]
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) &= \frac {1}{x}\\ q(x) &= -\frac {1}{2 x}\\ \end{align*}
Table 2.77: Table \(p(x),q(x)\) singularites.
| |
| \(p(x)=\frac {1}{x}\) |
| |
|
singularity | type |
| |
| \(x = 0\) | \(\text {``regular''}\) |
| |
| |
| \(q(x)=-\frac {1}{2 x}\) |
| |
|
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
\[ 2 x^{2} y^{\prime \prime }+2 y^{\prime } x -x y = 1 \]
Since this is an inhomogeneous, then let the solution be
\[ y = y_h + y_p \]
Where
\(y_h\) is the solution to the
homogeneous ode
\(2 x^{2} y^{\prime \prime }+2 y^{\prime } x -x y = 0\), and
\(y_p\) is a particular solution to the inhomogeneous ode.which is found using the
balance equation generated from indicial equation
First, we solve for \(y_h\) 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 (\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 ) x -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 }}2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \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 }}2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \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
\[ 2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+2 x^{n +r} a_{n} \left (n +r \right ) = 0 \]
When
\(n = 0\) the above
becomes
\[ 2 x^{r} a_{0} r \left (-1+r \right )+2 x^{r} a_{0} r = 0 \]
Or
\[ \left (2 x^{r} r \left (-1+r \right )+2 x^{r} r \right ) a_{0} = 0 \]
Since
\(a_{0}\neq 0\) then the above simplifies to
\[ 2 x^{r} r^{2} = 0 \]
Since the above is true for all
\(x\) then
the indicial equation becomes
\[ 2 r^{2} = 0 \]
Solving for
\(r\) gives the roots of the indicial equation as
\begin{align*} r_1 &= 0\\ r_2 &= 0 \end{align*}
The corresponding balance equation is found by replacing \(r\) by \(m\) and \(a\) by \(c\) to avoid confusing terms
between particular solution and the homogeneous solution. Hence the balance equation is
\begin{align*}\left (2 x^{m} m \left (-1+m \right )+2 x^{m} m \right ) c_{0} = 1 \end{align*}
This equation will used later to find the particular solution.
Since \(a_{0}\neq 0\) then the indicial equation becomes
\[ 2 x^{r} r^{2} = 0 \]
Solving for
\(r\) gives the roots of the indicial equation as
\([0, 0]\).
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. 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} \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
\[ a_{n} = \frac {a_{n -1}}{2 n^{2}+4 n r +2 r^{2}}\tag {4} \]
Which for the
root
\(r = 0\) becomes
\[ a_{n} = \frac {a_{n -1}}{2 n^{2}}\tag {5} \]
At this point, it is a good idea to keep track of
\(a_{n}\) in a table both before
substituting
\(r = 0\) 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
\[ a_{1}=\frac {1}{2 \left (r +1\right )^{2}} \]
Which for the root
\(r = 0\) becomes
\[ a_{1}={\frac {1}{2}} \]
And the table now
becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(\frac {1}{2 \left (r +1\right )^{2}}\) | \(\frac {1}{2}\) |
| | |
For \(n = 2\), using the above recursive equation gives
\[ a_{2}=\frac {1}{4 \left (r +1\right )^{2} \left (r +2\right )^{2}} \]
Which for the root
\(r = 0\) becomes
\[ a_{2}={\frac {1}{16}} \]
And the table now
becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(\frac {1}{2 \left (r +1\right )^{2}}\) | \(\frac {1}{2}\) |
| | |
| \(a_{2}\) | \(\frac {1}{4 \left (r +1\right )^{2} \left (r +2\right )^{2}}\) | \(\frac {1}{16}\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ a_{3}=\frac {1}{8 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}} \]
Which for the root
\(r = 0\) becomes
\[ a_{3}={\frac {1}{288}} \]
And the table now
becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {1}{2 \left (r +1\right )^{2}}\) | \(\frac {1}{2}\) |
| | |
| \(a_{2}\) | \(\frac {1}{4 \left (r +1\right )^{2} \left (r +2\right )^{2}}\) | \(\frac {1}{16}\) |
| | |
| \(a_{3}\) |
\(\frac {1}{8 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) |
\(\frac {1}{288}\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ a_{4}=\frac {1}{16 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}} \]
Which for the root
\(r = 0\) becomes
\[ a_{4}={\frac {1}{9216}} \]
And the table now
becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {1}{2 \left (r +1\right )^{2}}\) | \(\frac {1}{2}\) |
| | |
| \(a_{2}\) | \(\frac {1}{4 \left (r +1\right )^{2} \left (r +2\right )^{2}}\) | \(\frac {1}{16}\) |
| | |
| \(a_{3}\) | \(\frac {1}{8 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(\frac {1}{288}\) |
| | |
| \(a_{4}\) |
\(\frac {1}{16 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) |
\(\frac {1}{9216}\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ a_{5}=\frac {1}{32 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}} \]
Which for the root
\(r = 0\) becomes
\[ a_{5}={\frac {1}{460800}} \]
And the table now
becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {1}{2 \left (r +1\right )^{2}}\) |
\(\frac {1}{2}\) |
| | |
| \(a_{2}\) |
\(\frac {1}{4 \left (r +1\right )^{2} \left (r +2\right )^{2}}\) | \(\frac {1}{16}\) |
| | |
| \(a_{3}\) | \(\frac {1}{8 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(\frac {1}{288}\) |
| | |
| \(a_{4}\) |
\(\frac {1}{16 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) |
\(\frac {1}{9216}\) |
| | |
| \(a_{5}\) |
\(\frac {1}{32 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) |
\(\frac {1}{460800}\) |
| | |
Using the above table, then the first solution \(y_{1}\left (x \right )\) becomes
\begin{align*}
y_{1}\left (x \right )&= 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 \\
&= \frac {x}{2}+1+\frac {x^{2}}{16}+\frac {x^{3}}{288}+\frac {x^{4}}{9216}+\frac {x^{5}}{460800}+O\left (x^{6}\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
\(r = 0\). The above
table for
\(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| | | | |
| \(n\) |
\(b_{n ,r}\) |
\(a_{n}\) |
\(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) |
\(b_{n}\left (r =0\right )\) |
| | | | |
| \(b_{0}\) |
\(1\) |
\(1\) |
N/A since \(b_{n}\) starts from 1 |
N/A |
| | | | |
| \(b_{1}\) | \(\frac {1}{2 \left (r +1\right )^{2}}\) | \(\frac {1}{2}\) | \(-\frac {1}{\left (r +1\right )^{3}}\) | \(-1\) |
| | | | |
| \(b_{2}\) | \(\frac {1}{4 \left (r +1\right )^{2} \left (r +2\right )^{2}}\) | \(\frac {1}{16}\) | \(\frac {-2 r -3}{2 \left (r +1\right )^{3} \left (r +2\right )^{3}}\) | \(-{\frac {3}{16}}\) |
| | | | |
| \(b_{3}\) | \(\frac {1}{8 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(\frac {1}{288}\) | \(\frac {-3 r^{2}-12 r -11}{4 \left (r +1\right )^{3} \left (r +2\right )^{3} \left (r +3\right )^{3}}\) | \(-{\frac {11}{864}}\) |
| | | | |
| \(b_{4}\) |
\(\frac {1}{16 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) |
\(\frac {1}{9216}\) |
\(\frac {-2 r^{3}-15 r^{2}-35 r -25}{4 \left (r +1\right )^{3} \left (r +2\right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3}}\) |
\(-{\frac {25}{55296}}\) |
| | | | |
| \(b_{5}\) |
\(\frac {1}{32 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) |
\(\frac {1}{460800}\) |
\(\frac {-5 r^{4}-60 r^{3}-255 r^{2}-450 r -274}{16 \left (r +1\right )^{3} \left (r +2\right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3} \left (r +5\right )^{3}}\) |
\(-{\frac {137}{13824000}}\) |
| | | | |
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 \\
&= \left (\frac {x}{2}+1+\frac {x^{2}}{16}+\frac {x^{3}}{288}+\frac {x^{4}}{9216}+\frac {x^{5}}{460800}+O\left (x^{6}\right )\right ) \ln \left (x \right )-x -\frac {3 x^{2}}{16}-\frac {11 x^{3}}{864}-\frac {25 x^{4}}{55296}-\frac {137 x^{5}}{13824000}+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 \left (\frac {x}{2}+1+\frac {x^{2}}{16}+\frac {x^{3}}{288}+\frac {x^{4}}{9216}+\frac {x^{5}}{460800}+O\left (x^{6}\right )\right ) + c_2 \left (\left (\frac {x}{2}+1+\frac {x^{2}}{16}+\frac {x^{3}}{288}+\frac {x^{4}}{9216}+\frac {x^{5}}{460800}+O\left (x^{6}\right )\right ) \ln \left (x \right )-x -\frac {3 x^{2}}{16}-\frac {11 x^{3}}{864}-\frac {25 x^{4}}{55296}-\frac {137 x^{5}}{13824000}+O\left (x^{6}\right )\right ) \\
\end{align*}
The particular solution is found by solving for
\(c,m\) the balance equation
\begin{align*} \left (2 x^{m} m \left (-1+m \right )+2 x^{m} m \right ) c_{0}&=F \end{align*}
Where \(F(x)\) is the RHS of the ode. If \(F(x)\) has more than one term, then this is done for each term one at a
time and then all the particular solutions are added. The function \(F(x)\) will be converted to series if
needed. in order to solve for \(c_n,m\) for each term, the same recursive relation used to find \(y_h(x)\) is
used to find \(c_n,m\) which is used to find the particular solution \(\sum _{n=0} c_n x^{n+m}\) by replacing \(a_n\) by \(c_n\) and \(r\) by
\(m\).
The following are the values of \(a_n\) found in terms of the indicial root \(r\).
|
| \(a_{1} = \frac {a_{0}}{2 \left (r +1\right )^{2}}\) |
| \(a_{2} = \frac {a_{0}}{4 \left (r +1\right )^{2} \left (r +2\right )^{2}}\) |
| \(a_{3} = \frac {a_{0}}{8 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) |
| \(a_{4} = \frac {a_{0}}{16 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) |
| \(a_{5} = \frac {a_{0}}{32 \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) |
|
Unable to solve the balance equation \(\left (2 x^{m} m \left (-1+m \right )+2 x^{m} m \right ) c_{0}\) for \(c_{0}\) and \(x\). No particular solution exists.
Failed to convert RHS \(1\) to series in order to find particular solution. Unable to solve. Terminating
Unable to find the particular solution or no solution exists.
✗ Maple
Order:=6;
ode:=2*x^2*diff(diff(y(x),x),x)+2*diff(y(x),x)*x-y(x)*x = 1;
dsolve(ode,y(x),type='series',x=0);
\[ \text {No solution found} \]
Maple trace
Methods for second order ODEs:
--- Trying classification methods ---
trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
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
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful
<- solving first the homogeneous part of the ODE successful
✓ Mathematica. Time used: 0.094 (sec). Leaf size: 360
ode=2*x^2*D[y[x],{x,2}]+2*x*D[y[x],x]-x*y[x]==1;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
\[
y(x)\to c_2 \left (\frac {x^5}{460800}+\frac {x^4}{9216}+\frac {x^3}{288}+\frac {x^2}{16}+\frac {x}{2}+1\right )+c_1 \left (x^5 \left (\frac {\log (x)}{460800}-\frac {107}{13824000}\right )+x^4 \left (\frac {\log (x)}{9216}-\frac {19}{55296}\right )+x^3 \left (\frac {\log (x)}{288}-\frac {1}{108}\right )+x^2 \left (\frac {\log (x)}{16}-\frac {1}{8}\right )+x \left (\frac {\log (x)}{2}-\frac {1}{2}\right )+\log (x)+1\right )+\left (-\frac {137 x^6}{1990656000}+\frac {x^5}{4608000}+\frac {x^4}{73728}+\frac {x^3}{1728}+\frac {x^2}{64}+\frac {x}{4}+\frac {\log (x)}{2}\right ) \left (x^5 \left (\frac {\log (x)}{460800}-\frac {107}{13824000}\right )+x^4 \left (\frac {\log (x)}{9216}-\frac {19}{55296}\right )+x^3 \left (\frac {\log (x)}{288}-\frac {1}{108}\right )+x^2 \left (\frac {\log (x)}{16}-\frac {1}{8}\right )+x \left (\frac {\log (x)}{2}-\frac {1}{2}\right )+\log (x)+1\right )+\left (\frac {x^5}{460800}+\frac {x^4}{9216}+\frac {x^3}{288}+\frac {x^2}{16}+\frac {x}{2}+1\right ) \left (\frac {137 x^6 (6 \log (x)+5)}{11943936000}+\frac {x^5 (113-30 \log (x))}{138240000}+\frac {x^4 (41-12 \log (x))}{884736}+\frac {x^3 (3-\log (x))}{1728}+\frac {1}{128} x^2 (5-2 \log (x))+\frac {1}{4} x (2-\log (x))-\frac {1}{4} \log (x) (\log (x)+2)\right )
\]
✗ Sympy
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(2*x**2*Derivative(y(x), (x, 2)) - x*y(x) + 2*x*Derivative(y(x), x) - 1,0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
ValueError : ODE 2*x**2*Derivative(y(x), (x, 2)) - x*y(x) + 2*x*Derivative(y(x), x) - 1 does not match hint 2nd_power_series_regular