2.4.54 problem 51
Internal
problem
ID
[8369]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
4.0
Problem
number
:
51
Date
solved
:
Sunday, November 10, 2024 at 03:39:34 AM
CAS
classification
:
[[_Emden, _Fowler]]
Solve
\begin{align*} x^{2} y^{\prime \prime }-9 x y^{\prime }+25 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.
\[ x^{2} y^{\prime \prime }-9 x y^{\prime }+25 y = 0 \]
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 {9}{x}\\ q(x) &= \frac {25}{x^{2}}\\ \end{align*}
Table 2.100: Table \(p(x),q(x)\) singularites.
| |
\(p(x)=-\frac {9}{x}\) |
| |
singularity | type |
| |
\(x = 0\) | \(\text {``regular''}\) |
| |
| |
\(q(x)=\frac {25}{x^{2}}\) |
| |
singularity | type |
| |
\(x = 0\) | \(\text {``regular''}\) |
| |
Combining everything together gives the following summary of singularities for the ode
as
Regular singular points : \([0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to
be
\[ x^{2} y^{\prime \prime }-9 x y^{\prime }+25 y = 0 \]
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 )-9 x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+25 \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 (-9 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}25 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 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 =0}{\sum }}\left (-9 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}25 a_{n} x^{n +r}\right ) = 0
\end{equation}
The indicial
equation is obtained from \(n = 0\). From Eq (2B) this gives
\[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-9 x^{n +r} a_{n} \left (n +r \right )+25 a_{n} x^{n +r} = 0 \]
When \(n = 0\) the above becomes
\[ x^{r} a_{0} r \left (-1+r \right )-9 x^{r} a_{0} r +25 a_{0} x^{r} = 0 \]
Or
\[ \left (x^{r} r \left (-1+r \right )-9 x^{r} r +25 x^{r}\right ) a_{0} = 0 \]
Since \(a_{0}\neq 0\) then the above simplifies to
\[ \left (r -5\right )^{2} x^{r} = 0 \]
Since the above is true for all \(x\) then the
indicial equation becomes
\[ \left (r -5\right )^{2} = 0 \]
Solving for \(r\) gives the roots of the indicial equation as
\begin{align*} r_1 &= 5\\ r_2 &= 5 \end{align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes
\[ \left (r -5\right )^{2} x^{r} = 0 \]
Solving for \(r\) gives the roots of the indicial equation
as \([5, 5]\).
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, \(r = 5\), Eqs (1A,1B) become
\begin{align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +5}\\ 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 +5}\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 \(0\le n\) the recursive equation is
\begin{equation}
\tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )-9 a_{n} \left (n +r \right )+25 a_{n} = 0
\end{equation}
Solving for \(a_{n}\) from recursive equation
(4) gives
\[ a_{n} = 0\tag {4} \]
Which for the root \(r = 5\) becomes
\[ a_{n} = 0\tag {5} \]
At this point, it is a good idea to keep track of \(a_{n}\) in a
table both before substituting \(r = 5\) 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
\[ a_{1}=0 \]
And the table now becomes
| | |
\(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
\(a_{0}\) |
\(1\) | \(1\) |
| | |
\(a_{1}\) | \(0\) | \(0\) |
| | |
For \(n = 2\), using the above recursive equation gives
\[ a_{2}=0 \]
And the table now becomes
| | |
\(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
\(a_{0}\) |
\(1\) | \(1\) |
| | |
\(a_{1}\) | \(0\) | \(0\) |
| | |
\(a_{2}\) | \(0\) | \(0\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ a_{3}=0 \]
And the table now becomes
| | |
\(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
\(a_{0}\) |
\(1\) |
\(1\) |
| | |
\(a_{1}\) |
\(0\) | \(0\) |
| | |
\(a_{2}\) | \(0\) | \(0\) |
| | |
\(a_{3}\) |
\(0\) |
\(0\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ a_{4}=0 \]
And the table now becomes
| | |
\(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
\(a_{0}\) |
\(1\) |
\(1\) |
| | |
\(a_{1}\) |
\(0\) | \(0\) |
| | |
\(a_{2}\) | \(0\) | \(0\) |
| | |
\(a_{3}\) | \(0\) | \(0\) |
| | |
\(a_{4}\) |
\(0\) |
\(0\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ a_{5}=0 \]
And the table now becomes
| | |
\(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
\(a_{0}\) |
\(1\) |
\(1\) |
| | |
\(a_{1}\) |
\(0\) |
\(0\) |
| | |
\(a_{2}\) |
\(0\) | \(0\) |
| | |
\(a_{3}\) | \(0\) | \(0\) |
| | |
\(a_{4}\) |
\(0\) |
\(0\) |
| | |
\(a_{5}\) |
\(0\) |
\(0\) |
| | |
Using the above table, then the first solution \(y_{1}\left (x \right )\) is
\begin{align*}
y_{1}\left (x \right )&= x^{5} \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^{5} \left (1+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 \(r = 5\). The
above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| | | | |
\(n\) |
\(b_{n ,r}\) |
\(a_{n}\) |
\(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) |
\(b_{n}\left (r =5\right )\) |
| | | | |
\(b_{0}\) |
\(1\) |
\(1\) |
N/A since \(b_{n}\) starts from 1 |
N/A |
| | | | |
\(b_{1}\) | \(0\) | \(0\) | \(0\) | \(0\) |
| | | | |
\(b_{2}\) | \(0\) | \(0\) | \(0\) | \(0\) |
| | | | |
\(b_{3}\) | \(0\) | \(0\) | \(0\) | \(0\) |
| | | | |
\(b_{4}\) |
\(0\) |
\(0\) |
\(0\) |
\(0\) |
| | | | |
\(b_{5}\) |
\(0\) |
\(0\) |
\(0\) |
\(0\) |
| | | | |
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 \\
&= x^{5} \left (1+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{5} 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 \,x^{5} \left (1+O\left (x^{6}\right )\right ) + c_2 \left (x^{5} \left (1+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{5} O\left (x^{6}\right )\right ) \\
\end{align*}
Hence the final solution is
\begin{align*}
y &= y_h \\
&= c_1 \,x^{5} \left (1+O\left (x^{6}\right )\right )+c_2 \left (x^{5} \left (1+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{5} O\left (x^{6}\right )\right ) \\
\end{align*}
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )-9 x \left (\frac {d}{d x}y \left (x \right )\right )+25 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 {25 y \left (x \right )}{x^{2}}+\frac {9 \left (\frac {d}{d x}y \left (x \right )\right )}{x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (x \right )\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )-\frac {9 \left (\frac {d}{d x}y \left (x \right )\right )}{x}+\frac {25 y \left (x \right )}{x^{2}}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )-9 x \left (\frac {d}{d x}y \left (x \right )\right )+25 y \left (x \right )=0 \\ \bullet & {} & \textrm {Make a change of variables}\hspace {3pt} \\ {} & {} & t =\ln \left (x \right ) \\ \square & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {1st}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {x}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\left (\frac {d}{d t}y \left (t \right )\right ) \left (\frac {d}{d x}t \left (x \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {\frac {d}{d t}y \left (t \right )}{x} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {2nd}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {x}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=\left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) \left (\frac {d}{d x}t \left (x \right )\right )^{2}+\left (\frac {d^{2}}{d x^{2}}t \left (x \right )\right ) \left (\frac {d}{d t}y \left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=\frac {\frac {d^{2}}{d t^{2}}y \left (t \right )}{x^{2}}-\frac {\frac {d}{d t}y \left (t \right )}{x^{2}} \\ & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {\frac {d^{2}}{d t^{2}}y \left (t \right )}{x^{2}}-\frac {\frac {d}{d t}y \left (t \right )}{x^{2}}\right )-9 \frac {d}{d t}y \left (t \right )+25 y \left (t \right )=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right )-10 \frac {d}{d t}y \left (t \right )+25 y \left (t \right )=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}-10 r +25=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -5\right )^{2}=0 \\ \bullet & {} & \textrm {Root of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =5 \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{5 t} \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (t \right )\hspace {3pt}\textrm {by}\hspace {3pt} t \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=t \,{\mathrm e}^{5 t} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=\mathit {C1} y_{1}\left (t \right )+\mathit {C2} y_{2}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y \left (t \right )=\mathit {C1} \,{\mathrm e}^{5 t}+\mathit {C2} t \,{\mathrm e}^{5 t} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} t =\ln \left (x \right ) \\ {} & {} & y \left (x \right )=\mathit {C2} \ln \left (x \right ) x^{5}+\mathit {C1} \,x^{5} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y \left (x \right )=x^{5} \left (\mathit {C2} \ln \left (x \right )+\mathit {C1} \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
<- LODE of Euler type successful`
Maple dsolve solution
Solving time : 0.018
(sec)
Leaf size : 19
dsolve(x^2*diff(diff(y(x),x),x)-9*diff(y(x),x)*x+25*y(x) = 0,y(x),
series,x=0)
\[
y = x^{5} \left (c_{2} \ln \left (x \right )+c_{1} \right )+O\left (x^{6}\right )
\]
Mathematica DSolve solution
Solving time : 0.004
(sec)
Leaf size : 18
AsymptoticDSolveValue[{x^2*D[y[x],{x,2}]-9*x*D[y[x],x]+25*y[x]==0,{}},
y[x],{x,0,5}]
\[
y(x)\to c_1 x^5+c_2 x^5 \log (x)
\]