4.54 problem 51

Internal problem ID [7923]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 51
Date solved : Monday, October 21, 2024 at 04:35:00 PM
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 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*}

4.54.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )-9 x y^{\prime }+25 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 {25 y}{x^{2}}+\frac {9 y^{\prime }}{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 {9 y^{\prime }}{x}+\frac {25 y}{x^{2}}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )-9 x y^{\prime }+25 y=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} \\ {} & {} & y^{\prime }=\left (\frac {d}{d t}y \left (t \right )\right ) t^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\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}{d x}y^{\prime }=\left (\frac {d}{d t}\frac {d}{d t}y \left (t \right )\right ) {t^{\prime }\left (x \right )}^{2}+\left (\frac {d}{d x}t^{\prime }\left (x \right )\right ) \left (\frac {d}{d t}y \left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\frac {\frac {d}{d t}\frac {d}{d t}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}{d t}\frac {d}{d t}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}{d t}\frac {d}{d t}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=\mathit {C1} \,x^{5}+\mathit {C2} \,x^{5} \ln \left (x \right ) \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=x^{5} \left (\mathit {C1} +\mathit {C2} \ln \left (x \right )\right ) \end {array} \]

4.54.2 Maple trace

Methods for second order ODEs:
 

4.54.3 Maple dsolve solution

Solving time : 0.020 (sec)
Leaf size : 19

dsolve(x^2*diff(diff(y(x),x),x)-9*x*diff(y(x),x)+25*y(x) = 0,y(x), 
       series,x=0)
 

\[ y = x^{5} \left (c_1 +c_2 \ln \left (x \right )\right )+O\left (x^{6}\right ) \]

4.54.4 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}]
 

\[ y(x)\to c_1 x^5+c_2 x^5 \log (x) \]