15.30 problem 26

Internal problem ID [2028]
Book : Elementary differential equations with boundary value problems. William F. Trench. Brooks/Cole 2001
Section : Chapter 7 Series Solutions of Linear Second Equations. 7.6 THE METHOD OF FROBENIUS II. Exercises 7.6. Page 374
Problem number : 26
Date solved : Thursday, October 17, 2024 at 02:18:56 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

\begin{align*} x^{2} \left (1+x \right ) y^{\prime \prime }+x \left (1+2 x \right ) y^{\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.

\[ \left (x^{3}+x^{2}\right ) y^{\prime \prime }+\left (2 x^{2}+x \right ) y^{\prime }+x y = 0 \]

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) &= \frac {1+2 x}{x \left (1+x \right )}\\ q(x) &= \frac {1}{x \left (1+x \right )}\\ \end{align*}

Table 152: Table \(p(x),q(x)\) singularites.

\(p(x)=\frac {1+2 x}{x \left (1+x \right )}\)
singularity	type
\(x = -1\)	\(\text {``regular''}\)
\(x = 0\)	\(\text {``regular''}\)

\(q(x)=\frac {1}{x \left (1+x \right )}\)
singularity	type
\(x = -1\)	\(\text {``regular''}\)
\(x = 0\)	\(\text {``regular''}\)

Combining everything together gives the following summary of singularities for the ode as

Regular singular points : \([-1, 0, \infty ]\)

Irregular singular points : \([]\)

Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be

\[ x^{2} \left (1+x \right ) y^{\prime \prime }+\left (2 x^{2}+x \right ) y^{\prime }+x y = 0 \]

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 (1+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 (2 x^{2}+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\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^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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 }}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 }}2 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}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 }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r}\right ) = 0 \end{equation}

The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives

\[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right ) = 0 \]

When \(n = 0\) the above becomes

\[ x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r = 0 \]

Or

\[ \left (x^{r} r \left (-1+r \right )+x^{r} r \right ) a_{0} = 0 \]

Since \(a_{0}\neq 0\) then the above simplifies to

\[ x^{r} r^{2} = 0 \]

Since the above is true for all \(x\) then the indicial equation becomes

\[ r^{2} = 0 \]

Solving for \(r\) gives the roots of the indicial equation as

\begin{align*} r_1 &= 0\\ r_2 &= 0 \end{align*}

Since \(a_{0}\neq 0\) then the indicial equation becomes

\[ x^{r} r^{2} = 0 \]

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} a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+a_{n} \left (n +r \right ) \left (n +r -1\right )+2 a_{n -1} \left (n +r -1\right )+a_{n} \left (n +r \right )+a_{n -1} = 0 \end{equation}

Solving for \(a_{n}\) from recursive equation (4) gives

\[ a_{n} = -\frac {a_{n -1} \left (n^{2}+2 n r +r^{2}-n -r +1\right )}{n^{2}+2 n r +r^{2}}\tag {4} \]

Which for the root \(r = 0\) becomes

\[ a_{n} = -\frac {a_{n -1} \left (n^{2}-n +1\right )}{n^{2}}\tag {5} \]

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

\[ a_{1}=\frac {-r^{2}-r -1}{\left (1+r \right )^{2}} \]

Which for the root \(r = 0\) becomes

\[ a_{1}=-1 \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {-r^{2}-r -1}{\left (1+r \right )^{2}}\)	\(-1\)

For \(n = 2\), using the above recursive equation gives

\[ a_{2}=\frac {r^{4}+4 r^{3}+7 r^{2}+6 r +3}{\left (1+r \right )^{2} \left (r +2\right )^{2}} \]

Which for the root \(r = 0\) becomes

\[ a_{2}={\frac {3}{4}} \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {-r^{2}-r -1}{\left (1+r \right )^{2}}\)	\(-1\)
\(a_{2}\)	\(\frac {r^{4}+4 r^{3}+7 r^{2}+6 r +3}{\left (1+r \right )^{2} \left (r +2\right )^{2}}\)	\(\frac {3}{4}\)

For \(n = 3\), using the above recursive equation gives

\[ a_{3}=\frac {-r^{6}-9 r^{5}-34 r^{4}-69 r^{3}-82 r^{2}-57 r -21}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}} \]

Which for the root \(r = 0\) becomes

\[ a_{3}=-{\frac {7}{12}} \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {-r^{2}-r -1}{\left (1+r \right )^{2}}\)	\(-1\)
\(a_{2}\)	\(\frac {r^{4}+4 r^{3}+7 r^{2}+6 r +3}{\left (1+r \right )^{2} \left (r +2\right )^{2}}\)	\(\frac {3}{4}\)
\(a_{3}\)	\(\frac {-r^{6}-9 r^{5}-34 r^{4}-69 r^{3}-82 r^{2}-57 r -21}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\)	\(-{\frac {7}{12}}\)

For \(n = 4\), using the above recursive equation gives

\[ a_{4}=\frac {\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}} \]

Which for the root \(r = 0\) becomes

\[ a_{4}={\frac {91}{192}} \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {-r^{2}-r -1}{\left (1+r \right )^{2}}\)	\(-1\)
\(a_{2}\)	\(\frac {r^{4}+4 r^{3}+7 r^{2}+6 r +3}{\left (1+r \right )^{2} \left (r +2\right )^{2}}\)	\(\frac {3}{4}\)
\(a_{3}\)	\(\frac {-r^{6}-9 r^{5}-34 r^{4}-69 r^{3}-82 r^{2}-57 r -21}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\)	\(-{\frac {7}{12}}\)
\(a_{4}\)	\(\frac {\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\)	\(\frac {91}{192}\)

For \(n = 5\), using the above recursive equation gives

\[ a_{5}=-\frac {\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right ) \left (r^{2}+9 r +21\right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}} \]

Which for the root \(r = 0\) becomes

\[ a_{5}=-{\frac {637}{1600}} \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {-r^{2}-r -1}{\left (1+r \right )^{2}}\)	\(-1\)
\(a_{2}\)	\(\frac {r^{4}+4 r^{3}+7 r^{2}+6 r +3}{\left (1+r \right )^{2} \left (r +2\right )^{2}}\)	\(\frac {3}{4}\)
\(a_{3}\)	\(\frac {-r^{6}-9 r^{5}-34 r^{4}-69 r^{3}-82 r^{2}-57 r -21}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\)	\(-{\frac {7}{12}}\)
\(a_{4}\)	\(\frac {\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\)	\(\frac {91}{192}\)
\(a_{5}\)	\(-\frac {\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right ) \left (r^{2}+9 r +21\right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\)	\(-{\frac {637}{1600}}\)

For \(n = 6\), using the above recursive equation gives

\[ a_{6}=\frac {\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right ) \left (r^{2}+9 r +21\right ) \left (r^{2}+11 r +31\right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2}} \]

Which for the root \(r = 0\) becomes

\[ a_{6}={\frac {19747}{57600}} \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {-r^{2}-r -1}{\left (1+r \right )^{2}}\)	\(-1\)
\(a_{2}\)	\(\frac {r^{4}+4 r^{3}+7 r^{2}+6 r +3}{\left (1+r \right )^{2} \left (r +2\right )^{2}}\)	\(\frac {3}{4}\)
\(a_{3}\)	\(\frac {-r^{6}-9 r^{5}-34 r^{4}-69 r^{3}-82 r^{2}-57 r -21}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\)	\(-{\frac {7}{12}}\)
\(a_{4}\)	\(\frac {\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\)	\(\frac {91}{192}\)
\(a_{5}\)	\(-\frac {\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right ) \left (r^{2}+9 r +21\right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\)	\(-{\frac {637}{1600}}\)
\(a_{6}\)	\(\frac {\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right ) \left (r^{2}+9 r +21\right ) \left (r^{2}+11 r +31\right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2}}\)	\(\frac {19747}{57600}\)

For \(n = 7\), using the above recursive equation gives

\[ a_{7}=-\frac {\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right ) \left (r^{2}+9 r +21\right ) \left (r^{2}+11 r +31\right ) \left (r^{2}+13 r +43\right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2} \left (r +7\right )^{2}} \]

Which for the root \(r = 0\) becomes

\[ a_{7}=-{\frac {17329}{57600}} \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {-r^{2}-r -1}{\left (1+r \right )^{2}}\)	\(-1\)
\(a_{2}\)	\(\frac {r^{4}+4 r^{3}+7 r^{2}+6 r +3}{\left (1+r \right )^{2} \left (r +2\right )^{2}}\)	\(\frac {3}{4}\)
\(a_{3}\)	\(\frac {-r^{6}-9 r^{5}-34 r^{4}-69 r^{3}-82 r^{2}-57 r -21}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\)	\(-{\frac {7}{12}}\)
\(a_{4}\)	\(\frac {\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\)	\(\frac {91}{192}\)
\(a_{5}\)	\(-\frac {\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right ) \left (r^{2}+9 r +21\right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\)	\(-{\frac {637}{1600}}\)
\(a_{6}\)	\(\frac {\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right ) \left (r^{2}+9 r +21\right ) \left (r^{2}+11 r +31\right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2}}\)	\(\frac {19747}{57600}\)
\(a_{7}\)	\(-\frac {\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right ) \left (r^{2}+9 r +21\right ) \left (r^{2}+11 r +31\right ) \left (r^{2}+13 r +43\right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2} \left (r +7\right )^{2}}\)	\(-{\frac {17329}{57600}}\)

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}+a_{7} x^{7}+a_{8} x^{8}\dots \\ &= 1-x +\frac {3 x^{2}}{4}-\frac {7 x^{3}}{12}+\frac {91 x^{4}}{192}-\frac {637 x^{5}}{1600}+\frac {19747 x^{6}}{57600}-\frac {17329 x^{7}}{57600}+O\left (x^{8}\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
\(b_{1}\)	\(\frac {-r^{2}-r -1}{\left (1+r \right )^{2}}\)	\(-1\)	\(\frac {-r +1}{\left (1+r \right )^{3}}\)	\(1\)
\(b_{2}\)	\(\frac {r^{4}+4 r^{3}+7 r^{2}+6 r +3}{\left (1+r \right )^{2} \left (r +2\right )^{2}}\)	\(\frac {3}{4}\)	\(\frac {2 r^{4}+6 r^{3}+6 r^{2}-2 r -6}{\left (r +2\right )^{3} \left (1+r \right )^{3}}\)	\(-{\frac {3}{4}}\)
\(b_{3}\)	\(\frac {-r^{6}-9 r^{5}-34 r^{4}-69 r^{3}-82 r^{2}-57 r -21}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\)	\(-{\frac {7}{12}}\)	\(\frac {-3 r^{7}-30 r^{6}-126 r^{5}-276 r^{4}-306 r^{3}-90 r^{2}+147 r +120}{\left (r +3\right )^{3} \left (r +2\right )^{3} \left (1+r \right )^{3}}\)	\(\frac {5}{9}\)
\(b_{4}\)	\(\frac {\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\)	\(\frac {91}{192}\)	\(\frac {4 r^{10}+80 r^{9}+708 r^{8}+3624 r^{7}+11748 r^{6}+24684 r^{5}+32444 r^{4}+22468 r^{3}+396 r^{2}-11292 r -5988}{\left (r +4\right )^{3} \left (r +3\right )^{3} \left (r +2\right )^{3} \left (1+r \right )^{3}}\)	\(-{\frac {499}{1152}}\)
\(b_{5}\)	\(-\frac {\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right ) \left (r^{2}+9 r +21\right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\)	\(-{\frac {637}{1600}}\)	\(\frac {-5 r^{13}-165 r^{12}-2475 r^{11}-22285 r^{10}-133935 r^{9}-564765 r^{8}-1706725 r^{7}-3696675 r^{6}-5599065 r^{5}-5513735 r^{4}-2745165 r^{3}+483705 r^{2}+1469430 r +609084}{\left (1+r \right )^{3} \left (r +2\right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3} \left (r +5\right )^{3}}\)	\(\frac {16919}{48000}\)
\(b_{6}\)	\(\frac {\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right ) \left (r^{2}+9 r +21\right ) \left (r^{2}+11 r +31\right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2}}\)	\(\frac {19747}{57600}\)	\(\frac {6 r^{16}+294 r^{15}+6660 r^{14}+92460 r^{13}+879102 r^{12}+6058368 r^{11}[phone] r^{10}[phone] r^{9}[phone] r^{8}[phone] r^{7}[phone] r^{6}[phone] r^{5}[phone] r^{4}[phone] r^{3}-191517966 r^{2}-315293400 r -110537784}{\left (1+r \right )^{3} \left (r +2\right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3} \left (r +5\right )^{3} \left (r +6\right )^{3}}\)	\(-{\frac {56861}{192000}}\)
\(b_{7}\)	\(-\frac {\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right ) \left (r^{2}+9 r +21\right ) \left (r^{2}+11 r +31\right ) \left (r^{2}+13 r +43\right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2} \left (r +7\right )^{2}}\)	\(-{\frac {17329}{57600}}\)	\(-\frac {7 \left (r^{19}+68 r^{18}+2163 r^{17}+42744 r^{16}+587769 r^{15}+5969226 r^{14}[phone] r^{13}[phone] r^{12}[phone] r^{11}[phone] r^{10}[phone] r^{9}[phone] r^{8}[phone] r^{7}[phone] r^{6}[phone] r^{5}[phone] r^{4}[phone] r^{3}-12606541542 r^{2}-15062507748 r -4661724312\right )}{\left (1+r \right )^{3} \left (r +2\right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3} \left (r +5\right )^{3} \left (r +6\right )^{3} \left (r +7\right )^{3}}\)	\(\frac {1027717}{4032000}\)

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}+b_{7} x^{7}+b_{8} x^{8}\dots \\ &= \left (1-x +\frac {3 x^{2}}{4}-\frac {7 x^{3}}{12}+\frac {91 x^{4}}{192}-\frac {637 x^{5}}{1600}+\frac {19747 x^{6}}{57600}-\frac {17329 x^{7}}{57600}+O\left (x^{8}\right )\right ) \ln \left (x \right )+x -\frac {3 x^{2}}{4}+\frac {5 x^{3}}{9}-\frac {499 x^{4}}{1152}+\frac {16919 x^{5}}{48000}-\frac {56861 x^{6}}{192000}+\frac {1027717 x^{7}}{4032000}+O\left (x^{8}\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 (1-x +\frac {3 x^{2}}{4}-\frac {7 x^{3}}{12}+\frac {91 x^{4}}{192}-\frac {637 x^{5}}{1600}+\frac {19747 x^{6}}{57600}-\frac {17329 x^{7}}{57600}+O\left (x^{8}\right )\right ) + c_2 \left (\left (1-x +\frac {3 x^{2}}{4}-\frac {7 x^{3}}{12}+\frac {91 x^{4}}{192}-\frac {637 x^{5}}{1600}+\frac {19747 x^{6}}{57600}-\frac {17329 x^{7}}{57600}+O\left (x^{8}\right )\right ) \ln \left (x \right )+x -\frac {3 x^{2}}{4}+\frac {5 x^{3}}{9}-\frac {499 x^{4}}{1152}+\frac {16919 x^{5}}{48000}-\frac {56861 x^{6}}{192000}+\frac {1027717 x^{7}}{4032000}+O\left (x^{8}\right )\right ) \\ \end{align*}

Hence the final solution is

\begin{align*} y &= y_h \\ &= c_1 \left (1-x +\frac {3 x^{2}}{4}-\frac {7 x^{3}}{12}+\frac {91 x^{4}}{192}-\frac {637 x^{5}}{1600}+\frac {19747 x^{6}}{57600}-\frac {17329 x^{7}}{57600}+O\left (x^{8}\right )\right )+c_2 \left (\left (1-x +\frac {3 x^{2}}{4}-\frac {7 x^{3}}{12}+\frac {91 x^{4}}{192}-\frac {637 x^{5}}{1600}+\frac {19747 x^{6}}{57600}-\frac {17329 x^{7}}{57600}+O\left (x^{8}\right )\right ) \ln \left (x \right )+x -\frac {3 x^{2}}{4}+\frac {5 x^{3}}{9}-\frac {499 x^{4}}{1152}+\frac {16919 x^{5}}{48000}-\frac {56861 x^{6}}{192000}+\frac {1027717 x^{7}}{4032000}+O\left (x^{8}\right )\right ) \\ \end{align*}

15.30.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (x +1\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+x \left (2 x +1\right ) \left (\frac {d}{d x}y \left (x \right )\right )+x 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 {y \left (x \right )}{x \left (x +1\right )}-\frac {\left (2 x +1\right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x \left (x +1\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 {\left (2 x +1\right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x \left (x +1\right )}+\frac {y \left (x \right )}{x \left (x +1\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 {2 x +1}{x \left (x +1\right )}, P_{3}\left (x \right )=\frac {1}{x \left (x +1\right )}\right ] \\ {} & \circ & \left (x +1\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=1 \\ {} & \circ & \left (x +1\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=0 \\ {} & \circ & x =-1\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}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x \left (x +1\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (2 x +1\right ) \left (\frac {d}{d x}y \left (x \right )\right )+y \left (x \right )=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{2}-u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (2 u -1\right ) \left (\frac {d}{d u}y \left (u \right )\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 \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & 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..2 \\ {} & {} & 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} \\ {} & {} & -a_{0} r^{2} u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-a_{k +1} \left (k +1+r \right )^{2}+a_{k} \left (k^{2}+2 k r +r^{2}+k +r +1\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -a_{k +1} \left (k +1\right )^{2}+a_{k} \left (k^{2}+k +1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k^{2}+k +1\right )}{\left (k +1\right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k^{2}+k +1\right )}{\left (k +1\right )^{2}} \\ \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 +1}=\frac {a_{k} \left (k^{2}+k +1\right )}{\left (k +1\right )^{2}}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k}, a_{k +1}=\frac {a_{k} \left (k^{2}+k +1\right )}{\left (k +1\right )^{2}}\right ] \end {array} \]

15.30.2 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)] 
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 
   -> elliptic 
   -> Legendre 
   -> Kummer 
      -> hyper3: Equivalence to 1F1 under a power @ Moebius 
   -> hypergeometric 
      -> heuristic approach 
      <- heuristic approach successful 
      -> solution has integrals; searching for one without integrals... 
         -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
         <- hyper3 successful: received ODE is equivalent to the 2F1 ODE 
      <- hypergeometric solution without integrals succesful 
   <- hypergeometric successful 
<- special function solution successful`
 

15.30.3 Maple dsolve solution

Solving time : 0.011 (sec)
Leaf size : 52

dsolve(x^2*(x+1)*diff(diff(y(x),x),x)+x*(2*x+1)*diff(y(x),x)+x*y(x) = 0,y(x), 
       series,x=0)
 

\[ y = \left (\ln \left (x \right ) c_2 +c_1 \right ) \left (1-x +\frac {3}{4} x^{2}-\frac {7}{12} x^{3}+\frac {91}{192} x^{4}-\frac {637}{1600} x^{5}+\frac {19747}{57600} x^{6}-\frac {17329}{57600} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\left (x -\frac {3}{4} x^{2}+\frac {5}{9} x^{3}-\frac {499}{1152} x^{4}+\frac {16919}{48000} x^{5}-\frac {56861}{192000} x^{6}+\frac {1027717}{4032000} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) c_2 \]

15.30.4 Mathematica DSolve solution

Solving time : 0.014 (sec)
Leaf size : 151

AsymptoticDSolveValue[{x^2*(1+x)*D[y[x],{x,2}]+x*(1+2*x)*D[y[x],x]+x*y[x]==0,{}}, 
       y[x],{x,0,7}]
 

\[ y(x)\to c_1 \left (-\frac {17329 x^7}{57600}+\frac {19747 x^6}{57600}-\frac {637 x^5}{1600}+\frac {91 x^4}{192}-\frac {7 x^3}{12}+\frac {3 x^2}{4}-x+1\right )+c_2 \left (\frac {1027717 x^7}{4032000}-\frac {56861 x^6}{192000}+\frac {16919 x^5}{48000}-\frac {499 x^4}{1152}+\frac {5 x^3}{9}-\frac {3 x^2}{4}+\left (-\frac {17329 x^7}{57600}+\frac {19747 x^6}{57600}-\frac {637 x^5}{1600}+\frac {91 x^4}{192}-\frac {7 x^3}{12}+\frac {3 x^2}{4}-x+1\right ) \log (x)+x\right ) \]