15.48 problem 44

Internal problem ID [2046]
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 : 44
Date solved : Thursday, October 17, 2024 at 02:19:16 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

\begin{align*} x^{2} \left (1-2 x \right ) y^{\prime \prime }+3 x y^{\prime }+\left (1+4 x \right ) 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.

\[ \left (-2 x^{3}+x^{2}\right ) y^{\prime \prime }+3 x y^{\prime }+\left (1+4 x \right ) y = 0 \]

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 {3}{x \left (-1+2 x \right )}\\ q(x) &= -\frac {1+4 x}{x^{2} \left (-1+2 x \right )}\\ \end{align*}

Table 170: Table \(p(x),q(x)\) singularites.

\(p(x)=-\frac {3}{x \left (-1+2 x \right )}\)
singularity	type
\(x = 0\)	\(\text {``regular''}\)
\(x = {\frac {1}{2}}\)	\(\text {``regular''}\)

\(q(x)=-\frac {1+4 x}{x^{2} \left (-1+2 x \right )}\)
singularity	type
\(x = 0\)	\(\text {``regular''}\)
\(x = {\frac {1}{2}}\)	\(\text {``regular''}\)

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

Regular singular points : \(\left [0, {\frac {1}{2}}, \infty \right ]\)

Irregular singular points : \([]\)

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

\[ -x^{2} \left (-1+2 x \right ) y^{\prime \prime }+3 x y^{\prime }+\left (1+4 x \right ) y = 0 \]

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+2 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 )+3 x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (1+4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation}

Which simplifies to

\begin{equation} \tag{2A} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 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 }}3 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 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 (-2 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}4 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} \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 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 =0}{\sum }}3 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} x^{n +r}\right ) = 0 \end{equation}

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

\[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+3 x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \]

When \(n = 0\) the above becomes

\[ x^{r} a_{0} r \left (-1+r \right )+3 x^{r} a_{0} r +a_{0} x^{r} = 0 \]

Or

\[ \left (x^{r} r \left (-1+r \right )+3 x^{r} r +x^{r}\right ) a_{0} = 0 \]

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

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

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

\[ \left (r +1\right )^{2} = 0 \]

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

\begin{align*} r_1 &= -1\\ r_2 &= -1 \end{align*}

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

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

Solving for \(r\) gives the roots of the indicial equation as \([-1, -1]\).

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, \(r = -1\), Eqs (1A,1B) become

\begin{align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n -1}\\ 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 -1}\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 \(1\le n\) the recursive equation is

\begin{equation} \tag{3} -2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+a_{n} \left (n +r \right ) \left (n +r -1\right )+3 a_{n} \left (n +r \right )+a_{n}+4 a_{n -1} = 0 \end{equation}

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

\[ a_{n} = \frac {2 a_{n -1} \left (n^{2}+2 n r +r^{2}-3 n -3 r \right )}{n^{2}+2 n r +r^{2}+2 n +2 r +1}\tag {4} \]

Which for the root \(r = -1\) becomes

\[ a_{n} = \frac {2 a_{n -1} \left (n^{2}-5 n +4\right )}{n^{2}}\tag {5} \]

At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = -1\) 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

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

Which for the root \(r = -1\) becomes

\[ a_{1}=0 \]

And the table now becomes

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

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

\[ a_{2}=\frac {4 r^{3}-8 r^{2}-4 r +8}{\left (r +3\right )^{2} \left (r +2\right )} \]

Which for the root \(r = -1\) becomes

\[ a_{2}=0 \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {2 r^{2}-2 r -4}{\left (r +2\right )^{2}}\)	\(0\)
\(a_{2}\)	\(\frac {4 r^{3}-8 r^{2}-4 r +8}{\left (r +3\right )^{2} \left (r +2\right )}\)	\(0\)

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

\[ a_{3}=\frac {8 r^{4}-16 r^{3}-8 r^{2}+16 r}{\left (r +4\right )^{2} \left (r +2\right ) \left (r +3\right )} \]

Which for the root \(r = -1\) becomes

\[ a_{3}=0 \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {2 r^{2}-2 r -4}{\left (r +2\right )^{2}}\)	\(0\)
\(a_{2}\)	\(\frac {4 r^{3}-8 r^{2}-4 r +8}{\left (r +3\right )^{2} \left (r +2\right )}\)	\(0\)
\(a_{3}\)	\(\frac {8 r^{4}-16 r^{3}-8 r^{2}+16 r}{\left (r +4\right )^{2} \left (r +2\right ) \left (r +3\right )}\)	\(0\)

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

\[ a_{4}=\frac {16 \left (r +1\right )^{2} \left (-1+r \right ) \left (r -2\right ) r}{\left (r +5\right )^{2} \left (r +3\right ) \left (r +2\right ) \left (r +4\right )} \]

Which for the root \(r = -1\) becomes

\[ a_{4}=0 \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {2 r^{2}-2 r -4}{\left (r +2\right )^{2}}\)	\(0\)
\(a_{2}\)	\(\frac {4 r^{3}-8 r^{2}-4 r +8}{\left (r +3\right )^{2} \left (r +2\right )}\)	\(0\)
\(a_{3}\)	\(\frac {8 r^{4}-16 r^{3}-8 r^{2}+16 r}{\left (r +4\right )^{2} \left (r +2\right ) \left (r +3\right )}\)	\(0\)
\(a_{4}\)	\(\frac {16 \left (r +1\right )^{2} \left (-1+r \right ) \left (r -2\right ) r}{\left (r +5\right )^{2} \left (r +3\right ) \left (r +2\right ) \left (r +4\right )}\)	\(0\)

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

\[ a_{5}=\frac {32 \left (r +1\right )^{2} \left (-1+r \right ) \left (r -2\right ) r}{\left (r +6\right )^{2} \left (r +4\right ) \left (r +3\right ) \left (r +5\right )} \]

Which for the root \(r = -1\) becomes

\[ a_{5}=0 \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {2 r^{2}-2 r -4}{\left (r +2\right )^{2}}\)	\(0\)
\(a_{2}\)	\(\frac {4 r^{3}-8 r^{2}-4 r +8}{\left (r +3\right )^{2} \left (r +2\right )}\)	\(0\)
\(a_{3}\)	\(\frac {8 r^{4}-16 r^{3}-8 r^{2}+16 r}{\left (r +4\right )^{2} \left (r +2\right ) \left (r +3\right )}\)	\(0\)
\(a_{4}\)	\(\frac {16 \left (r +1\right )^{2} \left (-1+r \right ) \left (r -2\right ) r}{\left (r +5\right )^{2} \left (r +3\right ) \left (r +2\right ) \left (r +4\right )}\)	\(0\)
\(a_{5}\)	\(\frac {32 \left (r +1\right )^{2} \left (-1+r \right ) \left (r -2\right ) r}{\left (r +6\right )^{2} \left (r +4\right ) \left (r +3\right ) \left (r +5\right )}\)	\(0\)

Using the above table, then the first solution \(y_{1}\left (x \right )\) is

\begin{align*} y_{1}\left (x \right )&= \frac {1}{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 ) \\ &= \frac {1+O\left (x^{6}\right )}{x} \\ \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 \(r = -1\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table

\(n\)	\(b_{n ,r}\)	\(a_{n}\)	\(b_{n ,r} = \frac {d}{d r}a_{n ,r}\)	\(b_{n}\left (r =-1\right )\)
\(b_{0}\)	\(1\)	\(1\)	N/A since \(b_{n}\) starts from 1	N/A
\(b_{1}\)	\(\frac {2 r^{2}-2 r -4}{\left (r +2\right )^{2}}\)	\(0\)	\(\frac {10 r +4}{\left (r +2\right )^{3}}\)	\(-6\)
\(b_{2}\)	\(\frac {4 r^{3}-8 r^{2}-4 r +8}{\left (r +3\right )^{2} \left (r +2\right )}\)	\(0\)	\(\frac {40 r^{3}+56 r^{2}-112 r -80}{\left (r +3\right )^{3} \left (r +2\right )^{2}}\)	\(6\)
\(b_{3}\)	\(\frac {8 r^{4}-16 r^{3}-8 r^{2}+16 r}{\left (r +4\right )^{2} \left (r +2\right ) \left (r +3\right )}\)	\(0\)	\(\frac {120 r^{5}+528 r^{4}+24 r^{3}-1536 r^{2}-480 r +384}{\left (r +4\right )^{3} \left (r +2\right )^{2} \left (r +3\right )^{2}}\)	\(-{\frac {8}{3}}\)
\(b_{4}\)	\(\frac {16 \left (r +1\right )^{2} \left (-1+r \right ) \left (r -2\right ) r}{\left (r +5\right )^{2} \left (r +3\right ) \left (r +2\right ) \left (r +4\right )}\)	\(0\)	\(\frac {64 \left (r +1\right ) \left (5 r^{6}+42 r^{5}+83 r^{4}-84 r^{3}-274 r^{2}-12 r +60\right )}{\left (r +5\right )^{3} \left (r +3\right )^{2} \left (r +2\right )^{2} \left (r +4\right )^{2}}\)	\(0\)
\(b_{5}\)	\(\frac {32 \left (r +1\right )^{2} \left (-1+r \right ) \left (r -2\right ) r}{\left (r +6\right )^{2} \left (r +4\right ) \left (r +3\right ) \left (r +5\right )}\)	\(0\)	\(\frac {160 \left (r +1\right ) \left (5 r^{6}+57 r^{5}+173 r^{4}-57 r^{3}-634 r^{2}-24 r +144\right )}{\left (r +6\right )^{3} \left (r +4\right )^{2} \left (r +3\right )^{2} \left (r +5\right )^{2}}\)	\(0\)

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 \\ &= \frac {\left (1+O\left (x^{6}\right )\right ) \ln \left (x \right )}{x}+\frac {6 x^{2}-6 x -\frac {8 x^{3}}{3}+O\left (x^{6}\right )}{x} \\ \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 ) \\ &= \frac {c_1 \left (1+O\left (x^{6}\right )\right )}{x} + c_2 \left (\frac {\left (1+O\left (x^{6}\right )\right ) \ln \left (x \right )}{x}+\frac {6 x^{2}-6 x -\frac {8 x^{3}}{3}+O\left (x^{6}\right )}{x}\right ) \\ \end{align*}

Hence the final solution is

\begin{align*} y &= y_h \\ &= \frac {c_1 \left (1+O\left (x^{6}\right )\right )}{x}+c_2 \left (\frac {\left (1+O\left (x^{6}\right )\right ) \ln \left (x \right )}{x}+\frac {6 x^{2}-6 x -\frac {8 x^{3}}{3}+O\left (x^{6}\right )}{x}\right ) \\ \end{align*}

15.48.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (1-2 x \right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+3 x \left (\frac {d}{d x}y \left (x \right )\right )+\left (4 x +1\right ) 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 {\left (4 x +1\right ) y \left (x \right )}{x^{2} \left (2 x -1\right )}+\frac {3 \left (\frac {d}{d x}y \left (x \right )\right )}{x \left (2 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 {3 \left (\frac {d}{d x}y \left (x \right )\right )}{x \left (2 x -1\right )}-\frac {\left (4 x +1\right ) y \left (x \right )}{x^{2} \left (2 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 {3}{x \left (2 x -1\right )}, P_{3}\left (x \right )=-\frac {4 x +1}{x^{2} \left (2 x -1\right )}\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}}}=3 \\ {} & \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}}}=1 \\ {} & \circ & x =0\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}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{2} \left (2 x -1\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )-3 x \left (\frac {d}{d x}y \left (x \right )\right )+\left (-4 x -1\right ) y \left (x \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=\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^{m}\cdot y \left (x \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y \left (x \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot \left (\frac {d}{d x}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot \left (\frac {d}{d x}y \left (x \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..3 \\ {} & {} & x^{m}\cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \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 ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} \left (1+r \right )^{2} x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (-a_{k} \left (k +r +1\right )^{2}+2 a_{k -1} \left (k +r \right ) \left (k -3+r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -\left (1+r \right )^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =-1 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -a_{k} \left (k +r +1\right )^{2}+2 a_{k -1} \left (k +r \right ) \left (k -3+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & -a_{k +1} \left (k +2+r \right )^{2}+2 a_{k} \left (k +r +1\right ) \left (k +r -2\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {2 a_{k} \left (k +r +1\right ) \left (k +r -2\right )}{\left (k +2+r \right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =3 \\ {} & {} & a_{k +1}=\frac {2 a_{k} k \left (k -3\right )}{\left (k +1\right )^{2}} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=0 \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =1 \\ {} & {} & a_{2}=-a_{1} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{2}=0 \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =2 \\ {} & {} & a_{3}=-\frac {4 a_{2}}{9} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{3}=0 \\ \bullet & {} & \textrm {Terminating series solution of the ODE for}\hspace {3pt} r =-1\hspace {3pt}\textrm {. Use reduction of order to find the second linearly independent solution}\hspace {3pt} \\ {} & {} & y \left (x \right )=a_{0}\cdot 0 \end {array} \]

15.48.2 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 
   A Liouvillian solution exists 
   Reducible group (found an exponential solution) 
   Group is reducible, not completely reducible 
<- Kovacics algorithm successful`
 

15.48.3 Maple dsolve solution

Solving time : 0.008 (sec)
Leaf size : 34

dsolve(x^2*(-2*x+1)*diff(diff(y(x),x),x)+3*x*diff(y(x),x)+(4*x+1)*y(x) = 0,y(x), 
       series,x=0)
 

\[ y = \frac {\left (\ln \left (x \right ) c_2 +c_1 \right ) \left (1+\operatorname {O}\left (x^{6}\right )\right )+\left (\left (-6\right ) x +6 x^{2}-\frac {8}{3} x^{3}+\operatorname {O}\left (x^{6}\right )\right ) c_2}{x} \]

15.48.4 Mathematica DSolve solution

Solving time : 0.01 (sec)
Leaf size : 40

AsymptoticDSolveValue[{x^2*(1-2*x)*D[y[x],{x,2}]+3*x*D[y[x],x]+(1+4*x)*y[x]==0,{}}, 
       y[x],{x,0,5}]
 

\[ y(x)\to c_2 \left (\frac {-\frac {8 x^3}{3}+6 x^2-6 x}{x}+\frac {\log (x)}{x}\right )+\frac {c_1}{x} \]