4.39 problem 36

Internal problem ID [7908]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 36
Date solved : Monday, October 21, 2024 at 04:32:08 PM
CAS classification : [[_2nd_order, _exact, _linear, _homogeneous]]

Solve

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

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

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

Table 86: Table \(p(x),q(x)\) singularites.

\(p(x)=\frac {3}{x -1}\)
singularity	type
\(x = 1\)	\(\text {``regular''}\)

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

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

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

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 \left (x -1\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 (\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 (-x^{n +r -1} 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 ) = 0 \end{equation}

The next step is to make all powers of \(x\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r -1}\) and adjusting the power and the corresponding index gives

\begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}x^{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 -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r -1} \\ \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 \(n +r -1\).

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

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

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

When \(n = 0\) the above becomes

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

Or

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

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

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

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

\[ -r \left (-1+r \right ) = 0 \]

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

\begin{align*} r_1 &= 1\\ r_2 &= 0 \end{align*}

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

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

Solving for \(r\) gives the roots of the indicial equation as \([1, 0]\).

Since \(r_1 - r_2 = 1\) is an integer, then we can construct two linearly independent solutions

\begin{align*} y_{1}\left (x \right ) &= x^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+x^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end{align*}

Or

\begin{align*} y_{1}\left (x \right ) &= x \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end{align*}

Or

\begin{align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +1}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end{align*}

Where \(C\) above can be zero. We start by finding \(y_{1}\). 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 )+3 a_{n -1} \left (n +r -1\right )+a_{n -1} = 0 \end{equation}

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

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

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

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

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

\[ a_{1}=\frac {1+r}{r} \]

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

\[ a_{1}=2 \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {1+r}{r}\)	\(2\)

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

\[ a_{2}=\frac {2+r}{r} \]

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

\[ a_{2}=3 \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {1+r}{r}\)	\(2\)
\(a_{2}\)	\(\frac {2+r}{r}\)	\(3\)

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

\[ a_{3}=\frac {3+r}{r} \]

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

\[ a_{3}=4 \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {1+r}{r}\)	\(2\)
\(a_{2}\)	\(\frac {2+r}{r}\)	\(3\)
\(a_{3}\)	\(\frac {3+r}{r}\)	\(4\)

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

\[ a_{4}=\frac {4+r}{r} \]

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

\[ a_{4}=5 \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {1+r}{r}\)	\(2\)
\(a_{2}\)	\(\frac {2+r}{r}\)	\(3\)
\(a_{3}\)	\(\frac {3+r}{r}\)	\(4\)
\(a_{4}\)	\(\frac {4+r}{r}\)	\(5\)

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

\[ a_{5}=\frac {5+r}{r} \]

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

\[ a_{5}=6 \]

And the table now becomes

\(n\)	\(a_{n ,r}\)	\(a_{n}\)
\(a_{0}\)	\(1\)	\(1\)
\(a_{1}\)	\(\frac {1+r}{r}\)	\(2\)
\(a_{2}\)	\(\frac {2+r}{r}\)	\(3\)
\(a_{3}\)	\(\frac {3+r}{r}\)	\(4\)
\(a_{4}\)	\(\frac {4+r}{r}\)	\(5\)
\(a_{5}\)	\(\frac {5+r}{r}\)	\(6\)

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

\begin{align*} y_{1}\left (x \right )&= 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 ) \\ &= x \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right ) \end{align*}

Now the second solution \(y_{2}\left (x \right )\) is found. Let

\[ r_{1}-r_{2} = N \]

Where \(N\) is positive integer which is the difference between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=1\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{1}\left (r \right )\). If this limit exists, then \(C = 0\), else we need to keep the log term and \(C \neq 0\). The above table shows that

\begin{align*} a_N &= a_{1} \\ &= \frac {1+r}{r} \end{align*}

Therefore

\begin{align*} \lim _{r\rightarrow r_{2}}\frac {1+r}{r}&= \lim _{r\rightarrow 0}\frac {1+r}{r}\\ &= \textit {undefined} \end{align*}

Since the limit does not exist then the log term is needed. Therefore the second solution has the form

\[ y_{2}\left (x \right ) = C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) \]

Therefore

\begin{align*} \frac {d}{d x}y_{2}\left (x \right ) &= C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) \\ &= C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) \\ \frac {d^{2}}{d x^{2}}y_{2}\left (x \right ) &= C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right ) \\ &= C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n +r_{2}} b_{n} \left (n +r_{2}\right ) \left (-1+n +r_{2}\right )\right ) \\ \end{align*}

Substituting these back into the given ode \(x \left (x -1\right ) y^{\prime \prime }+3 x y^{\prime }+y = 0\) gives

\[ x \left (x -1\right ) \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+3 x \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right )+C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \]

Which can be written as

\begin{equation} \tag{7} \left (\left (x \left (x -1\right ) y_{1}^{\prime \prime }\left (x \right )+3 y_{1}^{\prime }\left (x \right ) x +y_{1}\left (x \right )\right ) \ln \left (x \right )+x \left (x -1\right ) \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+3 y_{1}\left (x \right )\right ) C +x \left (x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+3 x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation}

But since \(y_{1}\left (x \right )\) is a solution to the ode, then

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

Eq (7) simplifes to

\begin{equation} \tag{8} \left (x \left (x -1\right ) \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+3 y_{1}\left (x \right )\right ) C +x \left (x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+3 x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation}

Substituting \(y_{1} = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\) into the above gives

\begin{equation} \tag{9} \frac {\left (2 x \left (x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right )+\left (2 x +1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C}{x}+\frac {x^{2} \left (x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n +r_{2}} b_{n} \left (n +r_{2}\right ) \left (-1+n +r_{2}\right )\right )+3 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) x^{2}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) x}{x} = 0 \end{equation}

Since \(r_{1} = 1\) and \(r_{2} = 0\) then the above becomes

\begin{equation} \tag{10} \frac {\left (2 x \left (x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n} a_{n} \left (n +1\right )\right )+\left (2 x +1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +1}\right )\right ) C}{x}+\frac {x^{2} \left (x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n} b_{n} n \left (n -1\right )\right )+3 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -1} b_{n} n \right ) x^{2}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) x}{x} = 0 \end{equation}

Which simplifies to

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

The next step is to make all powers of \(x\) be \(n -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n -1}\) and adjusting the power and the corresponding index gives

\begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +1} a_{n} \left (n +1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}2 C a_{-2+n} \left (n -1\right ) x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n} a_{n} \left (n +1\right ) C \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 C a_{n -1} n \,x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +1} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}2 C a_{-2+n} x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n} C &= \moverset {\infty }{\munderset {n =1}{\sum }}C a_{n -1} x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{n} b_{n} n \left (n -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (n -1\right ) b_{n -1} \left (-2+n \right ) x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n} b_{n} n &= \moverset {\infty }{\munderset {n =1}{\sum }}3 \left (n -1\right ) b_{n -1} x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} x^{n -1} \\ \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 \(n -1\).

\begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =2}{\sum }}2 C a_{-2+n} \left (n -1\right ) x^{n -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 C a_{n -1} n \,x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}2 C a_{-2+n} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}C a_{n -1} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\left (n -1\right ) b_{n -1} \left (-2+n \right ) x^{n -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-n \,x^{n -1} b_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}3 \left (n -1\right ) b_{n -1} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} x^{n -1}\right ) = 0 \end{equation}

For \(n=0\) in Eq. (2B), we choose arbitray value for \(b_{0}\) as \(b_{0} = 1\). For \(n=N\), where \(N=1\) which is the difference between the two roots, we are free to choose \(b_{1} = 0\). Hence for \(n=1\), Eq (2B) gives

\[ -C +1 = 0 \]

Which is solved for \(C\). Solving for \(C\) gives

\[ C=1 \]

For \(n=2\), Eq (2B) gives

\[ \left (4 a_{0}-3 a_{1}\right ) C +4 b_{1}-2 b_{2} = 0 \]

Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives

\[ -2-2 b_{2} = 0 \]

Solving the above for \(b_{2}\) gives

\[ b_{2}=-1 \]

For \(n=3\), Eq (2B) gives

\[ \left (6 a_{1}-5 a_{2}\right ) C +9 b_{2}-6 b_{3} = 0 \]

Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives

\[ -12-6 b_{3} = 0 \]

Solving the above for \(b_{3}\) gives

\[ b_{3}=-2 \]

For \(n=4\), Eq (2B) gives

\[ \left (8 a_{2}-7 a_{3}\right ) C +16 b_{3}-12 b_{4} = 0 \]

Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives

\[ -36-12 b_{4} = 0 \]

Solving the above for \(b_{4}\) gives

\[ b_{4}=-3 \]

For \(n=5\), Eq (2B) gives

\[ \left (10 a_{3}-9 a_{4}\right ) C +25 b_{4}-20 b_{5} = 0 \]

Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives

\[ -80-20 b_{5} = 0 \]

Solving the above for \(b_{5}\) gives

\[ b_{5}=-4 \]

Now that we found all \(b_{n}\) and \(C\), we can calculate the second solution from

\[ y_{2}\left (x \right ) = C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) \]

Using the above value found for \(C=1\) and all \(b_{n}\), then the second solution becomes

\[ y_{2}\left (x \right )= 1\eslowast \left (x \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1-x^{2}-2 x^{3}-3 x^{4}-4 x^{5}+O\left (x^{6}\right ) \]

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 \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right ) + c_2 \left (1\eslowast \left (x \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1-x^{2}-2 x^{3}-3 x^{4}-4 x^{5}+O\left (x^{6}\right )\right ) \\ \end{align*}

Hence the final solution is

\begin{align*} y &= y_h \\ &= c_1 x \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right )+c_2 \left (x \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right ) \ln \left (x \right )+1-x^{2}-2 x^{3}-3 x^{4}-4 x^{5}+O\left (x^{6}\right )\right ) \\ \end{align*}

4.39.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (x -1\right ) \left (\frac {d}{d x}y^{\prime }\right )+3 x y^{\prime }+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 {y}{x \left (x -1\right )}-\frac {3 y^{\prime }}{x -1} \\ \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 {3 y^{\prime }}{x -1}+\frac {y}{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 {3}{x -1}, P_{3}\left (x \right )=\frac {1}{x \left (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}}}=0 \\ {} & \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}}}=0 \\ {} & \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 \left (x -1\right ) \left (\frac {d}{d x}y^{\prime }\right )+3 x y^{\prime }+y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\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 \cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime }=\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}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\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}{d x}y^{\prime }\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} r \left (-1+r \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-a_{k +1} \left (k +r +1\right ) \left (k +r \right )+a_{k} \left (k +r +1\right )^{2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +r +1\right ) \left (-a_{k +1} \left (k +r \right )+a_{k} \left (k +r +1\right )\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +r +1\right )}{k +r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +1\right )}{k} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +1}=\frac {a_{k} \left (k +1\right )}{k}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +2\right )}{k +1} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}, a_{k +1}=\frac {a_{k} \left (k +2\right )}{k +1}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +1}\right ), a_{k +1}=\frac {a_{k} \left (k +1\right )}{k}, b_{k +1}=\frac {b_{k} \left (k +2\right )}{k +1}\right ] \end {array} \]

4.39.2 Maple trace

Methods for second order ODEs:
 

4.39.3 Maple dsolve solution

Solving time : 0.014 (sec)
Leaf size : 60

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

\[ y = c_1 x \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (x +2 x^{2}+3 x^{3}+4 x^{4}+5 x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \ln \left (x \right ) c_2 +\left (1+3 x +5 x^{2}+7 x^{3}+9 x^{4}+11 x^{5}+\operatorname {O}\left (x^{6}\right )\right ) c_2 \]

4.39.4 Mathematica DSolve solution

Solving time : 0.044 (sec)
Leaf size : 63

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

\[ y(x)\to c_1 \left (x^4+x^3+x^2+\left (4 x^3+3 x^2+2 x+1\right ) x \log (x)+x+1\right )+c_2 \left (5 x^5+4 x^4+3 x^3+2 x^2+x\right ) \]