Internal problem ID [14824]
Internal file name [OUTPUT/14504_Monday_April_08_2024_06_26_15_AM_3095027/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.9, page 215
Problem number: 24.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Repeated root"
Maple gives the following as the ode type
[_Laguerre]
\[ \boxed {x y^{\prime \prime }+\left (1-x \right ) y^{\prime }+k y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ x y^{\prime \prime }+\left (1-x \right ) y^{\prime }+k y = 0 \] 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 {x -1}{x}\\ q(x) &= \frac {k}{x}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x y^{\prime \prime }+\left (1-x \right ) y^{\prime }+k y = 0 \] 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} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) x +\left (1-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+k \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 -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right )+\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 }}k 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 }}\left (-x^{n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}k a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}k 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 =0}{\sum }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}k a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+\left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ x^{-1+r} a_{0} r \left (-1+r \right )+r a_{0} x^{-1+r} = 0 \] Or \[ \left (x^{-1+r} r \left (-1+r \right )+r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ x^{-1+r} r^{2} = 0 \] 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 \[ x^{-1+r} r^{2} = 0 \] 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} \left (n +r \right ) \left (n +r -1\right )-a_{n -1} \left (n +r -1\right )+a_{n} \left (n +r \right )+k a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (k -n -r +1\right )}{n^{2}+2 n r +r^{2}}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = -\frac {a_{n -1} \left (k -n +1\right )}{n^{2}}\tag {5} \] 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 \[ a_{1}=\frac {-k +r}{\left (1+r \right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{1}=-k \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-k +r}{\left (1+r \right )^{2}}\) | \(-k\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {\left (k -1-r \right ) \left (k -r \right )}{\left (1+r \right )^{2} \left (r +2\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{2}=\frac {\left (k -1\right ) k}{4} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-k +r}{\left (1+r \right )^{2}}\) | \(-k\) |
| \(a_{2}\) | \(\frac {\left (k -1-r \right ) \left (k -r \right )}{\left (1+r \right )^{2} \left (r +2\right )^{2}}\) | \(\frac {\left (k -1\right ) k}{4}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {\left (k -2-r \right ) \left (k -1-r \right ) \left (k -r \right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (3+r \right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{3}=-\frac {\left (k -2\right ) \left (k -1\right ) k}{36} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-k +r}{\left (1+r \right )^{2}}\) | \(-k\) |
| \(a_{2}\) | \(\frac {\left (k -1-r \right ) \left (k -r \right )}{\left (1+r \right )^{2} \left (r +2\right )^{2}}\) | \(\frac {\left (k -1\right ) k}{4}\) |
| \(a_{3}\) | \(-\frac {\left (k -2-r \right ) \left (k -1-r \right ) \left (k -r \right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (3+r \right )^{2}}\) | \(-\frac {\left (k -2\right ) \left (k -1\right ) k}{36}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {\left (k -3-r \right ) \left (k -2-r \right ) \left (k -1-r \right ) \left (k -r \right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (3+r \right )^{2} \left (r +4\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{4}=\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{576} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-k +r}{\left (1+r \right )^{2}}\) | \(-k\) |
| \(a_{2}\) | \(\frac {\left (k -1-r \right ) \left (k -r \right )}{\left (1+r \right )^{2} \left (r +2\right )^{2}}\) | \(\frac {\left (k -1\right ) k}{4}\) |
| \(a_{3}\) | \(-\frac {\left (k -2-r \right ) \left (k -1-r \right ) \left (k -r \right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (3+r \right )^{2}}\) | \(-\frac {\left (k -2\right ) \left (k -1\right ) k}{36}\) |
| \(a_{4}\) | \(\frac {\left (k -3-r \right ) \left (k -2-r \right ) \left (k -1-r \right ) \left (k -r \right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (3+r \right )^{2} \left (r +4\right )^{2}}\) | \(\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{576}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {\left (k -4-r \right ) \left (k -3-r \right ) \left (k -2-r \right ) \left (k -1-r \right ) \left (k -r \right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (3+r \right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{5}=-\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{14400} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-k +r}{\left (1+r \right )^{2}}\) | \(-k\) |
| \(a_{2}\) | \(\frac {\left (k -1-r \right ) \left (k -r \right )}{\left (1+r \right )^{2} \left (r +2\right )^{2}}\) | \(\frac {\left (k -1\right ) k}{4}\) |
| \(a_{3}\) | \(-\frac {\left (k -2-r \right ) \left (k -1-r \right ) \left (k -r \right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (3+r \right )^{2}}\) | \(-\frac {\left (k -2\right ) \left (k -1\right ) k}{36}\) |
| \(a_{4}\) | \(\frac {\left (k -3-r \right ) \left (k -2-r \right ) \left (k -1-r \right ) \left (k -r \right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (3+r \right )^{2} \left (r +4\right )^{2}}\) | \(\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{576}\) |
| \(a_{5}\) | \(-\frac {\left (k -4-r \right ) \left (k -3-r \right ) \left (k -2-r \right ) \left (k -1-r \right ) \left (k -r \right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (3+r \right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(-\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{14400}\) |
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}\dots \\ &= -k x +1+\frac {\left (k -1\right ) k \,x^{2}}{4}-\frac {\left (k -2\right ) \left (k -1\right ) k \,x^{3}}{36}+\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k \,x^{4}}{576}-\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k \,x^{5}}{14400}+O\left (x^{6}\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 |
| \(b_{1}\) | \(\frac {-k +r}{\left (1+r \right )^{2}}\) | \(-k\) | \(\frac {1-r +2 k}{\left (1+r \right )^{3}}\) | \(2 k +1\) |
| \(b_{2}\) | \(\frac {\left (k -1-r \right ) \left (k -r \right )}{\left (1+r \right )^{2} \left (r +2\right )^{2}}\) | \(\frac {\left (k -1\right ) k}{4}\) | \(\frac {-2 r^{3}+\left (6 k -3\right ) r^{2}+\left (-4 k^{2}+10 k +1\right ) r -6 k^{2}+2 k +2}{\left (1+r \right )^{3} \left (r +2\right )^{3}}\) | \(-\frac {k}{2}+\frac {1}{4}-\frac {3 \left (k -1\right ) k}{4}\) |
| \(b_{3}\) | \(-\frac {\left (k -2-r \right ) \left (k -1-r \right ) \left (k -r \right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (3+r \right )^{2}}\) | \(-\frac {\left (k -2\right ) \left (k -1\right ) k}{36}\) | \(\frac {12-3 r^{5}+6 \left (-3+2 k \right ) r^{4}+\left (-15 k^{2}+66 k -35\right ) r^{3}+6 \left (k^{3}-12 k^{2}+20 k -3\right ) r^{2}+\left (24 k^{3}-105 k^{2}+78 k +14\right ) r +22 k^{3}-48 k^{2}+8 k}{\left (1+r \right )^{3} \left (r +2\right )^{3} \left (3+r \right )^{3}}\) | \(\frac {\left (k -1\right ) k}{36}+\frac {\left (k -2\right ) k}{36}+\frac {\left (k -2\right ) \left (k -1\right )}{36}+\frac {11 \left (k -2\right ) \left (k -1\right ) k}{108}\) |
| \(b_{4}\) | \(\frac {\left (k -3-r \right ) \left (k -2-r \right ) \left (k -1-r \right ) \left (k -r \right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (3+r \right )^{2} \left (r +4\right )^{2}}\) | \(\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{576}\) | \(\frac {-4 r^{7}+\left (20 k -50\right ) r^{6}+\left (-36 k^{2}+228 k -246\right ) r^{5}+\left (28 k^{3}-366 k^{2}+1014 k -592\right ) r^{4}+\left (-8 k^{4}+248 k^{3}-1408 k^{2}+2208 k -674\right ) r^{3}+\left (-60 k^{4}+780 k^{3}-2550 k^{2}+2382 k -198\right ) r^{2}+\left (-140 k^{4}+1040 k^{3}-2152 k^{2}+1076 k +228\right ) r -100 k^{4}+504 k^{3}-668 k^{2}+72 k +144}{\left (1+r \right )^{3} \left (r +2\right )^{3} \left (3+r \right )^{3} \left (r +4\right )^{3}}\) | \(-\frac {\left (k -2\right ) \left (k -1\right ) k}{576}-\frac {\left (k -3\right ) \left (k -1\right ) k}{576}-\frac {\left (k -3\right ) \left (k -2\right ) k}{576}-\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right )}{576}-\frac {25 \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{3456}\) |
| \(b_{5}\) | \(-\frac {\left (k -4-r \right ) \left (k -3-r \right ) \left (k -2-r \right ) \left (k -1-r \right ) \left (k -r \right )}{\left (1+r \right )^{2} \left (r +2\right )^{2} \left (3+r \right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(-\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{14400}\) | \(\frac {-5 r^{9}+\left (30 k -105\right ) r^{8}+\left (-70 k^{2}+580 k -930\right ) r^{7}+\left (80 k^{3}-1230 k^{2}+4690 k -4500\right ) r^{6}+\left (-45 k^{4}+1260 k^{3}-8895 k^{2}+20550 k -12819\right ) r^{5}+\left (10 k^{5}-625 k^{4}+7950 k^{3}-34175 k^{2}+52700 k -21315\right ) r^{4}+\left (120 k^{5}-3325 k^{4}+25700 k^{3}-74885 k^{2}+79270 k -18310\right ) r^{3}+\left (510 k^{5}-8475 k^{4}+44850 k^{3}-92775 k^{2}+65340 k -3600\right ) r^{2}+\left (900 k^{5}-10370 k^{4}+40060 k^{3}-59370 k^{2}+23800 k +5424\right ) r +548 k^{5}-4880 k^{4}+14380 k^{3}-14800 k^{2}+1152 k +2880}{\left (1+r \right )^{3} \left (r +2\right )^{3} \left (3+r \right )^{3} \left (r +4\right )^{3} \left (r +5\right )^{3}}\) | \(\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -2\right ) \left (k -1\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -3\right ) \left (k -1\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right )}{14400}+\frac {137 \left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{432000}\) |
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 \\ &= \left (-k x +1+\frac {\left (k -1\right ) k \,x^{2}}{4}-\frac {\left (k -2\right ) \left (k -1\right ) k \,x^{3}}{36}+\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k \,x^{4}}{576}-\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k \,x^{5}}{14400}+O\left (x^{6}\right )\right ) \ln \left (x \right )+\left (2 k +1\right ) x +\left (-\frac {k}{2}+\frac {1}{4}-\frac {3 \left (k -1\right ) k}{4}\right ) x^{2}+\left (\frac {\left (k -1\right ) k}{36}+\frac {\left (k -2\right ) k}{36}+\frac {\left (k -2\right ) \left (k -1\right )}{36}+\frac {11 \left (k -2\right ) \left (k -1\right ) k}{108}\right ) x^{3}+\left (-\frac {\left (k -2\right ) \left (k -1\right ) k}{576}-\frac {\left (k -3\right ) \left (k -1\right ) k}{576}-\frac {\left (k -3\right ) \left (k -2\right ) k}{576}-\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right )}{576}-\frac {25 \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{3456}\right ) x^{4}+\left (\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -2\right ) \left (k -1\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -3\right ) \left (k -1\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right )}{14400}+\frac {137 \left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{432000}\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} \left (-k x +1+\frac {\left (k -1\right ) k \,x^{2}}{4}-\frac {\left (k -2\right ) \left (k -1\right ) k \,x^{3}}{36}+\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k \,x^{4}}{576}-\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k \,x^{5}}{14400}+O\left (x^{6}\right )\right ) + c_{2} \left (\left (-k x +1+\frac {\left (k -1\right ) k \,x^{2}}{4}-\frac {\left (k -2\right ) \left (k -1\right ) k \,x^{3}}{36}+\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k \,x^{4}}{576}-\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k \,x^{5}}{14400}+O\left (x^{6}\right )\right ) \ln \left (x \right )+\left (2 k +1\right ) x +\left (-\frac {k}{2}+\frac {1}{4}-\frac {3 \left (k -1\right ) k}{4}\right ) x^{2}+\left (\frac {\left (k -1\right ) k}{36}+\frac {\left (k -2\right ) k}{36}+\frac {\left (k -2\right ) \left (k -1\right )}{36}+\frac {11 \left (k -2\right ) \left (k -1\right ) k}{108}\right ) x^{3}+\left (-\frac {\left (k -2\right ) \left (k -1\right ) k}{576}-\frac {\left (k -3\right ) \left (k -1\right ) k}{576}-\frac {\left (k -3\right ) \left (k -2\right ) k}{576}-\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right )}{576}-\frac {25 \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{3456}\right ) x^{4}+\left (\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -2\right ) \left (k -1\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -3\right ) \left (k -1\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right )}{14400}+\frac {137 \left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{432000}\right ) x^{5}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (-k x +1+\frac {\left (k -1\right ) k \,x^{2}}{4}-\frac {\left (k -2\right ) \left (k -1\right ) k \,x^{3}}{36}+\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k \,x^{4}}{576}-\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k \,x^{5}}{14400}+O\left (x^{6}\right )\right )+c_{2} \left (\left (-k x +1+\frac {\left (k -1\right ) k \,x^{2}}{4}-\frac {\left (k -2\right ) \left (k -1\right ) k \,x^{3}}{36}+\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k \,x^{4}}{576}-\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k \,x^{5}}{14400}+O\left (x^{6}\right )\right ) \ln \left (x \right )+\left (2 k +1\right ) x +\left (-\frac {k}{2}+\frac {1}{4}-\frac {3 \left (k -1\right ) k}{4}\right ) x^{2}+\left (\frac {\left (k -1\right ) k}{36}+\frac {\left (k -2\right ) k}{36}+\frac {\left (k -2\right ) \left (k -1\right )}{36}+\frac {11 \left (k -2\right ) \left (k -1\right ) k}{108}\right ) x^{3}+\left (-\frac {\left (k -2\right ) \left (k -1\right ) k}{576}-\frac {\left (k -3\right ) \left (k -1\right ) k}{576}-\frac {\left (k -3\right ) \left (k -2\right ) k}{576}-\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right )}{576}-\frac {25 \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{3456}\right ) x^{4}+\left (\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -2\right ) \left (k -1\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -3\right ) \left (k -1\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right )}{14400}+\frac {137 \left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{432000}\right ) x^{5}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (-k x +1+\frac {\left (k -1\right ) k \,x^{2}}{4}-\frac {\left (k -2\right ) \left (k -1\right ) k \,x^{3}}{36}+\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k \,x^{4}}{576}-\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k \,x^{5}}{14400}+O\left (x^{6}\right )\right )+c_{2} \left (\left (-k x +1+\frac {\left (k -1\right ) k \,x^{2}}{4}-\frac {\left (k -2\right ) \left (k -1\right ) k \,x^{3}}{36}+\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k \,x^{4}}{576}-\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k \,x^{5}}{14400}+O\left (x^{6}\right )\right ) \ln \left (x \right )+\left (2 k +1\right ) x +\left (-\frac {k}{2}+\frac {1}{4}-\frac {3 \left (k -1\right ) k}{4}\right ) x^{2}+\left (\frac {\left (k -1\right ) k}{36}+\frac {\left (k -2\right ) k}{36}+\frac {\left (k -2\right ) \left (k -1\right )}{36}+\frac {11 \left (k -2\right ) \left (k -1\right ) k}{108}\right ) x^{3}+\left (-\frac {\left (k -2\right ) \left (k -1\right ) k}{576}-\frac {\left (k -3\right ) \left (k -1\right ) k}{576}-\frac {\left (k -3\right ) \left (k -2\right ) k}{576}-\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right )}{576}-\frac {25 \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{3456}\right ) x^{4}+\left (\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -2\right ) \left (k -1\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -3\right ) \left (k -1\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right )}{14400}+\frac {137 \left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{432000}\right ) x^{5}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (-k x +1+\frac {\left (k -1\right ) k \,x^{2}}{4}-\frac {\left (k -2\right ) \left (k -1\right ) k \,x^{3}}{36}+\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k \,x^{4}}{576}-\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k \,x^{5}}{14400}+O\left (x^{6}\right )\right )+c_{2} \left (\left (-k x +1+\frac {\left (k -1\right ) k \,x^{2}}{4}-\frac {\left (k -2\right ) \left (k -1\right ) k \,x^{3}}{36}+\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k \,x^{4}}{576}-\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k \,x^{5}}{14400}+O\left (x^{6}\right )\right ) \ln \left (x \right )+\left (2 k +1\right ) x +\left (-\frac {k}{2}+\frac {1}{4}-\frac {3 \left (k -1\right ) k}{4}\right ) x^{2}+\left (\frac {\left (k -1\right ) k}{36}+\frac {\left (k -2\right ) k}{36}+\frac {\left (k -2\right ) \left (k -1\right )}{36}+\frac {11 \left (k -2\right ) \left (k -1\right ) k}{108}\right ) x^{3}+\left (-\frac {\left (k -2\right ) \left (k -1\right ) k}{576}-\frac {\left (k -3\right ) \left (k -1\right ) k}{576}-\frac {\left (k -3\right ) \left (k -2\right ) k}{576}-\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right )}{576}-\frac {25 \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{3456}\right ) x^{4}+\left (\frac {\left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -2\right ) \left (k -1\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -3\right ) \left (k -1\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) k}{14400}+\frac {\left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right )}{14400}+\frac {137 \left (k -4\right ) \left (k -3\right ) \left (k -2\right ) \left (k -1\right ) k}{432000}\right ) x^{5}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime } x +\left (1-x \right ) y^{\prime }+k y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {\left (x -1\right ) y^{\prime }}{x}-\frac {k y}{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} \\ {} & {} & y^{\prime \prime }-\frac {\left (x -1\right ) y^{\prime }}{x}+\frac {k y}{x}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {x -1}{x}, P_{3}\left (x \right )=\frac {k}{x}\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}}}=1 \\ {} & \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}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & y^{\prime \prime } x +\left (1-x \right ) y^{\prime }+k 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^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r^{2} x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right )^{2}+a_{k} \left (k -k -r \right )\right ) x^{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 -k \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k -k \right )}{\left (k +1\right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k -k \right )}{\left (k +1\right )^{2}} \\ \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 -k \right )}{\left (k +1\right )^{2}}\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 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 <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Kummer successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 309
Order:=6; dsolve(x*diff(y(x),x$2)+(1-x)*diff(y(x),x)+k*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \left (\left (2 k +1\right ) x +\left (\frac {1}{4} k +\frac {1}{4}-\frac {3}{4} k^{2}\right ) x^{2}+\left (-\frac {2}{9} k^{2}+\frac {1}{27} k +\frac {1}{18}+\frac {11}{108} k^{3}\right ) x^{3}+\left (\frac {7}{192} k^{3}-\frac {167}{3456} k^{2}+\frac {1}{192} k +\frac {1}{96}-\frac {25}{3456} k^{4}\right ) x^{4}+\left (\frac {1}{1500} k -\frac {61}{21600} k^{4}+\frac {1}{600}+\frac {719}{86400} k^{3}-\frac {37}{4320} k^{2}+\frac {137}{432000} k^{5}\right ) x^{5}+\operatorname {O}\left (x^{6}\right )\right ) c_{2} +\left (1-k x +\frac {1}{4} \left (-1+k \right ) k x^{2}-\frac {1}{36} \left (-2+k \right ) \left (-1+k \right ) k x^{3}+\frac {1}{576} \left (k -3\right ) \left (-2+k \right ) \left (-1+k \right ) k x^{4}-\frac {1}{14400} \left (-4+k \right ) \left (k -3\right ) \left (-2+k \right ) \left (-1+k \right ) k x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \left (c_{1} +c_{2} \ln \left (x \right )\right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 415
AsymptoticDSolveValue[x*y''[x]+(1-x)*y'[x]+k*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (-\frac {(k-4) (k-3) (k-2) (k-1) k x^5}{14400}+\frac {1}{576} (k-3) (k-2) (k-1) k x^4-\frac {1}{36} (k-2) (k-1) k x^3+\frac {1}{4} (k-1) k x^2-k x+1\right )+c_2 \left (\frac {(k-4) (k-3) (k-2) (k-1) x^5}{14400}+\frac {(k-4) (k-3) (k-2) k x^5}{14400}+\frac {(k-4) (k-3) (k-1) k x^5}{14400}+\frac {(k-4) (k-2) (k-1) k x^5}{14400}+\frac {137 (k-4) (k-3) (k-2) (k-1) k x^5}{432000}+\frac {(k-3) (k-2) (k-1) k x^5}{14400}-\frac {1}{576} (k-3) (k-2) (k-1) x^4-\frac {1}{576} (k-3) (k-2) k x^4-\frac {1}{576} (k-3) (k-1) k x^4-\frac {25 (k-3) (k-2) (k-1) k x^4}{3456}-\frac {1}{576} (k-2) (k-1) k x^4+\frac {1}{36} (k-2) (k-1) x^3+\frac {1}{36} (k-2) k x^3+\frac {11}{108} (k-2) (k-1) k x^3+\frac {1}{36} (k-1) k x^3-\frac {1}{4} (k-1) x^2-\frac {3}{4} (k-1) k x^2-\frac {k x^2}{4}+\left (-\frac {(k-4) (k-3) (k-2) (k-1) k x^5}{14400}+\frac {1}{576} (k-3) (k-2) (k-1) k x^4-\frac {1}{36} (k-2) (k-1) k x^3+\frac {1}{4} (k-1) k x^2-k x+1\right ) \log (x)+2 k x+x\right ) \]