Internal problem ID [1368]
Internal file name [OUTPUT/1369_Sunday_June_05_2022_02_13_32_AM_16321906/index.tex
]
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: 16.
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
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {25 x^{2} y^{\prime \prime }+x \left (15+x \right ) y^{\prime }+\left (x +1\right ) y=0} \] 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. \[ 25 x^{2} y^{\prime \prime }+\left (x^{2}+15 x \right ) y^{\prime }+\left (x +1\right ) y = 0 \] 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 {15+x}{25 x}\\ q(x) &= \frac {x +1}{25 x^{2}}\\ \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 \[ 25 x^{2} y^{\prime \prime }+\left (x^{2}+15 x \right ) y^{\prime }+\left (x +1\right ) y = 0 \] 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} 25 x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (x^{2}+15 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (x +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 }}25 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}15 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n}\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 \(n +r\) 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 ) &= \moverset {\infty }{\munderset {n =1}{\sum }}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 =0}{\sum }}25 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}15 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 25 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+15 x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 25 x^{r} a_{0} r \left (-1+r \right )+15 x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (25 x^{r} r \left (-1+r \right )+15 x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ x^{r} \left (5 r -1\right )^{2} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ \left (5 r -1\right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{5}}\\ r_2 &= {\frac {1}{5}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ x^{r} \left (5 r -1\right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{5}}, {\frac {1}{5}}\right ]\).
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, \(r = {\frac {1}{5}}\), Eqs (1A,1B) become \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {1}{5}}\\ y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +\frac {1}{5}}\right ) \end {align*}
We start by finding the first solution \(y_{1}\left (x \right )\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 25 a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -1} \left (n +r -1\right )+15 a_{n} \left (n +r \right )+a_{n -1}+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (n +r \right )}{25 n^{2}+50 n r +25 r^{2}-10 n -10 r +1}\tag {4} \] Which for the root \(r = {\frac {1}{5}}\) becomes \[ a_{n} = -\frac {a_{n -1} \left (5 n +1\right )}{125 n^{2}}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{5}}\) and after as more terms are found using the above recursive equation.
\(n\) | \(a_{n ,r}\) | \(a_{n}\) |
\(a_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ a_{1}=\frac {-1-r}{\left (5 r +4\right )^{2}} \] Which for the root \(r = {\frac {1}{5}}\) becomes \[ a_{1}=-{\frac {6}{125}} \] And the table now becomes
\(n\) | \(a_{n ,r}\) | \(a_{n}\) |
\(a_{0}\) | \(1\) | \(1\) |
\(a_{1}\) | \(\frac {-1-r}{\left (5 r +4\right )^{2}}\) | \(-{\frac {6}{125}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {\left (2+r \right ) \left (1+r \right )}{\left (5 r +4\right )^{2} \left (5 r +9\right )^{2}} \] Which for the root \(r = {\frac {1}{5}}\) becomes \[ a_{2}={\frac {33}{31250}} \] And the table now becomes
\(n\) | \(a_{n ,r}\) | \(a_{n}\) |
\(a_{0}\) | \(1\) | \(1\) |
\(a_{1}\) | \(\frac {-1-r}{\left (5 r +4\right )^{2}}\) | \(-{\frac {6}{125}}\) |
\(a_{2}\) | \(\frac {\left (2+r \right ) \left (1+r \right )}{\left (5 r +4\right )^{2} \left (5 r +9\right )^{2}}\) | \(\frac {33}{31250}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {\left (3+r \right ) \left (2+r \right ) \left (1+r \right )}{\left (5 r +4\right )^{2} \left (5 r +9\right )^{2} \left (5 r +14\right )^{2}} \] Which for the root \(r = {\frac {1}{5}}\) becomes \[ a_{3}=-{\frac {88}{5859375}} \] And the table now becomes
\(n\) | \(a_{n ,r}\) | \(a_{n}\) |
\(a_{0}\) | \(1\) | \(1\) |
\(a_{1}\) | \(\frac {-1-r}{\left (5 r +4\right )^{2}}\) | \(-{\frac {6}{125}}\) |
\(a_{2}\) | \(\frac {\left (2+r \right ) \left (1+r \right )}{\left (5 r +4\right )^{2} \left (5 r +9\right )^{2}}\) | \(\frac {33}{31250}\) |
\(a_{3}\) | \(-\frac {\left (3+r \right ) \left (2+r \right ) \left (1+r \right )}{\left (5 r +4\right )^{2} \left (5 r +9\right )^{2} \left (5 r +14\right )^{2}}\) | \(-{\frac {88}{5859375}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {\left (3+r \right ) \left (2+r \right ) \left (1+r \right ) \left (4+r \right )}{\left (5 r +4\right )^{2} \left (5 r +9\right )^{2} \left (5 r +14\right )^{2} \left (5 r +19\right )^{2}} \] Which for the root \(r = {\frac {1}{5}}\) becomes \[ a_{4}={\frac {77}{488281250}} \] And the table now becomes
\(n\) | \(a_{n ,r}\) | \(a_{n}\) |
\(a_{0}\) | \(1\) | \(1\) |
\(a_{1}\) | \(\frac {-1-r}{\left (5 r +4\right )^{2}}\) | \(-{\frac {6}{125}}\) |
\(a_{2}\) | \(\frac {\left (2+r \right ) \left (1+r \right )}{\left (5 r +4\right )^{2} \left (5 r +9\right )^{2}}\) | \(\frac {33}{31250}\) |
\(a_{3}\) | \(-\frac {\left (3+r \right ) \left (2+r \right ) \left (1+r \right )}{\left (5 r +4\right )^{2} \left (5 r +9\right )^{2} \left (5 r +14\right )^{2}}\) | \(-{\frac {88}{5859375}}\) |
\(a_{4}\) | \(\frac {\left (3+r \right ) \left (2+r \right ) \left (1+r \right ) \left (4+r \right )}{\left (5 r +4\right )^{2} \left (5 r +9\right )^{2} \left (5 r +14\right )^{2} \left (5 r +19\right )^{2}}\) | \(\frac {77}{488281250}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {\left (3+r \right ) \left (2+r \right ) \left (1+r \right ) \left (4+r \right ) \left (5+r \right )}{\left (5 r +4\right )^{2} \left (5 r +9\right )^{2} \left (5 r +14\right )^{2} \left (5 r +19\right )^{2} \left (5 r +24\right )^{2}} \] Which for the root \(r = {\frac {1}{5}}\) becomes \[ a_{5}=-{\frac {1001}{762939453125}} \] And the table now becomes
\(n\) | \(a_{n ,r}\) | \(a_{n}\) |
\(a_{0}\) | \(1\) | \(1\) |
\(a_{1}\) | \(\frac {-1-r}{\left (5 r +4\right )^{2}}\) | \(-{\frac {6}{125}}\) |
\(a_{2}\) | \(\frac {\left (2+r \right ) \left (1+r \right )}{\left (5 r +4\right )^{2} \left (5 r +9\right )^{2}}\) | \(\frac {33}{31250}\) |
\(a_{3}\) | \(-\frac {\left (3+r \right ) \left (2+r \right ) \left (1+r \right )}{\left (5 r +4\right )^{2} \left (5 r +9\right )^{2} \left (5 r +14\right )^{2}}\) | \(-{\frac {88}{5859375}}\) |
\(a_{4}\) | \(\frac {\left (3+r \right ) \left (2+r \right ) \left (1+r \right ) \left (4+r \right )}{\left (5 r +4\right )^{2} \left (5 r +9\right )^{2} \left (5 r +14\right )^{2} \left (5 r +19\right )^{2}}\) | \(\frac {77}{488281250}\) |
\(a_{5}\) | \(-\frac {\left (3+r \right ) \left (2+r \right ) \left (1+r \right ) \left (4+r \right ) \left (5+r \right )}{\left (5 r +4\right )^{2} \left (5 r +9\right )^{2} \left (5 r +14\right )^{2} \left (5 r +19\right )^{2} \left (5 r +24\right )^{2}}\) | \(-{\frac {1001}{762939453125}}\) |
Using the above table, then the first solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= x^{\frac {1}{5}} \left (a_{0}+a_{1} x +a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}+a_{6} x^{6}\dots \right ) \\ &= x^{\frac {1}{5}} \left (1-\frac {6 x}{125}+\frac {33 x^{2}}{31250}-\frac {88 x^{3}}{5859375}+\frac {77 x^{4}}{488281250}-\frac {1001 x^{5}}{762939453125}+O\left (x^{6}\right )\right ) \\ \end{align*} Now the second solution is found. The second solution is given by \[ y_{2}\left (x \right ) = y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right ) \] Where \(b_{n}\) is found using \[ b_{n} = \frac {d}{d r}a_{n ,r} \] And the above is then evaluated at \(r = {\frac {1}{5}}\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
\(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =\frac {1}{5}\right )\) |
\(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
\(b_{1}\) | \(\frac {-1-r}{\left (5 r +4\right )^{2}}\) | \(-{\frac {6}{125}}\) | \(\frac {5 r +6}{\left (5 r +4\right )^{3}}\) | \(\frac {7}{125}\) |
\(b_{2}\) | \(\frac {\left (2+r \right ) \left (1+r \right )}{\left (5 r +4\right )^{2} \left (5 r +9\right )^{2}}\) | \(\frac {33}{31250}\) | \(\frac {-50 r^{3}-225 r^{2}-323 r -152}{\left (5 r +4\right )^{3} \left (5 r +9\right )^{3}}\) | \(-{\frac {113}{62500}}\) |
\(b_{3}\) | \(-\frac {\left (3+r \right ) \left (2+r \right ) \left (1+r \right )}{\left (5 r +4\right )^{2} \left (5 r +9\right )^{2} \left (5 r +14\right )^{2}}\) | \(-{\frac {88}{5859375}}\) | \(\frac {375 r^{5}+3675 r^{4}+13885 r^{3}+25263 r^{2}+22142 r +7536}{\left (5 r +4\right )^{3} \left (5 r +9\right )^{3} \left (5 r +14\right )^{3}}\) | \(\frac {1091}{35156250}\) |
\(b_{4}\) | \(\frac {\left (3+r \right ) \left (2+r \right ) \left (1+r \right ) \left (4+r \right )}{\left (5 r +4\right )^{2} \left (5 r +9\right )^{2} \left (5 r +14\right )^{2} \left (5 r +19\right )^{2}}\) | \(\frac {77}{488281250}\) | \(\frac {-2500 r^{7}-42750 r^{6}-303750 r^{5}-1160040 r^{4}-2566146 r^{3}-3281970 r^{2}-2245580 r -636240}{\left (5 r +14\right )^{3} \left (5 r +4\right )^{3} \left (5 r +19\right )^{3} \left (5 r +9\right )^{3}}\) | \(-{\frac {1721}{4687500000}}\) |
\(b_{5}\) | \(-\frac {\left (3+r \right ) \left (2+r \right ) \left (1+r \right ) \left (4+r \right ) \left (5+r \right )}{\left (5 r +4\right )^{2} \left (5 r +9\right )^{2} \left (5 r +14\right )^{2} \left (5 r +19\right )^{2} \left (5 r +24\right )^{2}}\) | \(-{\frac {1001}{762939453125}}\) | \(\frac {15625 r^{9}+412500 r^{8}+4713750 r^{7}+30545250 r^{6}+123443175 r^{5}+321942630 r^{4}+540632810 r^{3}+562549380 r^{2}+328738800 r +82324224}{\left (5 r +4\right )^{3} \left (5 r +9\right )^{3} \left (5 r +14\right )^{3} \left (5 r +19\right )^{3} \left (5 r +24\right )^{3}}\) | \(\frac {609221}{183105468750000}\) |
The above table gives all values of \(b_{n}\) needed. Hence the second solution is \begin{align*} y_{2}\left (x \right )&=y_{1}\left (x \right ) \ln \left (x \right )+b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}\dots \\ &= x^{\frac {1}{5}} \left (1-\frac {6 x}{125}+\frac {33 x^{2}}{31250}-\frac {88 x^{3}}{5859375}+\frac {77 x^{4}}{488281250}-\frac {1001 x^{5}}{762939453125}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{\frac {1}{5}} \left (\frac {7 x}{125}-\frac {113 x^{2}}{62500}+\frac {1091 x^{3}}{35156250}-\frac {1721 x^{4}}{4687500000}+\frac {609221 x^{5}}{183105468750000}+O\left (x^{6}\right )\right ) \\ \end{align*} Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} x^{\frac {1}{5}} \left (1-\frac {6 x}{125}+\frac {33 x^{2}}{31250}-\frac {88 x^{3}}{5859375}+\frac {77 x^{4}}{488281250}-\frac {1001 x^{5}}{762939453125}+O\left (x^{6}\right )\right ) + c_{2} \left (x^{\frac {1}{5}} \left (1-\frac {6 x}{125}+\frac {33 x^{2}}{31250}-\frac {88 x^{3}}{5859375}+\frac {77 x^{4}}{488281250}-\frac {1001 x^{5}}{762939453125}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{\frac {1}{5}} \left (\frac {7 x}{125}-\frac {113 x^{2}}{62500}+\frac {1091 x^{3}}{35156250}-\frac {1721 x^{4}}{4687500000}+\frac {609221 x^{5}}{183105468750000}+O\left (x^{6}\right )\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {1}{5}} \left (1-\frac {6 x}{125}+\frac {33 x^{2}}{31250}-\frac {88 x^{3}}{5859375}+\frac {77 x^{4}}{488281250}-\frac {1001 x^{5}}{762939453125}+O\left (x^{6}\right )\right )+c_{2} \left (x^{\frac {1}{5}} \left (1-\frac {6 x}{125}+\frac {33 x^{2}}{31250}-\frac {88 x^{3}}{5859375}+\frac {77 x^{4}}{488281250}-\frac {1001 x^{5}}{762939453125}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{\frac {1}{5}} \left (\frac {7 x}{125}-\frac {113 x^{2}}{62500}+\frac {1091 x^{3}}{35156250}-\frac {1721 x^{4}}{4687500000}+\frac {609221 x^{5}}{183105468750000}+O\left (x^{6}\right )\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {1}{5}} \left (1-\frac {6 x}{125}+\frac {33 x^{2}}{31250}-\frac {88 x^{3}}{5859375}+\frac {77 x^{4}}{488281250}-\frac {1001 x^{5}}{762939453125}+O\left (x^{6}\right )\right )+c_{2} \left (x^{\frac {1}{5}} \left (1-\frac {6 x}{125}+\frac {33 x^{2}}{31250}-\frac {88 x^{3}}{5859375}+\frac {77 x^{4}}{488281250}-\frac {1001 x^{5}}{762939453125}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{\frac {1}{5}} \left (\frac {7 x}{125}-\frac {113 x^{2}}{62500}+\frac {1091 x^{3}}{35156250}-\frac {1721 x^{4}}{4687500000}+\frac {609221 x^{5}}{183105468750000}+O\left (x^{6}\right )\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {1}{5}} \left (1-\frac {6 x}{125}+\frac {33 x^{2}}{31250}-\frac {88 x^{3}}{5859375}+\frac {77 x^{4}}{488281250}-\frac {1001 x^{5}}{762939453125}+O\left (x^{6}\right )\right )+c_{2} \left (x^{\frac {1}{5}} \left (1-\frac {6 x}{125}+\frac {33 x^{2}}{31250}-\frac {88 x^{3}}{5859375}+\frac {77 x^{4}}{488281250}-\frac {1001 x^{5}}{762939453125}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{\frac {1}{5}} \left (\frac {7 x}{125}-\frac {113 x^{2}}{62500}+\frac {1091 x^{3}}{35156250}-\frac {1721 x^{4}}{4687500000}+\frac {609221 x^{5}}{183105468750000}+O\left (x^{6}\right )\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 25 x^{2} y^{\prime \prime }+\left (x^{2}+15 x \right ) y^{\prime }+\left (x +1\right ) 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}{25 x^{2}}-\frac {\left (15+x \right ) y^{\prime }}{25 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 (15+x \right ) y^{\prime }}{25 x}+\frac {\left (x +1\right ) y}{25 x^{2}}=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 {15+x}{25 x}, P_{3}\left (x \right )=\frac {x +1}{25 x^{2}}\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}}}=\frac {3}{5} \\ {} & \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}}}=\frac {1}{25} \\ {} & \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} \\ {} & {} & 25 x^{2} y^{\prime \prime }+x \left (15+x \right ) y^{\prime }+\left (x +1\right ) 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\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y=\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=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & 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^{2}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (-1+5 r \right )^{2} x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (5 k +5 r -1\right )^{2}+a_{k -1} \left (k +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+5 r \right )^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =\frac {1}{5} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (5 k +5 r -1\right )^{2}+a_{k -1} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +1} \left (5 k +4+5 r \right )^{2}+a_{k} \left (k +r +1\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 )}{\left (5 k +4+5 r \right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{5} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +\frac {6}{5}\right )}{\left (5 k +5\right )^{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{5} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{5}}, a_{k +1}=-\frac {a_{k} \left (k +\frac {6}{5}\right )}{\left (5 k +5\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 -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Whittaker successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 69
Order:=6; dsolve(25*x^2*diff(y(x),x$2)+x*(15+x)*diff(y(x),x)+(1+x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = x^{\frac {1}{5}} \left (\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1-\frac {6}{125} x +\frac {33}{31250} x^{2}-\frac {88}{5859375} x^{3}+\frac {77}{488281250} x^{4}-\frac {1001}{762939453125} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (\frac {7}{125} x -\frac {113}{62500} x^{2}+\frac {1091}{35156250} x^{3}-\frac {1721}{4687500000} x^{4}+\frac {609221}{183105468750000} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 134
AsymptoticDSolveValue[25*x^2*y''[x]+x*(15+x)*y'[x]+(1+x)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \sqrt [5]{x} \left (-\frac {1001 x^5}{762939453125}+\frac {77 x^4}{488281250}-\frac {88 x^3}{5859375}+\frac {33 x^2}{31250}-\frac {6 x}{125}+1\right )+c_2 \left (\sqrt [5]{x} \left (\frac {609221 x^5}{183105468750000}-\frac {1721 x^4}{4687500000}+\frac {1091 x^3}{35156250}-\frac {113 x^2}{62500}+\frac {7 x}{125}\right )+\sqrt [5]{x} \left (-\frac {1001 x^5}{762939453125}+\frac {77 x^4}{488281250}-\frac {88 x^3}{5859375}+\frac {33 x^2}{31250}-\frac {6 x}{125}+1\right ) \log (x)\right ) \]