Internal problem ID [1365]
Internal file name [OUTPUT/1366_Sunday_June_05_2022_02_13_23_AM_50001526/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: 13.
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]]
\[ \boxed {36 x^{2} \left (1-2 x \right ) y^{\prime \prime }+24 x \left (1-9 x \right ) y^{\prime }+\left (1-70 x \right ) y=0} \] 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. \[ \left (-72 x^{3}+36 x^{2}\right ) y^{\prime \prime }+\left (-216 x^{2}+24 x \right ) y^{\prime }+\left (1-70 x \right ) y = 0 \] 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 {6 x -\frac {2}{3}}{x \left (-1+2 x \right )}\\ q(x) &= \frac {70 x -1}{36 x^{2} \left (-1+2 x \right )}\\ \end {align*}
| \(p(x)=\frac {6 x -\frac {2}{3}}{x \left (-1+2 x \right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = {\frac {1}{2}}\) | \(\text {``regular''}\) |
| \(q(x)=\frac {70 x -1}{36 x^{2} \left (-1+2 x \right )}\) | |
| 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 \[ -36 y^{\prime \prime } x^{2} \left (-1+2 x \right )+\left (-216 x^{2}+24 x \right ) y^{\prime }+\left (1-70 x \right ) y = 0 \] 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} -36 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) x^{2} \left (-1+2 x \right )+\left (-216 x^{2}+24 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (1-70 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 (-72 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}36 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-216 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}24 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-70 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 (-72 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-72 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-216 x^{1+n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-216 a_{n -1} \left (n +r -1\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-70 x^{1+n +r} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-70 a_{n -1} x^{n +r}\right ) \\ \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 (-72 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}36 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-216 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}24 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-70 a_{n -1} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 36 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+24 x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 36 x^{r} a_{0} r \left (-1+r \right )+24 x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (36 x^{r} r \left (-1+r \right )+24 x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (6 r -1\right )^{2} x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ \left (6 r -1\right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{6}}\\ r_2 &= {\frac {1}{6}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (6 r -1\right )^{2} x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{6}}, {\frac {1}{6}}\right ]\).
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, \(r = {\frac {1}{6}}\), Eqs (1A,1B) become \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {1}{6}}\\ 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}{6}}\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} -72 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+36 a_{n} \left (n +r \right ) \left (n +r -1\right )-216 a_{n -1} \left (n +r -1\right )+24 a_{n} \left (n +r \right )+a_{n}-70 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {2 \left (6 n +6 r +1\right ) a_{n -1}}{-1+6 n +6 r}\tag {4} \] Which for the root \(r = {\frac {1}{6}}\) becomes \[ a_{n} = \frac {2 \left (3 n +1\right ) a_{n -1}}{3 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{6}}\) 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 \[ a_{1}=\frac {12 r +14}{5+6 r} \] Which for the root \(r = {\frac {1}{6}}\) becomes \[ a_{1}={\frac {8}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {12 r +14}{5+6 r}\) | \(\frac {8}{3}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {144 r^{2}+480 r +364}{36 r^{2}+96 r +55} \] Which for the root \(r = {\frac {1}{6}}\) becomes \[ a_{2}={\frac {56}{9}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {12 r +14}{5+6 r}\) | \(\frac {8}{3}\) |
| \(a_{2}\) | \(\frac {144 r^{2}+480 r +364}{36 r^{2}+96 r +55}\) | \(\frac {56}{9}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {1728 r^{3}+11232 r^{2}+22608 r +13832}{216 r^{3}+1188 r^{2}+1962 r +935} \] Which for the root \(r = {\frac {1}{6}}\) becomes \[ a_{3}={\frac {1120}{81}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {12 r +14}{5+6 r}\) | \(\frac {8}{3}\) |
| \(a_{2}\) | \(\frac {144 r^{2}+480 r +364}{36 r^{2}+96 r +55}\) | \(\frac {56}{9}\) |
| \(a_{3}\) | \(\frac {1728 r^{3}+11232 r^{2}+22608 r +13832}{216 r^{3}+1188 r^{2}+1962 r +935}\) | \(\frac {1120}{81}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {20736 r^{4}+221184 r^{3}+832896 r^{2}+1296384 r +691600}{1296 r^{4}+12096 r^{3}+39096 r^{2}+50736 r +21505} \] Which for the root \(r = {\frac {1}{6}}\) becomes \[ a_{4}={\frac {7280}{243}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {12 r +14}{5+6 r}\) | \(\frac {8}{3}\) |
| \(a_{2}\) | \(\frac {144 r^{2}+480 r +364}{36 r^{2}+96 r +55}\) | \(\frac {56}{9}\) |
| \(a_{3}\) | \(\frac {1728 r^{3}+11232 r^{2}+22608 r +13832}{216 r^{3}+1188 r^{2}+1962 r +935}\) | \(\frac {1120}{81}\) |
| \(a_{4}\) | \(\frac {20736 r^{4}+221184 r^{3}+832896 r^{2}+1296384 r +691600}{1296 r^{4}+12096 r^{3}+39096 r^{2}+50736 r +21505}\) | \(\frac {7280}{243}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {248832 r^{5}+3939840 r^{4}+23708160 r^{3}+67196160 r^{2}+88675008 r +42879200}{7776 r^{5}+110160 r^{4}+585360 r^{3}+1438200 r^{2}+1600374 r +623645} \] Which for the root \(r = {\frac {1}{6}}\) becomes \[ a_{5}={\frac {46592}{729}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {12 r +14}{5+6 r}\) | \(\frac {8}{3}\) |
| \(a_{2}\) | \(\frac {144 r^{2}+480 r +364}{36 r^{2}+96 r +55}\) | \(\frac {56}{9}\) |
| \(a_{3}\) | \(\frac {1728 r^{3}+11232 r^{2}+22608 r +13832}{216 r^{3}+1188 r^{2}+1962 r +935}\) | \(\frac {1120}{81}\) |
| \(a_{4}\) | \(\frac {20736 r^{4}+221184 r^{3}+832896 r^{2}+1296384 r +691600}{1296 r^{4}+12096 r^{3}+39096 r^{2}+50736 r +21505}\) | \(\frac {7280}{243}\) |
| \(a_{5}\) | \(\frac {248832 r^{5}+3939840 r^{4}+23708160 r^{3}+67196160 r^{2}+88675008 r +42879200}{7776 r^{5}+110160 r^{4}+585360 r^{3}+1438200 r^{2}+1600374 r +623645}\) | \(\frac {46592}{729}\) |
Using the above table, then the first solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= x^{\frac {1}{6}} \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}{6}} \left (1+\frac {8 x}{3}+\frac {56 x^{2}}{9}+\frac {1120 x^{3}}{81}+\frac {7280 x^{4}}{243}+\frac {46592 x^{5}}{729}+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 \(r = {\frac {1}{6}}\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =\frac {1}{6}\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(\frac {12 r +14}{5+6 r}\) | \(\frac {8}{3}\) | \(-\frac {24}{\left (5+6 r \right )^{2}}\) | \(-{\frac {2}{3}}\) |
| \(b_{2}\) | \(\frac {144 r^{2}+480 r +364}{36 r^{2}+96 r +55}\) | \(\frac {56}{9}\) | \(-\frac {96 \left (36 r^{2}+108 r +89\right )}{\left (36 r^{2}+96 r +55\right )^{2}}\) | \(-2\) |
| \(b_{3}\) | \(\frac {1728 r^{3}+11232 r^{2}+22608 r +13832}{216 r^{3}+1188 r^{2}+1962 r +935}\) | \(\frac {1120}{81}\) | \(-\frac {288 \left (1296 r^{4}+10368 r^{3}+31032 r^{2}+41184 r +20833\right )}{\left (216 r^{3}+1188 r^{2}+1962 r +935\right )^{2}}\) | \(-{\frac {1192}{243}}\) |
| \(b_{4}\) | \(\frac {20736 r^{4}+221184 r^{3}+832896 r^{2}+1296384 r +691600}{1296 r^{4}+12096 r^{3}+39096 r^{2}+50736 r +21505}\) | \(\frac {7280}{243}\) | \(-\frac {768 \left (46656 r^{6}+699840 r^{5}+4311792 r^{4}+13957920 r^{3}+25068636 r^{2}+23769180 r +9388385\right )}{\left (1296 r^{4}+12096 r^{3}+39096 r^{2}+50736 r +21505\right )^{2}}\) | \(-{\frac {8168}{729}}\) |
| \(b_{5}\) | \(\frac {248832 r^{5}+3939840 r^{4}+23708160 r^{3}+67196160 r^{2}+88675008 r +42879200}{7776 r^{5}+110160 r^{4}+585360 r^{3}+1438200 r^{2}+1600374 r +623645}\) | \(\frac {46592}{729}\) | \(-\frac {1920 \left (1679616 r^{8}+40310784 r^{7}+416358144 r^{6}+2415287808 r^{5}+8603000928 r^{4}+19268648064 r^{3}+26529316848 r^{2}+20585746080 r +6938037217\right )}{\left (7776 r^{5}+110160 r^{4}+585360 r^{3}+1438200 r^{2}+1600374 r +623645\right )^{2}}\) | \(-{\frac {270112}{10935}}\) |
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}{6}} \left (1+\frac {8 x}{3}+\frac {56 x^{2}}{9}+\frac {1120 x^{3}}{81}+\frac {7280 x^{4}}{243}+\frac {46592 x^{5}}{729}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{\frac {1}{6}} \left (-\frac {2 x}{3}-2 x^{2}-\frac {1192 x^{3}}{243}-\frac {8168 x^{4}}{729}-\frac {270112 x^{5}}{10935}+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}{6}} \left (1+\frac {8 x}{3}+\frac {56 x^{2}}{9}+\frac {1120 x^{3}}{81}+\frac {7280 x^{4}}{243}+\frac {46592 x^{5}}{729}+O\left (x^{6}\right )\right ) + c_{2} \left (x^{\frac {1}{6}} \left (1+\frac {8 x}{3}+\frac {56 x^{2}}{9}+\frac {1120 x^{3}}{81}+\frac {7280 x^{4}}{243}+\frac {46592 x^{5}}{729}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{\frac {1}{6}} \left (-\frac {2 x}{3}-2 x^{2}-\frac {1192 x^{3}}{243}-\frac {8168 x^{4}}{729}-\frac {270112 x^{5}}{10935}+O\left (x^{6}\right )\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {1}{6}} \left (1+\frac {8 x}{3}+\frac {56 x^{2}}{9}+\frac {1120 x^{3}}{81}+\frac {7280 x^{4}}{243}+\frac {46592 x^{5}}{729}+O\left (x^{6}\right )\right )+c_{2} \left (x^{\frac {1}{6}} \left (1+\frac {8 x}{3}+\frac {56 x^{2}}{9}+\frac {1120 x^{3}}{81}+\frac {7280 x^{4}}{243}+\frac {46592 x^{5}}{729}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{\frac {1}{6}} \left (-\frac {2 x}{3}-2 x^{2}-\frac {1192 x^{3}}{243}-\frac {8168 x^{4}}{729}-\frac {270112 x^{5}}{10935}+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}{6}} \left (1+\frac {8 x}{3}+\frac {56 x^{2}}{9}+\frac {1120 x^{3}}{81}+\frac {7280 x^{4}}{243}+\frac {46592 x^{5}}{729}+O\left (x^{6}\right )\right )+c_{2} \left (x^{\frac {1}{6}} \left (1+\frac {8 x}{3}+\frac {56 x^{2}}{9}+\frac {1120 x^{3}}{81}+\frac {7280 x^{4}}{243}+\frac {46592 x^{5}}{729}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{\frac {1}{6}} \left (-\frac {2 x}{3}-2 x^{2}-\frac {1192 x^{3}}{243}-\frac {8168 x^{4}}{729}-\frac {270112 x^{5}}{10935}+O\left (x^{6}\right )\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {1}{6}} \left (1+\frac {8 x}{3}+\frac {56 x^{2}}{9}+\frac {1120 x^{3}}{81}+\frac {7280 x^{4}}{243}+\frac {46592 x^{5}}{729}+O\left (x^{6}\right )\right )+c_{2} \left (x^{\frac {1}{6}} \left (1+\frac {8 x}{3}+\frac {56 x^{2}}{9}+\frac {1120 x^{3}}{81}+\frac {7280 x^{4}}{243}+\frac {46592 x^{5}}{729}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{\frac {1}{6}} \left (-\frac {2 x}{3}-2 x^{2}-\frac {1192 x^{3}}{243}-\frac {8168 x^{4}}{729}-\frac {270112 x^{5}}{10935}+O\left (x^{6}\right )\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -36 \left (\frac {d}{d x}y^{\prime }\right ) x^{2} \left (-1+2 x \right )+\left (-216 x^{2}+24 x \right ) y^{\prime }+\left (1-70 x \right ) 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 {\left (70 x -1\right ) y}{36 x^{2} \left (-1+2 x \right )}-\frac {2 \left (9 x -1\right ) y^{\prime }}{3 x \left (-1+2 x \right )} \\ \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 {2 \left (9 x -1\right ) y^{\prime }}{3 x \left (-1+2 x \right )}+\frac {\left (70 x -1\right ) y}{36 x^{2} \left (-1+2 x \right )}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {2 \left (9 x -1\right )}{3 x \left (-1+2 x \right )}, P_{3}\left (x \right )=\frac {70 x -1}{36 x^{2} \left (-1+2 x \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}}}=\frac {2}{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}}}=\frac {1}{36} \\ {} & \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} \\ {} & {} & 36 \left (\frac {d}{d x}y^{\prime }\right ) x^{2} \left (-1+2 x \right )+24 x \left (9 x -1\right ) y^{\prime }+y \left (70 x -1\right )=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^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..3 \\ {} & {} & 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} \left (-1+6 r \right )^{2} x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (-a_{k} \left (6 k +6 r -1\right )^{2}+2 a_{k -1} \left (6 k +1+6 r \right ) \left (6 k +6 r -1\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -\left (-1+6 r \right )^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =\frac {1}{6} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -36 \left (k +r -\frac {1}{6}\right ) \left (\left (-2 k -2 r -\frac {1}{3}\right ) a_{k -1}+a_{k} \left (k +r -\frac {1}{6}\right )\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & -36 \left (k +\frac {5}{6}+r \right ) \left (\left (-2 k -\frac {7}{3}-2 r \right ) a_{k}+a_{k +1} \left (k +\frac {5}{6}+r \right )\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {2 \left (6 k +6 r +7\right ) a_{k}}{6 k +6 r +5} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{6} \\ {} & {} & a_{k +1}=\frac {2 \left (6 k +8\right ) a_{k}}{6 k +6} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{6} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{6}}, a_{k +1}=\frac {2 \left (6 k +8\right ) a_{k}}{6 k +6}\right ] \end {array} \]
Maple trace Kovacic algorithm successful
`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 Solution has integrals. Trying a special function solution free of integrals... -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach <- heuristic approach successful -> solution has integrals; searching for one without integrals... -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 2F1 ODE <- hypergeometric solution without integrals succesful <- hypergeometric successful <- special function solution successful -> Trying to convert hypergeometric functions to elementary form... <- elementary form for at least one hypergeometric solution is achieved - returning with no uncomputed integrals <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 69
Order:=6; dsolve(36*x^2*(1-2*x)*diff(y(x),x$2)+24*x*(1-9*x)*diff(y(x),x)+(1-70*x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = x^{\frac {1}{6}} \left (\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1+\frac {8}{3} x +\frac {56}{9} x^{2}+\frac {1120}{81} x^{3}+\frac {7280}{243} x^{4}+\frac {46592}{729} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (-\frac {2}{3} x -2 x^{2}-\frac {1192}{243} x^{3}-\frac {8168}{729} x^{4}-\frac {270112}{10935} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 0.008 (sec). Leaf size: 132
AsymptoticDSolveValue[36*x^2*(1-2*x)*y''[x]+24*x*(1-9*x)*y'[x]+(1-70*x)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \sqrt [6]{x} \left (\frac {46592 x^5}{729}+\frac {7280 x^4}{243}+\frac {1120 x^3}{81}+\frac {56 x^2}{9}+\frac {8 x}{3}+1\right )+c_2 \left (\sqrt [6]{x} \left (-\frac {270112 x^5}{10935}-\frac {8168 x^4}{729}-\frac {1192 x^3}{243}-2 x^2-\frac {2 x}{3}\right )+\sqrt [6]{x} \left (\frac {46592 x^5}{729}+\frac {7280 x^4}{243}+\frac {1120 x^3}{81}+\frac {56 x^2}{9}+\frac {8 x}{3}+1\right ) \log (x)\right ) \]