Internal problem ID [1350]
Internal file name [OUTPUT/1351_Sunday_June_05_2022_02_12_22_AM_98132168/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: Example 7.6.2 page 369.
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]]
\[ \boxed {2 x^{2} \left (x +2\right ) y^{\prime \prime }+5 x^{2} y^{\prime }+\left (x +1\right ) y=0} \] 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. \[ \left (2 x^{3}+4 x^{2}\right ) y^{\prime \prime }+5 x^{2} y^{\prime }+\left (x +1\right ) y = 0 \] 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 {5}{2 \left (x +2\right )}\\ q(x) &= \frac {x +1}{2 x^{2} \left (x +2\right )}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-2, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 2 x^{2} \left (x +2\right ) y^{\prime \prime }+5 x^{2} y^{\prime }+\left (x +1\right ) y = 0 \] 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} 2 x^{2} \left (x +2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+5 x^{2} \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 }}2 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 x^{1+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 }}2 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}5 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}5 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 =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}5 a_{n -1} \left (n +r -1\right ) x^{n +r}\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 \[ 4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 4 x^{r} a_{0} r \left (-1+r \right )+a_{0} x^{r} = 0 \] Or \[ \left (4 x^{r} r \left (-1+r \right )+x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ x^{r} \left (2 r -1\right )^{2} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ \left (2 r -1\right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{2}}\\ r_2 &= {\frac {1}{2}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ x^{r} \left (2 r -1\right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{2}}, {\frac {1}{2}}\right ]\).
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, \(r = {\frac {1}{2}}\), Eqs (1A,1B) become \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {1}{2}}\\ 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}{2}}\right ) \end {align*}
We start by finding the first solution \(y_{1}\left (x \right )\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+4 a_{n} \left (n +r \right ) \left (n +r -1\right )+5 a_{n -1} \left (n +r -1\right )+a_{n -1}+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {\left (n +r \right ) a_{n -1}}{-1+2 n +2 r}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{n} = -\frac {\left (2 n +1\right ) a_{n -1}}{4 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{2}}\) 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 \[ a_{1}=\frac {-1-r}{1+2 r} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{1}=-{\frac {3}{4}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-r}{1+2 r}\) | \(-{\frac {3}{4}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {r^{2}+3 r +2}{4 r^{2}+8 r +3} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{2}={\frac {15}{32}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-r}{1+2 r}\) | \(-{\frac {3}{4}}\) |
| \(a_{2}\) | \(\frac {r^{2}+3 r +2}{4 r^{2}+8 r +3}\) | \(\frac {15}{32}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-r^{3}-6 r^{2}-11 r -6}{8 r^{3}+36 r^{2}+46 r +15} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{3}=-{\frac {35}{128}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-r}{1+2 r}\) | \(-{\frac {3}{4}}\) |
| \(a_{2}\) | \(\frac {r^{2}+3 r +2}{4 r^{2}+8 r +3}\) | \(\frac {15}{32}\) |
| \(a_{3}\) | \(\frac {-r^{3}-6 r^{2}-11 r -6}{8 r^{3}+36 r^{2}+46 r +15}\) | \(-{\frac {35}{128}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r^{4}+10 r^{3}+35 r^{2}+50 r +24}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{4}={\frac {315}{2048}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-r}{1+2 r}\) | \(-{\frac {3}{4}}\) |
| \(a_{2}\) | \(\frac {r^{2}+3 r +2}{4 r^{2}+8 r +3}\) | \(\frac {15}{32}\) |
| \(a_{3}\) | \(\frac {-r^{3}-6 r^{2}-11 r -6}{8 r^{3}+36 r^{2}+46 r +15}\) | \(-{\frac {35}{128}}\) |
| \(a_{4}\) | \(\frac {r^{4}+10 r^{3}+35 r^{2}+50 r +24}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) | \(\frac {315}{2048}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-r^{5}-15 r^{4}-85 r^{3}-225 r^{2}-274 r -120}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{5}=-{\frac {693}{8192}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-r}{1+2 r}\) | \(-{\frac {3}{4}}\) |
| \(a_{2}\) | \(\frac {r^{2}+3 r +2}{4 r^{2}+8 r +3}\) | \(\frac {15}{32}\) |
| \(a_{3}\) | \(\frac {-r^{3}-6 r^{2}-11 r -6}{8 r^{3}+36 r^{2}+46 r +15}\) | \(-{\frac {35}{128}}\) |
| \(a_{4}\) | \(\frac {r^{4}+10 r^{3}+35 r^{2}+50 r +24}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) | \(\frac {315}{2048}\) |
| \(a_{5}\) | \(\frac {-r^{5}-15 r^{4}-85 r^{3}-225 r^{2}-274 r -120}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945}\) | \(-{\frac {693}{8192}}\) |
Using the above table, then the first solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= \sqrt {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 ) \\ &= \sqrt {x}\, \left (1-\frac {3 x}{4}+\frac {15 x^{2}}{32}-\frac {35 x^{3}}{128}+\frac {315 x^{4}}{2048}-\frac {693 x^{5}}{8192}+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 \(r = {\frac {1}{2}}\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =\frac {1}{2}\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(\frac {-1-r}{1+2 r}\) | \(-{\frac {3}{4}}\) | \(\frac {1}{\left (1+2 r \right )^{2}}\) | \(\frac {1}{4}\) |
| \(b_{2}\) | \(\frac {r^{2}+3 r +2}{4 r^{2}+8 r +3}\) | \(\frac {15}{32}\) | \(\frac {-4 r^{2}-10 r -7}{\left (4 r^{2}+8 r +3\right )^{2}}\) | \(-{\frac {13}{64}}\) |
| \(b_{3}\) | \(\frac {-r^{3}-6 r^{2}-11 r -6}{8 r^{3}+36 r^{2}+46 r +15}\) | \(-{\frac {35}{128}}\) | \(\frac {12 r^{4}+84 r^{3}+219 r^{2}+252 r +111}{\left (8 r^{3}+36 r^{2}+46 r +15\right )^{2}}\) | \(\frac {101}{768}\) |
| \(b_{4}\) | \(\frac {r^{4}+10 r^{3}+35 r^{2}+50 r +24}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) | \(\frac {315}{2048}\) | \(\frac {-32 r^{6}-432 r^{5}-2384 r^{4}-6876 r^{3}-10946 r^{2}-9162 r -3198}{\left (16 r^{4}+128 r^{3}+344 r^{2}+352 r +105\right )^{2}}\) | \(-{\frac {641}{8192}}\) |
| \(b_{5}\) | \(\frac {-r^{5}-15 r^{4}-85 r^{3}-225 r^{2}-274 r -120}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945}\) | \(-{\frac {693}{8192}}\) | \(\frac {80 r^{8}+1760 r^{7}+16600 r^{6}+87560 r^{5}+282265 r^{4}+569360 r^{3}+702575 r^{2}+486750 r +146430}{\left (32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945\right )^{2}}\) | \(\frac {7303}{163840}\) |
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 \\ &= \sqrt {x}\, \left (1-\frac {3 x}{4}+\frac {15 x^{2}}{32}-\frac {35 x^{3}}{128}+\frac {315 x^{4}}{2048}-\frac {693 x^{5}}{8192}+O\left (x^{6}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (\frac {x}{4}-\frac {13 x^{2}}{64}+\frac {101 x^{3}}{768}-\frac {641 x^{4}}{8192}+\frac {7303 x^{5}}{163840}+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} \sqrt {x}\, \left (1-\frac {3 x}{4}+\frac {15 x^{2}}{32}-\frac {35 x^{3}}{128}+\frac {315 x^{4}}{2048}-\frac {693 x^{5}}{8192}+O\left (x^{6}\right )\right ) + c_{2} \left (\sqrt {x}\, \left (1-\frac {3 x}{4}+\frac {15 x^{2}}{32}-\frac {35 x^{3}}{128}+\frac {315 x^{4}}{2048}-\frac {693 x^{5}}{8192}+O\left (x^{6}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (\frac {x}{4}-\frac {13 x^{2}}{64}+\frac {101 x^{3}}{768}-\frac {641 x^{4}}{8192}+\frac {7303 x^{5}}{163840}+O\left (x^{6}\right )\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \sqrt {x}\, \left (1-\frac {3 x}{4}+\frac {15 x^{2}}{32}-\frac {35 x^{3}}{128}+\frac {315 x^{4}}{2048}-\frac {693 x^{5}}{8192}+O\left (x^{6}\right )\right )+c_{2} \left (\sqrt {x}\, \left (1-\frac {3 x}{4}+\frac {15 x^{2}}{32}-\frac {35 x^{3}}{128}+\frac {315 x^{4}}{2048}-\frac {693 x^{5}}{8192}+O\left (x^{6}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (\frac {x}{4}-\frac {13 x^{2}}{64}+\frac {101 x^{3}}{768}-\frac {641 x^{4}}{8192}+\frac {7303 x^{5}}{163840}+O\left (x^{6}\right )\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \sqrt {x}\, \left (1-\frac {3 x}{4}+\frac {15 x^{2}}{32}-\frac {35 x^{3}}{128}+\frac {315 x^{4}}{2048}-\frac {693 x^{5}}{8192}+O\left (x^{6}\right )\right )+c_{2} \left (\sqrt {x}\, \left (1-\frac {3 x}{4}+\frac {15 x^{2}}{32}-\frac {35 x^{3}}{128}+\frac {315 x^{4}}{2048}-\frac {693 x^{5}}{8192}+O\left (x^{6}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (\frac {x}{4}-\frac {13 x^{2}}{64}+\frac {101 x^{3}}{768}-\frac {641 x^{4}}{8192}+\frac {7303 x^{5}}{163840}+O\left (x^{6}\right )\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \sqrt {x}\, \left (1-\frac {3 x}{4}+\frac {15 x^{2}}{32}-\frac {35 x^{3}}{128}+\frac {315 x^{4}}{2048}-\frac {693 x^{5}}{8192}+O\left (x^{6}\right )\right )+c_{2} \left (\sqrt {x}\, \left (1-\frac {3 x}{4}+\frac {15 x^{2}}{32}-\frac {35 x^{3}}{128}+\frac {315 x^{4}}{2048}-\frac {693 x^{5}}{8192}+O\left (x^{6}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (\frac {x}{4}-\frac {13 x^{2}}{64}+\frac {101 x^{3}}{768}-\frac {641 x^{4}}{8192}+\frac {7303 x^{5}}{163840}+O\left (x^{6}\right )\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x^{2} \left (x +2\right ) \left (\frac {d}{d x}y^{\prime }\right )+5 x^{2} y^{\prime }+\left (x +1\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 (x +1\right ) y}{2 x^{2} \left (x +2\right )}-\frac {5 y^{\prime }}{2 \left (x +2\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 {5 y^{\prime }}{2 \left (x +2\right )}+\frac {\left (x +1\right ) y}{2 x^{2} \left (x +2\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 {5}{2 \left (x +2\right )}, P_{3}\left (x \right )=\frac {x +1}{2 x^{2} \left (x +2\right )}\right ] \\ {} & \circ & \left (x +2\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-2 \\ {} & {} & \left (\left (x +2\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-2}}}=\frac {5}{2} \\ {} & \circ & \left (x +2\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-2 \\ {} & {} & \left (\left (x +2\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-2}}}=0 \\ {} & \circ & x =-2\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}=-2 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 2 x^{2} \left (x +2\right ) \left (\frac {d}{d x}y^{\prime }\right )+5 x^{2} y^{\prime }+\left (x +1\right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -2\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (2 u^{3}-8 u^{2}+8 u \right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (5 u^{2}-20 u +20\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (u -1\right ) y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 4 a_{0} r \left (3+2 r \right ) u^{-1+r}+\left (4 a_{1} \left (1+r \right ) \left (5+2 r \right )-a_{0} \left (8 r^{2}+12 r +1\right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (4 a_{k +1} \left (k +r +1\right ) \left (2 k +5+2 r \right )-a_{k} \left (8 k^{2}+16 k r +8 r^{2}+12 k +12 r +1\right )+a_{k -1} \left (k +r \right ) \left (2 k -1+2 r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 4 r \left (3+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -\frac {3}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 4 a_{1} \left (1+r \right ) \left (5+2 r \right )-a_{0} \left (8 r^{2}+12 r +1\right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 2 \left (-4 a_{k}+a_{k -1}+4 a_{k +1}\right ) k^{2}+\left (4 \left (-4 a_{k}+a_{k -1}+4 a_{k +1}\right ) r -12 a_{k}-a_{k -1}+28 a_{k +1}\right ) k +2 \left (-4 a_{k}+a_{k -1}+4 a_{k +1}\right ) r^{2}+\left (-12 a_{k}-a_{k -1}+28 a_{k +1}\right ) r -a_{k}+20 a_{k +1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 2 \left (-4 a_{k +1}+a_{k}+4 a_{k +2}\right ) \left (k +1\right )^{2}+\left (4 \left (-4 a_{k +1}+a_{k}+4 a_{k +2}\right ) r -12 a_{k +1}-a_{k}+28 a_{k +2}\right ) \left (k +1\right )+2 \left (-4 a_{k +1}+a_{k}+4 a_{k +2}\right ) r^{2}+\left (-12 a_{k +1}-a_{k}+28 a_{k +2}\right ) r -a_{k +1}+20 a_{k +2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}+4 k r a_{k}-16 k r a_{k +1}+2 r^{2} a_{k}-8 r^{2} a_{k +1}+3 k a_{k}-28 k a_{k +1}+3 r a_{k}-28 r a_{k +1}+a_{k}-21 a_{k +1}}{4 \left (2 k^{2}+4 k r +2 r^{2}+11 k +11 r +14\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}+3 k a_{k}-28 k a_{k +1}+a_{k}-21 a_{k +1}}{4 \left (2 k^{2}+11 k +14\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}+3 k a_{k}-28 k a_{k +1}+a_{k}-21 a_{k +1}}{4 \left (2 k^{2}+11 k +14\right )}, 20 a_{1}-a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +2\right )^{k}, a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}+3 k a_{k}-28 k a_{k +1}+a_{k}-21 a_{k +1}}{4 \left (2 k^{2}+11 k +14\right )}, 20 a_{1}-a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {3}{2} \\ {} & {} & a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}-3 k a_{k}-4 k a_{k +1}+a_{k}+3 a_{k +1}}{4 \left (2 k^{2}+5 k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {3}{2} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -\frac {3}{2}}, a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}-3 k a_{k}-4 k a_{k +1}+a_{k}+3 a_{k +1}}{4 \left (2 k^{2}+5 k +2\right )}, -4 a_{1}-a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +2\right )^{k -\frac {3}{2}}, a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}-3 k a_{k}-4 k a_{k +1}+a_{k}+3 a_{k +1}}{4 \left (2 k^{2}+5 k +2\right )}, -4 a_{1}-a_{0}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +2\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +2\right )^{k -\frac {3}{2}}\right ), a_{k +2}=-\frac {2 k^{2} a_{k}-8 k^{2} a_{k +1}+3 k a_{k}-28 k a_{k +1}+a_{k}-21 a_{k +1}}{4 \left (2 k^{2}+11 k +14\right )}, 20 a_{1}-a_{0}=0, b_{k +2}=-\frac {2 k^{2} b_{k}-8 k^{2} b_{k +1}-3 k b_{k}-4 k b_{k +1}+b_{k}+3 b_{k +1}}{4 \left (2 k^{2}+5 k +2\right )}, -4 b_{1}-b_{0}=0\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 <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 69
Order:=6; dsolve(2*x^2*(2+x)*diff(y(x),x$2)+5*x^2*diff(y(x),x)+(1+x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \sqrt {x}\, \left (\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1-\frac {3}{4} x +\frac {15}{32} x^{2}-\frac {35}{128} x^{3}+\frac {315}{2048} x^{4}-\frac {693}{8192} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (\frac {1}{4} x -\frac {13}{64} x^{2}+\frac {101}{768} x^{3}-\frac {641}{8192} x^{4}+\frac {7303}{163840} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 0.008 (sec). Leaf size: 134
AsymptoticDSolveValue[2*x^2*(2+x)*y''[x]+5*x^2*y'[x]+(1+x)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \sqrt {x} \left (-\frac {693 x^5}{8192}+\frac {315 x^4}{2048}-\frac {35 x^3}{128}+\frac {15 x^2}{32}-\frac {3 x}{4}+1\right )+c_2 \left (\sqrt {x} \left (\frac {7303 x^5}{163840}-\frac {641 x^4}{8192}+\frac {101 x^3}{768}-\frac {13 x^2}{64}+\frac {x}{4}\right )+\sqrt {x} \left (-\frac {693 x^5}{8192}+\frac {315 x^4}{2048}-\frac {35 x^3}{128}+\frac {15 x^2}{32}-\frac {3 x}{4}+1\right ) \log (x)\right ) \]