Internal problem ID [6961]
Internal file name [OUTPUT/6204_Friday_August_12_2022_11_05_17_PM_55982926/index.tex]
Book: Elementary differential equations. Rainville, Bedient, Bedient. Prentice Hall. NJ. 8th
edition. 1997.
Section: CHAPTER 18. Power series solutions. 18.6. Indicial Equation with Equal Roots.
Exercises page 373
Problem number: 14.
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]]
\[ \boxed {x^{2} y^{\prime \prime }+x \left (3+2 x \right ) y^{\prime }+\left (1+3 x \right ) y=0} \] 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. \[ x^{2} y^{\prime \prime }+\left (2 x^{2}+3 x \right ) y^{\prime }+\left (1+3 x \right ) y = 0 \] 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 {3+2 x}{x}\\ q(x) &= \frac {1+3 x}{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 \[ x^{2} y^{\prime \prime }+\left (2 x^{2}+3 x \right ) y^{\prime }+\left (1+3 x \right ) y = 0 \] 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} 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 (2 x^{2}+3 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (1+3 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} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 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 }}2 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}3 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 }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+3 x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+3 x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+3 x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r +1\right )^{2} x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ \left (r +1\right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= -1\\ r_2 &= -1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r +1\right )^{2} x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([-1, -1]\).
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, \(r = -1\), Eqs (1A,1B) become \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n -1}\\ 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 -1}\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} a_{n} \left (n +r \right ) \left (n +r -1\right )+2 a_{n -1} \left (n +r -1\right )+3 a_{n} \left (n +r \right )+a_{n}+3 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (2 n +2 r +1\right )}{n^{2}+2 n r +r^{2}+2 n +2 r +1}\tag {4} \] Which for the root \(r = -1\) becomes \[ a_{n} = \frac {a_{n -1} \left (-2 n +1\right )}{n^{2}}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = -1\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ a_{1}=\frac {-3-2 r}{\left (2+r \right )^{2}} \] Which for the root \(r = -1\) becomes \[ a_{1}=-1 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-2 r}{\left (2+r \right )^{2}}\) | \(-1\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {4 r^{2}+16 r +15}{\left (2+r \right )^{2} \left (3+r \right )^{2}} \] Which for the root \(r = -1\) becomes \[ a_{2}={\frac {3}{4}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-2 r}{\left (2+r \right )^{2}}\) | \(-1\) |
| \(a_{2}\) | \(\frac {4 r^{2}+16 r +15}{\left (2+r \right )^{2} \left (3+r \right )^{2}}\) | \(\frac {3}{4}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-8 r^{3}-60 r^{2}-142 r -105}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2}} \] Which for the root \(r = -1\) becomes \[ a_{3}=-{\frac {5}{12}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-2 r}{\left (2+r \right )^{2}}\) | \(-1\) |
| \(a_{2}\) | \(\frac {4 r^{2}+16 r +15}{\left (2+r \right )^{2} \left (3+r \right )^{2}}\) | \(\frac {3}{4}\) |
| \(a_{3}\) | \(\frac {-8 r^{3}-60 r^{2}-142 r -105}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2}}\) | \(-{\frac {5}{12}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {16 r^{4}+192 r^{3}+824 r^{2}+1488 r +945}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2}} \] Which for the root \(r = -1\) becomes \[ a_{4}={\frac {35}{192}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-2 r}{\left (2+r \right )^{2}}\) | \(-1\) |
| \(a_{2}\) | \(\frac {4 r^{2}+16 r +15}{\left (2+r \right )^{2} \left (3+r \right )^{2}}\) | \(\frac {3}{4}\) |
| \(a_{3}\) | \(\frac {-8 r^{3}-60 r^{2}-142 r -105}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2}}\) | \(-{\frac {5}{12}}\) |
| \(a_{4}\) | \(\frac {16 r^{4}+192 r^{3}+824 r^{2}+1488 r +945}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2}}\) | \(\frac {35}{192}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-32 r^{5}-560 r^{4}-3760 r^{3}-12040 r^{2}-18258 r -10395}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2}} \] Which for the root \(r = -1\) becomes \[ a_{5}=-{\frac {21}{320}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-2 r}{\left (2+r \right )^{2}}\) | \(-1\) |
| \(a_{2}\) | \(\frac {4 r^{2}+16 r +15}{\left (2+r \right )^{2} \left (3+r \right )^{2}}\) | \(\frac {3}{4}\) |
| \(a_{3}\) | \(\frac {-8 r^{3}-60 r^{2}-142 r -105}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2}}\) | \(-{\frac {5}{12}}\) |
| \(a_{4}\) | \(\frac {16 r^{4}+192 r^{3}+824 r^{2}+1488 r +945}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2}}\) | \(\frac {35}{192}\) |
| \(a_{5}\) | \(\frac {-32 r^{5}-560 r^{4}-3760 r^{3}-12040 r^{2}-18258 r -10395}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2}}\) | \(-{\frac {21}{320}}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {64 r^{6}+1536 r^{5}+14800 r^{4}+72960 r^{3}+193036 r^{2}+258144 r +135135}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2} \left (7+r \right )^{2}} \] Which for the root \(r = -1\) becomes \[ a_{6}={\frac {77}{3840}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-2 r}{\left (2+r \right )^{2}}\) | \(-1\) |
| \(a_{2}\) | \(\frac {4 r^{2}+16 r +15}{\left (2+r \right )^{2} \left (3+r \right )^{2}}\) | \(\frac {3}{4}\) |
| \(a_{3}\) | \(\frac {-8 r^{3}-60 r^{2}-142 r -105}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2}}\) | \(-{\frac {5}{12}}\) |
| \(a_{4}\) | \(\frac {16 r^{4}+192 r^{3}+824 r^{2}+1488 r +945}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2}}\) | \(\frac {35}{192}\) |
| \(a_{5}\) | \(\frac {-32 r^{5}-560 r^{4}-3760 r^{3}-12040 r^{2}-18258 r -10395}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2}}\) | \(-{\frac {21}{320}}\) |
| \(a_{6}\) | \(\frac {64 r^{6}+1536 r^{5}+14800 r^{4}+72960 r^{3}+193036 r^{2}+258144 r +135135}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2} \left (7+r \right )^{2}}\) | \(\frac {77}{3840}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {-128 r^{7}-4032 r^{6}-52640 r^{5}-367920 r^{4}-1480472 r^{3}-3411828 r^{2}-4142430 r -2027025}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2} \left (7+r \right )^{2} \left (8+r \right )^{2}} \] Which for the root \(r = -1\) becomes \[ a_{7}=-{\frac {143}{26880}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-2 r}{\left (2+r \right )^{2}}\) | \(-1\) |
| \(a_{2}\) | \(\frac {4 r^{2}+16 r +15}{\left (2+r \right )^{2} \left (3+r \right )^{2}}\) | \(\frac {3}{4}\) |
| \(a_{3}\) | \(\frac {-8 r^{3}-60 r^{2}-142 r -105}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2}}\) | \(-{\frac {5}{12}}\) |
| \(a_{4}\) | \(\frac {16 r^{4}+192 r^{3}+824 r^{2}+1488 r +945}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2}}\) | \(\frac {35}{192}\) |
| \(a_{5}\) | \(\frac {-32 r^{5}-560 r^{4}-3760 r^{3}-12040 r^{2}-18258 r -10395}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2}}\) | \(-{\frac {21}{320}}\) |
| \(a_{6}\) | \(\frac {64 r^{6}+1536 r^{5}+14800 r^{4}+72960 r^{3}+193036 r^{2}+258144 r +135135}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2} \left (7+r \right )^{2}}\) | \(\frac {77}{3840}\) |
| \(a_{7}\) | \(\frac {-128 r^{7}-4032 r^{6}-52640 r^{5}-367920 r^{4}-1480472 r^{3}-3411828 r^{2}-4142430 r -2027025}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2} \left (7+r \right )^{2} \left (8+r \right )^{2}}\) | \(-{\frac {143}{26880}}\) |
Using the above table, then the first solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= \frac {1}{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}+a_{7} x^{7}+a_{8} x^{8}\dots \right ) \\ &= \frac {1-x +\frac {3 x^{2}}{4}-\frac {5 x^{3}}{12}+\frac {35 x^{4}}{192}-\frac {21 x^{5}}{320}+\frac {77 x^{6}}{3840}-\frac {143 x^{7}}{26880}+O\left (x^{8}\right )}{x} \\ \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 \(r = -1\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =-1\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(\frac {-3-2 r}{\left (2+r \right )^{2}}\) | \(-1\) | \(\frac {2+2 r}{\left (2+r \right )^{3}}\) | \(0\) |
| \(b_{2}\) | \(\frac {4 r^{2}+16 r +15}{\left (2+r \right )^{2} \left (3+r \right )^{2}}\) | \(\frac {3}{4}\) | \(\frac {-8 r^{3}-48 r^{2}-92 r -54}{\left (2+r \right )^{3} \left (3+r \right )^{3}}\) | \(-{\frac {1}{4}}\) |
| \(b_{3}\) | \(\frac {-8 r^{3}-60 r^{2}-142 r -105}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2}}\) | \(-{\frac {5}{12}}\) | \(\frac {24 r^{5}+312 r^{4}+1582 r^{3}+3888 r^{2}+4592 r +2052}{\left (2+r \right )^{3} \left (3+r \right )^{3} \left (4+r \right )^{3}}\) | \(\frac {1}{4}\) |
| \(b_{4}\) | \(\frac {16 r^{4}+192 r^{3}+824 r^{2}+1488 r +945}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2}}\) | \(\frac {35}{192}\) | \(-\frac {4 \left (16 r^{7}+352 r^{6}+3252 r^{5}+16316 r^{4}+47870 r^{3}+81801 r^{2}+74943 r +28125\right )}{\left (2+r \right )^{3} \left (3+r \right )^{3} \left (4+r \right )^{3} \left (5+r \right )^{3}}\) | \(-{\frac {19}{128}}\) |
| \(b_{5}\) | \(\frac {-32 r^{5}-560 r^{4}-3760 r^{3}-12040 r^{2}-18258 r -10395}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2}}\) | \(-{\frac {21}{320}}\) | \(\frac {160 r^{9}+5280 r^{8}+76080 r^{7}+627360 r^{6}+3257298 r^{5}+11021190 r^{4}+24241310 r^{3}+33314670 r^{2}+25840152 r +8559000}{\left (2+r \right )^{3} \left (3+r \right )^{3} \left (4+r \right )^{3} \left (5+r \right )^{3} \left (6+r \right )^{3}}\) | \(\frac {25}{384}\) |
| \(b_{6}\) | \(\frac {64 r^{6}+1536 r^{5}+14800 r^{4}+72960 r^{3}+193036 r^{2}+258144 r +135135}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2} \left (7+r \right )^{2}}\) | \(\frac {77}{3840}\) | \(\frac {-384 r^{11}-17664 r^{10}-363520 r^{9}-4413600 r^{8}-35087928 r^{7}-191540448 r^{6}-731531148 r^{5}-1951225410 r^{4}-3554388008 r^{3}-4199544378 r^{2}-2885493312 r -868681800}{\left (2+r \right )^{3} \left (3+r \right )^{3} \left (4+r \right )^{3} \left (5+r \right )^{3} \left (6+r \right )^{3} \left (7+r \right )^{3}}\) | \(-{\frac {317}{13824}}\) |
| \(b_{7}\) | \(\frac {-128 r^{7}-4032 r^{6}-52640 r^{5}-367920 r^{4}-1480472 r^{3}-3411828 r^{2}-4142430 r -2027025}{\left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2} \left (7+r \right )^{2} \left (8+r \right )^{2}}\) | \(-{\frac {143}{26880}}\) | \(\frac {896 r^{13}+54656 r^{12}+1516704 r^{11}+25332608 r^{10}+283897320 r^{9}+2252199096 r^{8}+12998655094 r^{7}+55197175012 r^{6}+172213088334 r^{5}+389265555448 r^{4}+618393138552 r^{3}+652194947880 r^{2}+407953227600 r +113776941600}{\left (2+r \right )^{3} \left (3+r \right )^{3} \left (4+r \right )^{3} \left (5+r \right )^{3} \left (6+r \right )^{3} \left (7+r \right )^{3} \left (8+r \right )^{3}}\) | \(\frac {469}{69120}\) |
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}+b_{7} x^{7}+b_{8} x^{8}\dots \\ &= \frac {\left (1-x +\frac {3 x^{2}}{4}-\frac {5 x^{3}}{12}+\frac {35 x^{4}}{192}-\frac {21 x^{5}}{320}+\frac {77 x^{6}}{3840}-\frac {143 x^{7}}{26880}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x}+\frac {-\frac {x^{2}}{4}+\frac {x^{3}}{4}-\frac {19 x^{4}}{128}+\frac {25 x^{5}}{384}-\frac {317 x^{6}}{13824}+\frac {469 x^{7}}{69120}+O\left (x^{8}\right )}{x} \\ \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 ) \\ &= \frac {c_{1} \left (1-x +\frac {3 x^{2}}{4}-\frac {5 x^{3}}{12}+\frac {35 x^{4}}{192}-\frac {21 x^{5}}{320}+\frac {77 x^{6}}{3840}-\frac {143 x^{7}}{26880}+O\left (x^{8}\right )\right )}{x} + c_{2} \left (\frac {\left (1-x +\frac {3 x^{2}}{4}-\frac {5 x^{3}}{12}+\frac {35 x^{4}}{192}-\frac {21 x^{5}}{320}+\frac {77 x^{6}}{3840}-\frac {143 x^{7}}{26880}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x}+\frac {-\frac {x^{2}}{4}+\frac {x^{3}}{4}-\frac {19 x^{4}}{128}+\frac {25 x^{5}}{384}-\frac {317 x^{6}}{13824}+\frac {469 x^{7}}{69120}+O\left (x^{8}\right )}{x}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= \frac {c_{1} \left (1-x +\frac {3 x^{2}}{4}-\frac {5 x^{3}}{12}+\frac {35 x^{4}}{192}-\frac {21 x^{5}}{320}+\frac {77 x^{6}}{3840}-\frac {143 x^{7}}{26880}+O\left (x^{8}\right )\right )}{x}+c_{2} \left (\frac {\left (1-x +\frac {3 x^{2}}{4}-\frac {5 x^{3}}{12}+\frac {35 x^{4}}{192}-\frac {21 x^{5}}{320}+\frac {77 x^{6}}{3840}-\frac {143 x^{7}}{26880}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x}+\frac {-\frac {x^{2}}{4}+\frac {x^{3}}{4}-\frac {19 x^{4}}{128}+\frac {25 x^{5}}{384}-\frac {317 x^{6}}{13824}+\frac {469 x^{7}}{69120}+O\left (x^{8}\right )}{x}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} \left (1-x +\frac {3 x^{2}}{4}-\frac {5 x^{3}}{12}+\frac {35 x^{4}}{192}-\frac {21 x^{5}}{320}+\frac {77 x^{6}}{3840}-\frac {143 x^{7}}{26880}+O\left (x^{8}\right )\right )}{x}+c_{2} \left (\frac {\left (1-x +\frac {3 x^{2}}{4}-\frac {5 x^{3}}{12}+\frac {35 x^{4}}{192}-\frac {21 x^{5}}{320}+\frac {77 x^{6}}{3840}-\frac {143 x^{7}}{26880}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x}+\frac {-\frac {x^{2}}{4}+\frac {x^{3}}{4}-\frac {19 x^{4}}{128}+\frac {25 x^{5}}{384}-\frac {317 x^{6}}{13824}+\frac {469 x^{7}}{69120}+O\left (x^{8}\right )}{x}\right ) \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} \left (1-x +\frac {3 x^{2}}{4}-\frac {5 x^{3}}{12}+\frac {35 x^{4}}{192}-\frac {21 x^{5}}{320}+\frac {77 x^{6}}{3840}-\frac {143 x^{7}}{26880}+O\left (x^{8}\right )\right )}{x}+c_{2} \left (\frac {\left (1-x +\frac {3 x^{2}}{4}-\frac {5 x^{3}}{12}+\frac {35 x^{4}}{192}-\frac {21 x^{5}}{320}+\frac {77 x^{6}}{3840}-\frac {143 x^{7}}{26880}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x}+\frac {-\frac {x^{2}}{4}+\frac {x^{3}}{4}-\frac {19 x^{4}}{128}+\frac {25 x^{5}}{384}-\frac {317 x^{6}}{13824}+\frac {469 x^{7}}{69120}+O\left (x^{8}\right )}{x}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (2 x^{2}+3 x \right ) y^{\prime }+\left (1+3 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 (1+3 x \right ) y}{x^{2}}-\frac {\left (3+2 x \right ) y^{\prime }}{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} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {\left (3+2 x \right ) y^{\prime }}{x}+\frac {\left (1+3 x \right ) y}{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 {3+2 x}{x}, P_{3}\left (x \right )=\frac {1+3 x}{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}}}=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}}}=1 \\ {} & \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} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )+x \left (3+2 x \right ) y^{\prime }+\left (1+3 x \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 \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\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} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (1+r \right )^{2} x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (k +r +1\right )^{2}+a_{k -1} \left (2 k +1+2 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+r \right )^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =-1 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (k +r +1\right )^{2}+a_{k -1} \left (2 k +1+2 r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +1} \left (k +2+r \right )^{2}+a_{k} \left (2 k +2 r +3\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (2 k +2 r +3\right )}{\left (k +2+r \right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (2 k +1\right )}{\left (k +1\right )^{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -1}, a_{k +1}=-\frac {a_{k} \left (2 k +1\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 <- Bessel successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 79
Order:=8; dsolve(x^2*diff(y(x),x$2)+x*(3+2*x)*diff(y(x),x)+(1+3*x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1-x +\frac {3}{4} x^{2}-\frac {5}{12} x^{3}+\frac {35}{192} x^{4}-\frac {21}{320} x^{5}+\frac {77}{3840} x^{6}-\frac {143}{26880} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\left (-\frac {1}{4} x^{2}+\frac {1}{4} x^{3}-\frac {19}{128} x^{4}+\frac {25}{384} x^{5}-\frac {317}{13824} x^{6}+\frac {469}{69120} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) c_{2}}{x} \]
✓ Solution by Mathematica
Time used: 0.006 (sec). Leaf size: 161
AsymptoticDSolveValue[x^2*y''[x]+x*(3+2*x)*y'[x]+(1+3*x)*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to \frac {c_1 \left (-\frac {143 x^7}{26880}+\frac {77 x^6}{3840}-\frac {21 x^5}{320}+\frac {35 x^4}{192}-\frac {5 x^3}{12}+\frac {3 x^2}{4}-x+1\right )}{x}+c_2 \left (\frac {\frac {469 x^7}{69120}-\frac {317 x^6}{13824}+\frac {25 x^5}{384}-\frac {19 x^4}{128}+\frac {x^3}{4}-\frac {x^2}{4}}{x}+\frac {\left (-\frac {143 x^7}{26880}+\frac {77 x^6}{3840}-\frac {21 x^5}{320}+\frac {35 x^4}{192}-\frac {5 x^3}{12}+\frac {3 x^2}{4}-x+1\right ) \log (x)}{x}\right ) \]