Internal problem ID [7017]
Internal file name [OUTPUT/6260_Thursday_August_18_2022_07_12_07_AM_64728628/index.tex]
Book: Elementary differential equations. Rainville, Bedient, Bedient. Prentice Hall. NJ. 8th
edition. 1997.
Section: CHAPTER 18. Power series solutions. Miscellaneous Exercises. page 394
Problem number: 15.
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]]
\[ \boxed {4 x^{2} y^{\prime \prime }-y^{\prime } x^{2}+y=0} \] 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. \[ 4 x^{2} y^{\prime \prime }-y^{\prime } x^{2}+y = 0 \] 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 {1}{4}}\\ q(x) &= \frac {1}{4 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 \[ 4 x^{2} y^{\prime \prime }-y^{\prime } x^{2}+y = 0 \] 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} 4 \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 (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x^{2}+\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 }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n +r} a_{n} \left (n +r \right )\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 }}\left (-x^{1+n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) x^{n +r}\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} \left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) x^{n +r}\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 \[ x^{r} \left (-1+2 r \right )^{2} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ \left (-1+2 r \right )^{2} = 0 \] 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 \[ x^{r} \left (-1+2 r \right )^{2} = 0 \] 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} 4 a_{n} \left (n +r \right ) \left (n +r -1\right )-a_{n -1} \left (n +r -1\right )+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -1} \left (n +r -1\right )}{4 n^{2}+8 n r +4 r^{2}-4 n -4 r +1}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{n} = \frac {a_{n -1} \left (2 n -1\right )}{8 n^{2}}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{2}}\) 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 \[ a_{1}=\frac {r}{\left (2 r +1\right )^{2}} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{1}={\frac {1}{8}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{\left (2 r +1\right )^{2}}\) | \(\frac {1}{8}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {r \left (1+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2}} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{2}={\frac {3}{256}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{\left (2 r +1\right )^{2}}\) | \(\frac {1}{8}\) |
| \(a_{2}\) | \(\frac {r \left (1+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2}}\) | \(\frac {3}{256}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {r \left (1+r \right ) \left (2+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2}} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{3}={\frac {5}{6144}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{\left (2 r +1\right )^{2}}\) | \(\frac {1}{8}\) |
| \(a_{2}\) | \(\frac {r \left (1+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2}}\) | \(\frac {3}{256}\) |
| \(a_{3}\) | \(\frac {r \left (1+r \right ) \left (2+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2}}\) | \(\frac {5}{6144}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2}} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{4}={\frac {35}{786432}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{\left (2 r +1\right )^{2}}\) | \(\frac {1}{8}\) |
| \(a_{2}\) | \(\frac {r \left (1+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2}}\) | \(\frac {3}{256}\) |
| \(a_{3}\) | \(\frac {r \left (1+r \right ) \left (2+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2}}\) | \(\frac {5}{6144}\) |
| \(a_{4}\) | \(\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2}}\) | \(\frac {35}{786432}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2}} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{5}={\frac {21}{10485760}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{\left (2 r +1\right )^{2}}\) | \(\frac {1}{8}\) |
| \(a_{2}\) | \(\frac {r \left (1+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2}}\) | \(\frac {3}{256}\) |
| \(a_{3}\) | \(\frac {r \left (1+r \right ) \left (2+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2}}\) | \(\frac {5}{6144}\) |
| \(a_{4}\) | \(\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2}}\) | \(\frac {35}{786432}\) |
| \(a_{5}\) | \(\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2}}\) | \(\frac {21}{10485760}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2} \left (2 r +11\right )^{2}} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{6}={\frac {77}{1006632960}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{\left (2 r +1\right )^{2}}\) | \(\frac {1}{8}\) |
| \(a_{2}\) | \(\frac {r \left (1+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2}}\) | \(\frac {3}{256}\) |
| \(a_{3}\) | \(\frac {r \left (1+r \right ) \left (2+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2}}\) | \(\frac {5}{6144}\) |
| \(a_{4}\) | \(\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2}}\) | \(\frac {35}{786432}\) |
| \(a_{5}\) | \(\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2}}\) | \(\frac {21}{10485760}\) |
| \(a_{6}\) | \(\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2} \left (2 r +11\right )^{2}}\) | \(\frac {77}{1006632960}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right ) \left (6+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2} \left (2 r +11\right )^{2} \left (2 r +13\right )^{2}} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{7}={\frac {143}{56371445760}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{\left (2 r +1\right )^{2}}\) | \(\frac {1}{8}\) |
| \(a_{2}\) | \(\frac {r \left (1+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2}}\) | \(\frac {3}{256}\) |
| \(a_{3}\) | \(\frac {r \left (1+r \right ) \left (2+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2}}\) | \(\frac {5}{6144}\) |
| \(a_{4}\) | \(\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2}}\) | \(\frac {35}{786432}\) |
| \(a_{5}\) | \(\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2}}\) | \(\frac {21}{10485760}\) |
| \(a_{6}\) | \(\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2} \left (2 r +11\right )^{2}}\) | \(\frac {77}{1006632960}\) |
| \(a_{7}\) | \(\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right ) \left (6+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2} \left (2 r +11\right )^{2} \left (2 r +13\right )^{2}}\) | \(\frac {143}{56371445760}\) |
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}+a_{7} x^{7}+a_{8} x^{8}\dots \right ) \\ &= \sqrt {x}\, \left (1+\frac {x}{8}+\frac {3 x^{2}}{256}+\frac {5 x^{3}}{6144}+\frac {35 x^{4}}{786432}+\frac {21 x^{5}}{10485760}+\frac {77 x^{6}}{1006632960}+\frac {143 x^{7}}{56371445760}+O\left (x^{8}\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 |
| \(b_{1}\) | \(\frac {r}{\left (2 r +1\right )^{2}}\) | \(\frac {1}{8}\) | \(\frac {-2 r +1}{\left (2 r +1\right )^{3}}\) | \(0\) |
| \(b_{2}\) | \(\frac {r \left (1+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2}}\) | \(\frac {3}{256}\) | \(\frac {-8 r^{3}-12 r^{2}-2 r +3}{\left (2 r +1\right )^{3} \left (3+2 r \right )^{3}}\) | \(-{\frac {1}{256}}\) |
| \(b_{3}\) | \(\frac {r \left (1+r \right ) \left (2+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2}}\) | \(\frac {5}{6144}\) | \(\frac {-24 r^{5}-132 r^{4}-250 r^{3}-171 r^{2}-2 r +30}{\left (2 r +1\right )^{3} \left (3+2 r \right )^{3} \left (5+2 r \right )^{3}}\) | \(-{\frac {1}{2048}}\) |
| \(b_{4}\) | \(\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2}}\) | \(\frac {35}{786432}\) | \(\frac {-64 r^{7}-736 r^{6}-3360 r^{5}-7664 r^{4}-8876 r^{3}-4302 r^{2}+198 r +630}{\left (2 r +1\right )^{3} \left (3+2 r \right )^{3} \left (5+2 r \right )^{3} \left (7+2 r \right )^{3}}\) | \(-{\frac {19}{524288}}\) |
| \(b_{5}\) | \(\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2}}\) | \(\frac {21}{10485760}\) | \(\frac {-160 r^{9}-3120 r^{8}-25680 r^{7}-115800 r^{6}-309978 r^{5}-495915 r^{4}-444770 r^{3}-174375 r^{2}+13428 r +22680}{\left (2 r +1\right )^{3} \left (3+2 r \right )^{3} \left (5+2 r \right )^{3} \left (7+2 r \right )^{3} \left (2 r +9\right )^{3}}\) | \(-{\frac {25}{12582912}}\) |
| \(b_{6}\) | \(\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2} \left (2 r +11\right )^{2}}\) | \(\frac {77}{1006632960}\) | \(\frac {-384 r^{11}-11328 r^{10}-146080 r^{9}-1080720 r^{8}-5057448 r^{7}-15547932 r^{6}-31500150 r^{5}-40807155 r^{4}-30992588 r^{3}-10463535 r^{2}+1010700 r +1247400}{\left (2 r +1\right )^{3} \left (3+2 r \right )^{3} \left (5+2 r \right )^{3} \left (7+2 r \right )^{3} \left (2 r +9\right )^{3} \left (2 r +11\right )^{3}}\) | \(-{\frac {317}{3623878656}}\) |
| \(b_{7}\) | \(\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right ) \left (6+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2} \left (2 r +11\right )^{2} \left (2 r +13\right )^{2}}\) | \(\frac {143}{56371445760}\) | \(\frac {-896 r^{13}-37184 r^{12}-690144 r^{11}-7558544 r^{10}-54261480 r^{9}-268105068 r^{8}-930884890 r^{7}-2274587371 r^{6}-3844781304 r^{5}-4305889063 r^{4}-2899569036 r^{3}-873762120 r^{2}+96298200 r +97297200}{\left (2 r +1\right )^{3} \left (3+2 r \right )^{3} \left (5+2 r \right )^{3} \left (7+2 r \right )^{3} \left (2 r +9\right )^{3} \left (2 r +11\right )^{3} \left (2 r +13\right )^{3}}\) | \(-{\frac {469}{144955146240}}\) |
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 \\ &= \sqrt {x}\, \left (1+\frac {x}{8}+\frac {3 x^{2}}{256}+\frac {5 x^{3}}{6144}+\frac {35 x^{4}}{786432}+\frac {21 x^{5}}{10485760}+\frac {77 x^{6}}{1006632960}+\frac {143 x^{7}}{56371445760}+O\left (x^{8}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (-\frac {x^{2}}{256}-\frac {x^{3}}{2048}-\frac {19 x^{4}}{524288}-\frac {25 x^{5}}{12582912}-\frac {317 x^{6}}{3623878656}-\frac {469 x^{7}}{144955146240}+O\left (x^{8}\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 {x}{8}+\frac {3 x^{2}}{256}+\frac {5 x^{3}}{6144}+\frac {35 x^{4}}{786432}+\frac {21 x^{5}}{10485760}+\frac {77 x^{6}}{1006632960}+\frac {143 x^{7}}{56371445760}+O\left (x^{8}\right )\right ) + c_{2} \left (\sqrt {x}\, \left (1+\frac {x}{8}+\frac {3 x^{2}}{256}+\frac {5 x^{3}}{6144}+\frac {35 x^{4}}{786432}+\frac {21 x^{5}}{10485760}+\frac {77 x^{6}}{1006632960}+\frac {143 x^{7}}{56371445760}+O\left (x^{8}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (-\frac {x^{2}}{256}-\frac {x^{3}}{2048}-\frac {19 x^{4}}{524288}-\frac {25 x^{5}}{12582912}-\frac {317 x^{6}}{3623878656}-\frac {469 x^{7}}{144955146240}+O\left (x^{8}\right )\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \sqrt {x}\, \left (1+\frac {x}{8}+\frac {3 x^{2}}{256}+\frac {5 x^{3}}{6144}+\frac {35 x^{4}}{786432}+\frac {21 x^{5}}{10485760}+\frac {77 x^{6}}{1006632960}+\frac {143 x^{7}}{56371445760}+O\left (x^{8}\right )\right )+c_{2} \left (\sqrt {x}\, \left (1+\frac {x}{8}+\frac {3 x^{2}}{256}+\frac {5 x^{3}}{6144}+\frac {35 x^{4}}{786432}+\frac {21 x^{5}}{10485760}+\frac {77 x^{6}}{1006632960}+\frac {143 x^{7}}{56371445760}+O\left (x^{8}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (-\frac {x^{2}}{256}-\frac {x^{3}}{2048}-\frac {19 x^{4}}{524288}-\frac {25 x^{5}}{12582912}-\frac {317 x^{6}}{3623878656}-\frac {469 x^{7}}{144955146240}+O\left (x^{8}\right )\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \sqrt {x}\, \left (1+\frac {x}{8}+\frac {3 x^{2}}{256}+\frac {5 x^{3}}{6144}+\frac {35 x^{4}}{786432}+\frac {21 x^{5}}{10485760}+\frac {77 x^{6}}{1006632960}+\frac {143 x^{7}}{56371445760}+O\left (x^{8}\right )\right )+c_{2} \left (\sqrt {x}\, \left (1+\frac {x}{8}+\frac {3 x^{2}}{256}+\frac {5 x^{3}}{6144}+\frac {35 x^{4}}{786432}+\frac {21 x^{5}}{10485760}+\frac {77 x^{6}}{1006632960}+\frac {143 x^{7}}{56371445760}+O\left (x^{8}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (-\frac {x^{2}}{256}-\frac {x^{3}}{2048}-\frac {19 x^{4}}{524288}-\frac {25 x^{5}}{12582912}-\frac {317 x^{6}}{3623878656}-\frac {469 x^{7}}{144955146240}+O\left (x^{8}\right )\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \sqrt {x}\, \left (1+\frac {x}{8}+\frac {3 x^{2}}{256}+\frac {5 x^{3}}{6144}+\frac {35 x^{4}}{786432}+\frac {21 x^{5}}{10485760}+\frac {77 x^{6}}{1006632960}+\frac {143 x^{7}}{56371445760}+O\left (x^{8}\right )\right )+c_{2} \left (\sqrt {x}\, \left (1+\frac {x}{8}+\frac {3 x^{2}}{256}+\frac {5 x^{3}}{6144}+\frac {35 x^{4}}{786432}+\frac {21 x^{5}}{10485760}+\frac {77 x^{6}}{1006632960}+\frac {143 x^{7}}{56371445760}+O\left (x^{8}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (-\frac {x^{2}}{256}-\frac {x^{3}}{2048}-\frac {19 x^{4}}{524288}-\frac {25 x^{5}}{12582912}-\frac {317 x^{6}}{3623878656}-\frac {469 x^{7}}{144955146240}+O\left (x^{8}\right )\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 y^{\prime \prime } x^{2}-y^{\prime } x^{2}+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 {y^{\prime }}{4}-\frac {y}{4 x^{2}} \\ \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 {y^{\prime }}{4}+\frac {y}{4 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 {1}{4}, P_{3}\left (x \right )=\frac {1}{4 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}}}=0 \\ {} & \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}{4} \\ {} & \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} \\ {} & {} & 4 y^{\prime \prime } x^{2}-y^{\prime } x^{2}+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^{2}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r +1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -1 \\ {} & {} & x^{2}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} \left (k -1+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 -1+r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (-1+2 r \right )^{2} x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (2 k +2 r -1\right )^{2}-a_{k -1} \left (k -1+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+2 r \right )^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =\frac {1}{2} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (2 k +2 r -1\right )^{2}-a_{k -1} \left (k -1+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +1} \left (2 k +1+2 r \right )^{2}-a_{k} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +r \right )}{\left (2 k +1+2 r \right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +\frac {1}{2}\right )}{\left (2 k +2\right )^{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}}, a_{k +1}=\frac {a_{k} \left (k +\frac {1}{2}\right )}{\left (2 k +2\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: 0.015 (sec). Leaf size: 79
Order:=8; dsolve(4*x^2*diff(y(x),x$2)-x^2*diff(y(x),x)+y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \sqrt {x}\, \left (\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1+\frac {1}{8} x +\frac {3}{256} x^{2}+\frac {5}{6144} x^{3}+\frac {35}{786432} x^{4}+\frac {21}{10485760} x^{5}+\frac {77}{1006632960} x^{6}+\frac {143}{56371445760} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\left (-\frac {1}{256} x^{2}-\frac {1}{2048} x^{3}-\frac {19}{524288} x^{4}-\frac {25}{12582912} x^{5}-\frac {317}{3623878656} x^{6}-\frac {469}{144955146240} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 0.007 (sec). Leaf size: 171
AsymptoticDSolveValue[4*x^2*y''[x]-x^2*y'[x]+y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \sqrt {x} \left (\frac {143 x^7}{56371445760}+\frac {77 x^6}{1006632960}+\frac {21 x^5}{10485760}+\frac {35 x^4}{786432}+\frac {5 x^3}{6144}+\frac {3 x^2}{256}+\frac {x}{8}+1\right )+c_2 \left (\sqrt {x} \left (-\frac {469 x^7}{144955146240}-\frac {317 x^6}{3623878656}-\frac {25 x^5}{12582912}-\frac {19 x^4}{524288}-\frac {x^3}{2048}-\frac {x^2}{256}\right )+\sqrt {x} \left (\frac {143 x^7}{56371445760}+\frac {77 x^6}{1006632960}+\frac {21 x^5}{10485760}+\frac {35 x^4}{786432}+\frac {5 x^3}{6144}+\frac {3 x^2}{256}+\frac {x}{8}+1\right ) \log (x)\right ) \]