Internal problem ID [6451]
Internal file name [OUTPUT/5699_Sunday_June_05_2022_03_47_40_PM_73099135/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 4. Power Series Solutions and Special Functions. Section 4.4. REGULAR
SINGULAR POINTS. Page 175
Problem number: 3(b).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference not integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {4 x^{2} y^{\prime \prime }+\left (2 x^{4}-5 x \right ) y^{\prime }+\left (3 x^{2}+2\right ) y=0} \] 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. \[ 4 x^{2} y^{\prime \prime }+\left (2 x^{4}-5 x \right ) y^{\prime }+\left (3 x^{2}+2\right ) y = 0 \] 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 {2 x^{3}-5}{4 x}\\ q(x) &= \frac {3 x^{2}+2}{4 x^{2}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, \infty , -\infty ]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 4 x^{2} y^{\prime \prime }+\left (2 x^{4}-5 x \right ) y^{\prime }+\left (3 x^{2}+2\right ) y = 0 \] 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} 4 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^{4}-5 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (3 x^{2}+2\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 }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r +3} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 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^{n +r +3} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}2 a_{n -3} \left (n -3+r \right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}3 a_{n -2} 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 }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}2 a_{n -3} \left (n -3+r \right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}3 a_{n -2} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-5 x^{n +r} a_{n} \left (n +r \right )+2 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 4 x^{r} a_{0} r \left (-1+r \right )-5 x^{r} a_{0} r +2 a_{0} x^{r} = 0 \] Or \[ \left (4 x^{r} r \left (-1+r \right )-5 x^{r} r +2 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (4 r^{2}-9 r +2\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 4 r^{2}-9 r +2 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 2\\ r_2 &= {\frac {1}{4}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (4 r^{2}-9 r +2\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [2, {\frac {1}{4}}\right ]\).
Since \(r_1 - r_2 = {\frac {7}{4}}\) is not an integer, then we can construct two linearly independent solutions \begin {align*} y_{1}\left (x \right ) &= x^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= x^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +2}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {1}{4}} \end {align*}
We start by finding \(y_{1}\left (x \right )\). 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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = 0 \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = -\frac {3}{r \left (4 r +7\right )} \] For \(3\le n\) the recursive equation is \begin{equation} \tag{3} 4 a_{n} \left (n +r \right ) \left (n +r -1\right )+2 a_{n -3} \left (n -3+r \right )-5 a_{n} \left (n +r \right )+3 a_{n -2}+2 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {2 n a_{n -3}+2 r a_{n -3}-6 a_{n -3}+3 a_{n -2}}{4 n^{2}+8 n r +4 r^{2}-9 n -9 r +2}\tag {4} \] Which for the root \(r = 2\) becomes \[ a_{n} = \frac {-2 n a_{n -3}+2 a_{n -3}-3 a_{n -2}}{n \left (4 n +7\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 2\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {3}{r \left (4 r +7\right )}\) | \(-{\frac {1}{10}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {2 r}{4 r^{2}+15 r +11} \] Which for the root \(r = 2\) becomes \[ a_{3}=-{\frac {4}{57}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {3}{r \left (4 r +7\right )}\) | \(-{\frac {1}{10}}\) |
| \(a_{3}\) | \(-\frac {2 r}{4 r^{2}+15 r +11}\) | \(-{\frac {4}{57}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {9}{r \left (4 r +7\right ) \left (4 r^{2}+23 r +30\right )} \] Which for the root \(r = 2\) becomes \[ a_{4}={\frac {3}{920}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {3}{r \left (4 r +7\right )}\) | \(-{\frac {1}{10}}\) |
| \(a_{3}\) | \(-\frac {2 r}{4 r^{2}+15 r +11}\) | \(-{\frac {4}{57}}\) |
| \(a_{4}\) | \(\frac {9}{r \left (4 r +7\right ) \left (4 r^{2}+23 r +30\right )}\) | \(\frac {3}{920}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {48 r^{3}+180 r^{2}+246 r +132}{\left (4 r +7\right ) r \left (4 r^{2}+15 r +11\right ) \left (4 r^{2}+31 r +57\right )} \] Which for the root \(r = 2\) becomes \[ a_{5}={\frac {32}{4275}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {3}{r \left (4 r +7\right )}\) | \(-{\frac {1}{10}}\) |
| \(a_{3}\) | \(-\frac {2 r}{4 r^{2}+15 r +11}\) | \(-{\frac {4}{57}}\) |
| \(a_{4}\) | \(\frac {9}{r \left (4 r +7\right ) \left (4 r^{2}+23 r +30\right )}\) | \(\frac {3}{920}\) |
| \(a_{5}\) | \(\frac {48 r^{3}+180 r^{2}+246 r +132}{\left (4 r +7\right ) r \left (4 r^{2}+15 r +11\right ) \left (4 r^{2}+31 r +57\right )}\) | \(\frac {32}{4275}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {64 r^{6}+672 r^{5}+2564 r^{4}+4212 r^{3}+2412 r^{2}-405 r -297}{\left (4 r^{2}+15 r +11\right ) r \left (4 r +7\right ) \left (4 r^{2}+23 r +30\right ) \left (4 r^{2}+39 r +92\right )} \] Which for the root \(r = 2\) becomes \[ a_{6}={\frac {36287}{9753840}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {3}{r \left (4 r +7\right )}\) | \(-{\frac {1}{10}}\) |
| \(a_{3}\) | \(-\frac {2 r}{4 r^{2}+15 r +11}\) | \(-{\frac {4}{57}}\) |
| \(a_{4}\) | \(\frac {9}{r \left (4 r +7\right ) \left (4 r^{2}+23 r +30\right )}\) | \(\frac {3}{920}\) |
| \(a_{5}\) | \(\frac {48 r^{3}+180 r^{2}+246 r +132}{\left (4 r +7\right ) r \left (4 r^{2}+15 r +11\right ) \left (4 r^{2}+31 r +57\right )}\) | \(\frac {32}{4275}\) |
| \(a_{6}\) | \(\frac {64 r^{6}+672 r^{5}+2564 r^{4}+4212 r^{3}+2412 r^{2}-405 r -297}{\left (4 r^{2}+15 r +11\right ) r \left (4 r +7\right ) \left (4 r^{2}+23 r +30\right ) \left (4 r^{2}+39 r +92\right )}\) | \(\frac {36287}{9753840}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=-\frac {18 \left (48 r^{5}+552 r^{4}+2567 r^{3}+6075 r^{2}+7147 r +3168\right )}{\left (4 r +7\right ) \left (4 r^{2}+23 r +30\right ) r \left (4 r^{2}+15 r +11\right ) \left (4 r^{2}+31 r +57\right ) \left (4 r^{2}+47 r +135\right )} \] Which for the root \(r = 2\) becomes \[ a_{7}=-{\frac {4037}{16059750}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {3}{r \left (4 r +7\right )}\) | \(-{\frac {1}{10}}\) |
| \(a_{3}\) | \(-\frac {2 r}{4 r^{2}+15 r +11}\) | \(-{\frac {4}{57}}\) |
| \(a_{4}\) | \(\frac {9}{r \left (4 r +7\right ) \left (4 r^{2}+23 r +30\right )}\) | \(\frac {3}{920}\) |
| \(a_{5}\) | \(\frac {48 r^{3}+180 r^{2}+246 r +132}{\left (4 r +7\right ) r \left (4 r^{2}+15 r +11\right ) \left (4 r^{2}+31 r +57\right )}\) | \(\frac {32}{4275}\) |
| \(a_{6}\) | \(\frac {64 r^{6}+672 r^{5}+2564 r^{4}+4212 r^{3}+2412 r^{2}-405 r -297}{\left (4 r^{2}+15 r +11\right ) r \left (4 r +7\right ) \left (4 r^{2}+23 r +30\right ) \left (4 r^{2}+39 r +92\right )}\) | \(\frac {36287}{9753840}\) |
| \(a_{7}\) | \(-\frac {18 \left (48 r^{5}+552 r^{4}+2567 r^{3}+6075 r^{2}+7147 r +3168\right )}{\left (4 r +7\right ) \left (4 r^{2}+23 r +30\right ) r \left (4 r^{2}+15 r +11\right ) \left (4 r^{2}+31 r +57\right ) \left (4 r^{2}+47 r +135\right )}\) | \(-{\frac {4037}{16059750}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{2} \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 ) \\ &= x^{2} \left (1-\frac {x^{2}}{10}-\frac {4 x^{3}}{57}+\frac {3 x^{4}}{920}+\frac {32 x^{5}}{4275}+\frac {36287 x^{6}}{9753840}-\frac {4037 x^{7}}{16059750}+O\left (x^{8}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Eq (2B) derived above is now used to find all \(b_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ b_{1} = 0 \] Substituting \(n = 2\) in Eq. (2B) gives \[ b_{2} = -\frac {3}{r \left (4 r +7\right )} \] For \(3\le n\) the recursive equation is \begin{equation} \tag{3} 4 b_{n} \left (n +r \right ) \left (n +r -1\right )+2 b_{n -3} \left (n -3+r \right )-5 b_{n} \left (n +r \right )+3 b_{n -2}+2 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {2 n b_{n -3}+2 r b_{n -3}-6 b_{n -3}+3 b_{n -2}}{4 n^{2}+8 n r +4 r^{2}-9 n -9 r +2}\tag {4} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ b_{n} = \frac {-4 n b_{n -3}+11 b_{n -3}-6 b_{n -2}}{8 n^{2}-14 n}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = {\frac {1}{4}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {3}{r \left (4 r +7\right )}\) | \(-{\frac {3}{2}}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {2 r}{4 r^{2}+15 r +11} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ b_{3}=-{\frac {1}{30}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {3}{r \left (4 r +7\right )}\) | \(-{\frac {3}{2}}\) |
| \(b_{3}\) | \(-\frac {2 r}{4 r^{2}+15 r +11}\) | \(-{\frac {1}{30}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {9}{r \left (4 r +7\right ) \left (4 r^{2}+23 r +30\right )} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ b_{4}={\frac {1}{8}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {3}{r \left (4 r +7\right )}\) | \(-{\frac {3}{2}}\) |
| \(b_{3}\) | \(-\frac {2 r}{4 r^{2}+15 r +11}\) | \(-{\frac {1}{30}}\) |
| \(b_{4}\) | \(\frac {9}{r \left (4 r +7\right ) \left (4 r^{2}+23 r +30\right )}\) | \(\frac {1}{8}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {48 r^{3}+180 r^{2}+246 r +132}{\left (4 r +7\right ) r \left (4 r^{2}+15 r +11\right ) \left (4 r^{2}+31 r +57\right )} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ b_{5}={\frac {137}{1300}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {3}{r \left (4 r +7\right )}\) | \(-{\frac {3}{2}}\) |
| \(b_{3}\) | \(-\frac {2 r}{4 r^{2}+15 r +11}\) | \(-{\frac {1}{30}}\) |
| \(b_{4}\) | \(\frac {9}{r \left (4 r +7\right ) \left (4 r^{2}+23 r +30\right )}\) | \(\frac {1}{8}\) |
| \(b_{5}\) | \(\frac {48 r^{3}+180 r^{2}+246 r +132}{\left (4 r +7\right ) r \left (4 r^{2}+15 r +11\right ) \left (4 r^{2}+31 r +57\right )}\) | \(\frac {137}{1300}\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=\frac {64 r^{6}+672 r^{5}+2564 r^{4}+4212 r^{3}+2412 r^{2}-405 r -297}{\left (4 r^{2}+15 r +11\right ) r \left (4 r +7\right ) \left (4 r^{2}+23 r +30\right ) \left (4 r^{2}+39 r +92\right )} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ b_{6}=-{\frac {19}{12240}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {3}{r \left (4 r +7\right )}\) | \(-{\frac {3}{2}}\) |
| \(b_{3}\) | \(-\frac {2 r}{4 r^{2}+15 r +11}\) | \(-{\frac {1}{30}}\) |
| \(b_{4}\) | \(\frac {9}{r \left (4 r +7\right ) \left (4 r^{2}+23 r +30\right )}\) | \(\frac {1}{8}\) |
| \(b_{5}\) | \(\frac {48 r^{3}+180 r^{2}+246 r +132}{\left (4 r +7\right ) r \left (4 r^{2}+15 r +11\right ) \left (4 r^{2}+31 r +57\right )}\) | \(\frac {137}{1300}\) |
| \(b_{6}\) | \(\frac {64 r^{6}+672 r^{5}+2564 r^{4}+4212 r^{3}+2412 r^{2}-405 r -297}{\left (4 r^{2}+15 r +11\right ) r \left (4 r +7\right ) \left (4 r^{2}+23 r +30\right ) \left (4 r^{2}+39 r +92\right )}\) | \(-{\frac {19}{12240}}\) |
For \(n = 7\), using the above recursive equation gives \[ b_{7}=-\frac {18 \left (48 r^{5}+552 r^{4}+2567 r^{3}+6075 r^{2}+7147 r +3168\right )}{\left (4 r +7\right ) \left (4 r^{2}+23 r +30\right ) r \left (4 r^{2}+15 r +11\right ) \left (4 r^{2}+31 r +57\right ) \left (4 r^{2}+47 r +135\right )} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ b_{7}=-{\frac {7169}{764400}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {3}{r \left (4 r +7\right )}\) | \(-{\frac {3}{2}}\) |
| \(b_{3}\) | \(-\frac {2 r}{4 r^{2}+15 r +11}\) | \(-{\frac {1}{30}}\) |
| \(b_{4}\) | \(\frac {9}{r \left (4 r +7\right ) \left (4 r^{2}+23 r +30\right )}\) | \(\frac {1}{8}\) |
| \(b_{5}\) | \(\frac {48 r^{3}+180 r^{2}+246 r +132}{\left (4 r +7\right ) r \left (4 r^{2}+15 r +11\right ) \left (4 r^{2}+31 r +57\right )}\) | \(\frac {137}{1300}\) |
| \(b_{6}\) | \(\frac {64 r^{6}+672 r^{5}+2564 r^{4}+4212 r^{3}+2412 r^{2}-405 r -297}{\left (4 r^{2}+15 r +11\right ) r \left (4 r +7\right ) \left (4 r^{2}+23 r +30\right ) \left (4 r^{2}+39 r +92\right )}\) | \(-{\frac {19}{12240}}\) |
| \(b_{7}\) | \(-\frac {18 \left (48 r^{5}+552 r^{4}+2567 r^{3}+6075 r^{2}+7147 r +3168\right )}{\left (4 r +7\right ) \left (4 r^{2}+23 r +30\right ) r \left (4 r^{2}+15 r +11\right ) \left (4 r^{2}+31 r +57\right ) \left (4 r^{2}+47 r +135\right )}\) | \(-{\frac {7169}{764400}}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{2} \left (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 \right ) \\ &= x^{\frac {1}{4}} \left (1-\frac {3 x^{2}}{2}-\frac {x^{3}}{30}+\frac {x^{4}}{8}+\frac {137 x^{5}}{1300}-\frac {19 x^{6}}{12240}-\frac {7169 x^{7}}{764400}+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} x^{2} \left (1-\frac {x^{2}}{10}-\frac {4 x^{3}}{57}+\frac {3 x^{4}}{920}+\frac {32 x^{5}}{4275}+\frac {36287 x^{6}}{9753840}-\frac {4037 x^{7}}{16059750}+O\left (x^{8}\right )\right ) + c_{2} x^{\frac {1}{4}} \left (1-\frac {3 x^{2}}{2}-\frac {x^{3}}{30}+\frac {x^{4}}{8}+\frac {137 x^{5}}{1300}-\frac {19 x^{6}}{12240}-\frac {7169 x^{7}}{764400}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{2} \left (1-\frac {x^{2}}{10}-\frac {4 x^{3}}{57}+\frac {3 x^{4}}{920}+\frac {32 x^{5}}{4275}+\frac {36287 x^{6}}{9753840}-\frac {4037 x^{7}}{16059750}+O\left (x^{8}\right )\right )+c_{2} x^{\frac {1}{4}} \left (1-\frac {3 x^{2}}{2}-\frac {x^{3}}{30}+\frac {x^{4}}{8}+\frac {137 x^{5}}{1300}-\frac {19 x^{6}}{12240}-\frac {7169 x^{7}}{764400}+O\left (x^{8}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{2} \left (1-\frac {x^{2}}{10}-\frac {4 x^{3}}{57}+\frac {3 x^{4}}{920}+\frac {32 x^{5}}{4275}+\frac {36287 x^{6}}{9753840}-\frac {4037 x^{7}}{16059750}+O\left (x^{8}\right )\right )+c_{2} x^{\frac {1}{4}} \left (1-\frac {3 x^{2}}{2}-\frac {x^{3}}{30}+\frac {x^{4}}{8}+\frac {137 x^{5}}{1300}-\frac {19 x^{6}}{12240}-\frac {7169 x^{7}}{764400}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{2} \left (1-\frac {x^{2}}{10}-\frac {4 x^{3}}{57}+\frac {3 x^{4}}{920}+\frac {32 x^{5}}{4275}+\frac {36287 x^{6}}{9753840}-\frac {4037 x^{7}}{16059750}+O\left (x^{8}\right )\right )+c_{2} x^{\frac {1}{4}} \left (1-\frac {3 x^{2}}{2}-\frac {x^{3}}{30}+\frac {x^{4}}{8}+\frac {137 x^{5}}{1300}-\frac {19 x^{6}}{12240}-\frac {7169 x^{7}}{764400}+O\left (x^{8}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 x^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (2 x^{4}-5 x \right ) y^{\prime }+\left (3 x^{2}+2\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 (3 x^{2}+2\right ) y}{4 x^{2}}-\frac {\left (2 x^{3}-5\right ) y^{\prime }}{4 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 (2 x^{3}-5\right ) y^{\prime }}{4 x}+\frac {\left (3 x^{2}+2\right ) 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 {2 x^{3}-5}{4 x}, P_{3}\left (x \right )=\frac {3 x^{2}+2}{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}}}=-\frac {5}{4} \\ {} & \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}{2} \\ {} & \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 x^{2} \left (\frac {d}{d x}y^{\prime }\right )+x \left (2 x^{3}-5\right ) y^{\prime }+\left (3 x^{2}+2\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..2 \\ {} & {} & 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..4 \\ {} & {} & 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+4 r \right ) \left (-2+r \right ) x^{r}+a_{1} \left (3+4 r \right ) \left (-1+r \right ) x^{1+r}+\left (a_{2} \left (7+4 r \right ) r +3 a_{0}\right ) x^{2+r}+\left (\moverset {\infty }{\munderset {k =3}{\sum }}\left (a_{k} \left (4 k +4 r -1\right ) \left (k +r -2\right )+3 a_{k -2}+2 a_{k -3} \left (k -3+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+4 r \right ) \left (-2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{2, \frac {1}{4}\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [a_{1} \left (3+4 r \right ) \left (-1+r \right )=0, a_{2} \left (7+4 r \right ) r +3 a_{0}=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=0, a_{2}=-\frac {3 a_{0}}{r \left (7+4 r \right )}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (4 k +4 r -1\right ) \left (k +r -2\right )+3 a_{k -2}+2 a_{k -3} \left (k -3+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +3 \\ {} & {} & a_{k +3} \left (4 k +11+4 r \right ) \left (k +1+r \right )+3 a_{k +1}+2 a_{k} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=-\frac {2 k a_{k}+2 r a_{k}+3 a_{k +1}}{\left (4 k +11+4 r \right ) \left (k +1+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +3}=-\frac {2 k a_{k}+4 a_{k}+3 a_{k +1}}{\left (4 k +19\right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2}, a_{k +3}=-\frac {2 k a_{k}+4 a_{k}+3 a_{k +1}}{\left (4 k +19\right ) \left (k +3\right )}, a_{1}=0, a_{2}=-\frac {a_{0}}{10}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{4} \\ {} & {} & a_{k +3}=-\frac {2 k a_{k}+\frac {1}{2} a_{k}+3 a_{k +1}}{\left (4 k +12\right ) \left (k +\frac {5}{4}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{4} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{4}}, a_{k +3}=-\frac {2 k a_{k}+\frac {1}{2} a_{k}+3 a_{k +1}}{\left (4 k +12\right ) \left (k +\frac {5}{4}\right )}, a_{1}=0, a_{2}=-\frac {3 a_{0}}{2}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{4}}\right ), a_{k +3}=-\frac {2 k a_{k}+4 a_{k}+3 a_{k +1}}{\left (4 k +19\right ) \left (k +3\right )}, a_{1}=0, a_{2}=-\frac {a_{0}}{10}, b_{k +3}=-\frac {2 k b_{k}+\frac {1}{2} b_{k}+3 b_{k +1}}{\left (4 k +12\right ) \left (k +\frac {5}{4}\right )}, b_{1}=0, b_{2}=-\frac {3 b_{0}}{2}\right ] \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius trying a solution in terms of MeijerG functions -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying differential order: 2; exact nonlinear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying 2nd order, integrating factor of the form mu(x,y) -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying to convert to an ODE of Bessel type trying to convert to an ODE of Bessel type -> trying reduction of order to Riccati trying Riccati sub-methods: -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 3`[0, y]
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 51
Order:=8; dsolve(4*x^2*diff(y(x),x$2)+(2*x^4-5*x)*diff(y(x),x)+(3*x^2+2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{\frac {1}{4}} \left (1-\frac {3}{2} x^{2}-\frac {1}{30} x^{3}+\frac {1}{8} x^{4}+\frac {137}{1300} x^{5}-\frac {19}{12240} x^{6}-\frac {7169}{764400} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+c_{2} x^{2} \left (1-\frac {1}{10} x^{2}-\frac {4}{57} x^{3}+\frac {3}{920} x^{4}+\frac {32}{4275} x^{5}+\frac {36287}{9753840} x^{6}-\frac {4037}{16059750} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 106
AsymptoticDSolveValue[4*x^2*y''[x]+(2*x^4-5*x)*y'[x]+(3*x^2+2)*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \left (-\frac {4037 x^7}{16059750}+\frac {36287 x^6}{9753840}+\frac {32 x^5}{4275}+\frac {3 x^4}{920}-\frac {4 x^3}{57}-\frac {x^2}{10}+1\right ) x^2+c_2 \left (-\frac {7169 x^7}{764400}-\frac {19 x^6}{12240}+\frac {137 x^5}{1300}+\frac {x^4}{8}-\frac {x^3}{30}-\frac {3 x^2}{2}+1\right ) \sqrt [4]{x} \]