Internal problem ID [1431]
Internal file name [OUTPUT/1432_Sunday_June_05_2022_02_17_03_AM_1594100/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 III. Exercises 7.7. Page 389
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. Difference is integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {4 x^{2} \left (1+2 x \right ) y^{\prime \prime }-2 x \left (-x +4\right ) y^{\prime }-\left (7+5 x \right ) y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \left (8 x^{3}+4 x^{2}\right ) y^{\prime \prime }+\left (2 x^{2}-8 x \right ) y^{\prime }+\left (-5 x -7\right ) y = 0 \] 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 {x -4}{2 x \left (1+2 x \right )}\\ q(x) &= -\frac {7+5 x}{4 x^{2} \left (1+2 x \right )}\\ \end {align*}
| \(p(x)=\frac {x -4}{2 x \left (1+2 x \right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -{\frac {1}{2}}\) | \(\text {``regular''}\) |
| \(q(x)=-\frac {7+5 x}{4 x^{2} \left (1+2 x \right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -{\frac {1}{2}}\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, -{\frac {1}{2}}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 4 x^{2} \left (1+2 x \right ) y^{\prime \prime }+\left (2 x^{2}-8 x \right ) y^{\prime }+\left (-5 x -7\right ) y = 0 \] 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 (1+2 x \right ) \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}-8 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (-5 x -7\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 }}8 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 }}2 x^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-8 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-7 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 }}8 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}8 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\ \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 }}\left (-5 x^{1+n +r} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-5 a_{n -1} x^{n +r}\right ) \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =1}{\sum }}8 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 }}2 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-8 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-5 a_{n -1} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-7 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-8 x^{n +r} a_{n} \left (n +r \right )-7 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 4 x^{r} a_{0} r \left (-1+r \right )-8 x^{r} a_{0} r -7 a_{0} x^{r} = 0 \] Or \[ \left (4 x^{r} r \left (-1+r \right )-8 x^{r} r -7 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (4 r^{2}-12 r -7\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 4 r^{2}-12 r -7 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {7}{2}}\\ r_2 &= -{\frac {1}{2}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (4 r^{2}-12 r -7\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {7}{2}}, -{\frac {1}{2}}\right ]\).
Since \(r_1 - r_2 = 4\) is 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 ) &= C y_{1}\left (x \right ) \ln \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 ) &= x^{{7}/{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\frac {\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}}{\sqrt {x}} \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {7}{2}}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {1}{2}}\right ) \end {align*}
Where \(C\) above can be zero. We start by finding \(y_{1}\). 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} 8 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 )+2 a_{n -1} \left (n +r -1\right )-8 a_{n} \left (n +r \right )-5 a_{n -1}-7 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (8 n^{2}+16 n r +8 r^{2}-22 n -22 r +9\right )}{4 n^{2}+8 n r +4 r^{2}-12 n -12 r -7}\tag {4} \] Which for the root \(r = {\frac {7}{2}}\) becomes \[ a_{n} = -\frac {a_{n -1} \left (4 n^{2}+17 n +15\right )}{2 n \left (n +4\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {7}{2}}\) 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 \[ a_{1}=\frac {-8 r^{2}+6 r +5}{4 r^{2}-4 r -15} \] Which for the root \(r = {\frac {7}{2}}\) becomes \[ a_{1}=-{\frac {18}{5}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-8 r^{2}+6 r +5}{4 r^{2}-4 r -15}\) | \(-{\frac {18}{5}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {32 r^{3}-32 r^{2}-14 r +5}{8 r^{3}-12 r^{2}-50 r +75} \] Which for the root \(r = {\frac {7}{2}}\) becomes \[ a_{2}={\frac {39}{4}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-8 r^{2}+6 r +5}{4 r^{2}-4 r -15}\) | \(-{\frac {18}{5}}\) |
| \(a_{2}\) | \(\frac {32 r^{3}-32 r^{2}-14 r +5}{8 r^{3}-12 r^{2}-50 r +75}\) | \(\frac {39}{4}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-128 r^{4}+32 r^{3}+152 r^{2}+22 r -15}{\left (4 r^{2}+12 r -7\right ) \left (4 r^{2}-16 r +15\right )} \] Which for the root \(r = {\frac {7}{2}}\) becomes \[ a_{3}=-{\frac {663}{28}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-8 r^{2}+6 r +5}{4 r^{2}-4 r -15}\) | \(-{\frac {18}{5}}\) |
| \(a_{2}\) | \(\frac {32 r^{3}-32 r^{2}-14 r +5}{8 r^{3}-12 r^{2}-50 r +75}\) | \(\frac {39}{4}\) |
| \(a_{3}\) | \(\frac {-128 r^{4}+32 r^{3}+152 r^{2}+22 r -15}{\left (4 r^{2}+12 r -7\right ) \left (4 r^{2}-16 r +15\right )}\) | \(-{\frac {663}{28}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {256 r^{4}+256 r^{3}-544 r^{2}-304 r +105}{16 r^{4}-232 r^{2}+384 r -135} \] Which for the root \(r = {\frac {7}{2}}\) becomes \[ a_{4}={\frac {13923}{256}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-8 r^{2}+6 r +5}{4 r^{2}-4 r -15}\) | \(-{\frac {18}{5}}\) |
| \(a_{2}\) | \(\frac {32 r^{3}-32 r^{2}-14 r +5}{8 r^{3}-12 r^{2}-50 r +75}\) | \(\frac {39}{4}\) |
| \(a_{3}\) | \(\frac {-128 r^{4}+32 r^{3}+152 r^{2}+22 r -15}{\left (4 r^{2}+12 r -7\right ) \left (4 r^{2}-16 r +15\right )}\) | \(-{\frac {663}{28}}\) |
| \(a_{4}\) | \(\frac {256 r^{4}+256 r^{3}-544 r^{2}-304 r +105}{16 r^{4}-232 r^{2}+384 r -135}\) | \(\frac {13923}{256}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-1024 r^{5}-3840 r^{4}-640 r^{3}+7200 r^{2}+2924 r -1155}{32 r^{5}+80 r^{4}-560 r^{3}+40 r^{2}+1098 r -495} \] Which for the root \(r = {\frac {7}{2}}\) becomes \[ a_{5}=-{\frac {7735}{64}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-8 r^{2}+6 r +5}{4 r^{2}-4 r -15}\) | \(-{\frac {18}{5}}\) |
| \(a_{2}\) | \(\frac {32 r^{3}-32 r^{2}-14 r +5}{8 r^{3}-12 r^{2}-50 r +75}\) | \(\frac {39}{4}\) |
| \(a_{3}\) | \(\frac {-128 r^{4}+32 r^{3}+152 r^{2}+22 r -15}{\left (4 r^{2}+12 r -7\right ) \left (4 r^{2}-16 r +15\right )}\) | \(-{\frac {663}{28}}\) |
| \(a_{4}\) | \(\frac {256 r^{4}+256 r^{3}-544 r^{2}-304 r +105}{16 r^{4}-232 r^{2}+384 r -135}\) | \(\frac {13923}{256}\) |
| \(a_{5}\) | \(\frac {-1024 r^{5}-3840 r^{4}-640 r^{3}+7200 r^{2}+2924 r -1155}{32 r^{5}+80 r^{4}-560 r^{3}+40 r^{2}+1098 r -495}\) | \(-{\frac {7735}{64}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{{7}/{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}\dots \right ) \\ &= x^{{7}/{2}} \left (1-\frac {18 x}{5}+\frac {39 x^{2}}{4}-\frac {663 x^{3}}{28}+\frac {13923 x^{4}}{256}-\frac {7735 x^{5}}{64}+O\left (x^{6}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Let \[ r_{1}-r_{2} = N \] Where \(N\) is positive integer which is the difference between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=4\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{4}\left (r \right )\). If this limit exists, then \(C = 0\), else we need to keep the log term and \(C \neq 0\). The above table shows that \begin {align*} a_N &= a_{4} \\ &= \frac {256 r^{4}+256 r^{3}-544 r^{2}-304 r +105}{16 r^{4}-232 r^{2}+384 r -135} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {256 r^{4}+256 r^{3}-544 r^{2}-304 r +105}{16 r^{4}-232 r^{2}+384 r -135}&= \lim _{r\rightarrow -{\frac {1}{2}}}\frac {256 r^{4}+256 r^{3}-544 r^{2}-304 r +105}{16 r^{4}-232 r^{2}+384 r -135}\\ &= -{\frac {35}{128}} \end {align*}
The limit is \(-{\frac {35}{128}}\). Since the limit exists then the log term is not needed and we can set \(C = 0\). Therefore the second solution has the form \begin {align*} y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r}\\ &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {1}{2}} \end {align*}
Eq (3) derived above is 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 \(b_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{4} 8 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+4 b_{n} \left (n +r \right ) \left (n +r -1\right )+2 b_{n -1} \left (n +r -1\right )-8 b_{n} \left (n +r \right )-5 b_{n -1}-7 b_{n} = 0 \end{equation} Which for for the root \(r = -{\frac {1}{2}}\) becomes \begin{equation} \tag{4A} 8 b_{n -1} \left (n -\frac {3}{2}\right ) \left (n -\frac {5}{2}\right )+4 b_{n} \left (n -\frac {1}{2}\right ) \left (n -\frac {3}{2}\right )+2 b_{n -1} \left (n -\frac {3}{2}\right )-8 b_{n} \left (n -\frac {1}{2}\right )-5 b_{n -1}-7 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = -\frac {b_{n -1} \left (8 n^{2}+16 n r +8 r^{2}-22 n -22 r +9\right )}{4 n^{2}+8 n r +4 r^{2}-12 n -12 r -7}\tag {5} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{n} = -\frac {b_{n -1} \left (8 n^{2}-30 n +22\right )}{4 n^{2}-16 n}\tag {6} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -{\frac {1}{2}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ b_{1}=-\frac {8 r^{2}-6 r -5}{4 r^{2}-4 r -15} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{1}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-8 r^{2}+6 r +5}{4 r^{2}-4 r -15}\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {32 r^{3}-32 r^{2}-14 r +5}{\left (4 r^{2}+4 r -15\right ) \left (2 r -5\right )} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{2}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-8 r^{2}+6 r +5}{4 r^{2}-4 r -15}\) | \(0\) |
| \(b_{2}\) | \(\frac {32 r^{3}-32 r^{2}-14 r +5}{8 r^{3}-12 r^{2}-50 r +75}\) | \(0\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {128 r^{4}-32 r^{3}-152 r^{2}-22 r +15}{\left (2 r -5\right ) \left (2 r -3\right ) \left (4 r^{2}+12 r -7\right )} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{3}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-8 r^{2}+6 r +5}{4 r^{2}-4 r -15}\) | \(0\) |
| \(b_{2}\) | \(\frac {32 r^{3}-32 r^{2}-14 r +5}{8 r^{3}-12 r^{2}-50 r +75}\) | \(0\) |
| \(b_{3}\) | \(\frac {-128 r^{4}+32 r^{3}+152 r^{2}+22 r -15}{16 r^{4}-16 r^{3}-160 r^{2}+292 r -105}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {256 r^{4}+256 r^{3}-544 r^{2}-304 r +105}{\left (2 r +9\right ) \left (2 r -1\right ) \left (2 r -5\right ) \left (2 r -3\right )} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{4}=-{\frac {35}{128}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-8 r^{2}+6 r +5}{4 r^{2}-4 r -15}\) | \(0\) |
| \(b_{2}\) | \(\frac {32 r^{3}-32 r^{2}-14 r +5}{8 r^{3}-12 r^{2}-50 r +75}\) | \(0\) |
| \(b_{3}\) | \(\frac {-128 r^{4}+32 r^{3}+152 r^{2}+22 r -15}{16 r^{4}-16 r^{3}-160 r^{2}+292 r -105}\) | \(0\) |
| \(b_{4}\) | \(\frac {256 r^{4}+256 r^{3}-544 r^{2}-304 r +105}{16 r^{4}-232 r^{2}+384 r -135}\) | \(-{\frac {35}{128}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {1024 r^{5}+3840 r^{4}+640 r^{3}-7200 r^{2}-2924 r +1155}{\left (2 r -1\right ) \left (2 r -5\right ) \left (2 r -3\right ) \left (4 r^{2}+28 r +33\right )} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{5}={\frac {63}{64}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-8 r^{2}+6 r +5}{4 r^{2}-4 r -15}\) | \(0\) |
| \(b_{2}\) | \(\frac {32 r^{3}-32 r^{2}-14 r +5}{8 r^{3}-12 r^{2}-50 r +75}\) | \(0\) |
| \(b_{3}\) | \(\frac {-128 r^{4}+32 r^{3}+152 r^{2}+22 r -15}{16 r^{4}-16 r^{3}-160 r^{2}+292 r -105}\) | \(0\) |
| \(b_{4}\) | \(\frac {256 r^{4}+256 r^{3}-544 r^{2}-304 r +105}{16 r^{4}-232 r^{2}+384 r -135}\) | \(-{\frac {35}{128}}\) |
| \(b_{5}\) | \(\frac {-1024 r^{5}-3840 r^{4}-640 r^{3}+7200 r^{2}+2924 r -1155}{32 r^{5}+80 r^{4}-560 r^{3}+40 r^{2}+1098 r -495}\) | \(\frac {63}{64}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{{7}/{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}\dots \right ) \\ &= \frac {1-\frac {35 x^{4}}{128}+\frac {63 x^{5}}{64}+O\left (x^{6}\right )}{\sqrt {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 ) \\ &= c_{1} x^{{7}/{2}} \left (1-\frac {18 x}{5}+\frac {39 x^{2}}{4}-\frac {663 x^{3}}{28}+\frac {13923 x^{4}}{256}-\frac {7735 x^{5}}{64}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1-\frac {35 x^{4}}{128}+\frac {63 x^{5}}{64}+O\left (x^{6}\right )\right )}{\sqrt {x}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{{7}/{2}} \left (1-\frac {18 x}{5}+\frac {39 x^{2}}{4}-\frac {663 x^{3}}{28}+\frac {13923 x^{4}}{256}-\frac {7735 x^{5}}{64}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {35 x^{4}}{128}+\frac {63 x^{5}}{64}+O\left (x^{6}\right )\right )}{\sqrt {x}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{{7}/{2}} \left (1-\frac {18 x}{5}+\frac {39 x^{2}}{4}-\frac {663 x^{3}}{28}+\frac {13923 x^{4}}{256}-\frac {7735 x^{5}}{64}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {35 x^{4}}{128}+\frac {63 x^{5}}{64}+O\left (x^{6}\right )\right )}{\sqrt {x}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{{7}/{2}} \left (1-\frac {18 x}{5}+\frac {39 x^{2}}{4}-\frac {663 x^{3}}{28}+\frac {13923 x^{4}}{256}-\frac {7735 x^{5}}{64}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {35 x^{4}}{128}+\frac {63 x^{5}}{64}+O\left (x^{6}\right )\right )}{\sqrt {x}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 x^{2} \left (1+2 x \right ) y^{\prime \prime }+\left (2 x^{2}-8 x \right ) y^{\prime }+\left (-5 x -7\right ) 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 {\left (7+5 x \right ) y}{4 x^{2} \left (1+2 x \right )}-\frac {\left (x -4\right ) y^{\prime }}{2 x \left (1+2 x \right )} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+\frac {\left (x -4\right ) y^{\prime }}{2 x \left (1+2 x \right )}-\frac {\left (7+5 x \right ) y}{4 x^{2} \left (1+2 x \right )}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {x -4}{2 x \left (1+2 x \right )}, P_{3}\left (x \right )=-\frac {7+5 x}{4 x^{2} \left (1+2 x \right )}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=-2 \\ {} & \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 {7}{4} \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 4 x^{2} \left (1+2 x \right ) y^{\prime \prime }+2 x \left (x -4\right ) y^{\prime }+\left (-5 x -7\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^{m}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..3 \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (1+2 r \right ) \left (-7+2 r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (2 k +2 r +1\right ) \left (2 k +2 r -7\right )+a_{k -1} \left (2 k -1+2 r \right ) \left (4 k -9+4 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 ) \left (-7+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {1}{2}, \frac {7}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 8 \left (k -\frac {1}{2}+r \right ) \left (k -\frac {9}{4}+r \right ) a_{k -1}+4 \left (k +r +\frac {1}{2}\right ) \left (k +r -\frac {7}{2}\right ) a_{k}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 8 \left (k +r +\frac {1}{2}\right ) \left (k -\frac {5}{4}+r \right ) a_{k}+4 \left (k +\frac {3}{2}+r \right ) \left (k -\frac {5}{2}+r \right ) a_{k +1}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {\left (2 k +2 r +1\right ) \left (4 k +4 r -5\right ) a_{k}}{\left (2 k +3+2 r \right ) \left (2 k -5+2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{2} \\ {} & {} & a_{k +1}=-\frac {2 k \left (4 k -7\right ) a_{k}}{\left (2 k +2\right ) \left (2 k -6\right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =-\frac {1}{2}\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =3 \\ {} & {} & a_{k +1}=-\frac {2 k \left (4 k -7\right ) a_{k}}{\left (2 k +2\right ) \left (2 k -6\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {7}{2} \\ {} & {} & a_{k +1}=-\frac {\left (2 k +8\right ) \left (4 k +9\right ) a_{k}}{\left (2 k +10\right ) \left (2 k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {7}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {7}{2}}, a_{k +1}=-\frac {\left (2 k +8\right ) \left (4 k +9\right ) a_{k}}{\left (2 k +10\right ) \left (2 k +2\right )}\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) Reducible group (found another exponential solution) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 41
Order:=6; dsolve(4*x^2*(1+2*x)*diff(y(x),x$2)-2*x*(4-x)*diff(y(x),x)-(7+5*x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{1} x^{4} \left (1-\frac {18}{5} x +\frac {39}{4} x^{2}-\frac {663}{28} x^{3}+\frac {13923}{256} x^{4}-\frac {7735}{64} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (-144-\frac {405}{8} x^{4}+\frac {729}{4} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{\sqrt {x}} \]
✓ Solution by Mathematica
Time used: 0.069 (sec). Leaf size: 67
AsymptoticDSolveValue[4*x^2*(1+2*x)*y''[x]-2*x*(4-x)*y'[x]-(7+5*x)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {1}{\sqrt {x}}-\frac {35 x^{7/2}}{128}\right )+c_2 \left (\frac {13923 x^{15/2}}{256}-\frac {663 x^{13/2}}{28}+\frac {39 x^{11/2}}{4}-\frac {18 x^{9/2}}{5}+x^{7/2}\right ) \]