Internal problem ID [6969]
Internal file name [OUTPUT/6212_Friday_August_12_2022_11_05_34_PM_18887529/index.tex]
Book: Elementary differential equations. Rainville, Bedient, Bedient. Prentice Hall. NJ. 8th
edition. 1997.
Section: CHAPTER 18. Power series solutions. 18.8 Indicial Equation with Difference of Roots a
Positive Integer: Nonlogarithmic Case. Exercises page 380
Problem number: 5.
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]]
\[ \boxed {x \left (1+x \right ) y^{\prime \prime }+\left (x +5\right ) y^{\prime }-4 y=0} \] 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. \[ \left (x^{2}+x \right ) y^{\prime \prime }+\left (x +5\right ) y^{\prime }-4 y = 0 \] 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 +5}{x \left (1+x \right )}\\ q(x) &= -\frac {4}{x \left (1+x \right )}\\ \end {align*}
| \(p(x)=\frac {x +5}{x \left (1+x \right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=-\frac {4}{x \left (1+x \right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-1, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x \left (1+x \right ) y^{\prime \prime }+\left (x +5\right ) y^{\prime }-4 y = 0 \] 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 \left (1+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 (x +5\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )-4 \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 }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 a_{n} x^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 a_{n} x^{n +r}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 a_{n -1} x^{n +r -1}\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 \(n +r -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+5 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ x^{-1+r} a_{0} r \left (-1+r \right )+5 r a_{0} x^{-1+r} = 0 \] Or \[ \left (x^{-1+r} r \left (-1+r \right )+5 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (4+r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r \left (4+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= -4 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (4+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([0, -4]\).
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 ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\\ 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}}{x^{4}} \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\\ 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 -4}\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} a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -1} \left (n +r -1\right )+5 a_{n} \left (n +r \right )-4 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (n^{2}+2 n r +r^{2}-2 n -2 r -3\right )}{n^{2}+2 n r +r^{2}+4 n +4 r}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = -\frac {a_{n -1} \left (n^{2}-2 n -3\right )}{n \left (n +4\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 0\) 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 \[ a_{1}=\frac {-r^{2}+4}{r^{2}+6 r +5} \] Which for the root \(r = 0\) becomes \[ a_{1}={\frac {4}{5}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+4}{r^{2}+6 r +5}\) | \(\frac {4}{5}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {r^{3}-7 r +6}{r^{3}+12 r^{2}+41 r +30} \] Which for the root \(r = 0\) becomes \[ a_{2}={\frac {1}{5}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+4}{r^{2}+6 r +5}\) | \(\frac {4}{5}\) |
| \(a_{2}\) | \(\frac {r^{3}-7 r +6}{r^{3}+12 r^{2}+41 r +30}\) | \(\frac {1}{5}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {\left (4+r \right ) r \left (-1+r \right ) \left (r -2\right )}{\left (r +7\right ) \left (r +6\right ) \left (r +5\right ) \left (r +1\right )} \] Which for the root \(r = 0\) becomes \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+4}{r^{2}+6 r +5}\) | \(\frac {4}{5}\) |
| \(a_{2}\) | \(\frac {r^{3}-7 r +6}{r^{3}+12 r^{2}+41 r +30}\) | \(\frac {1}{5}\) |
| \(a_{3}\) | \(-\frac {\left (4+r \right ) r \left (-1+r \right ) \left (r -2\right )}{\left (r +7\right ) \left (r +6\right ) \left (r +5\right ) \left (r +1\right )}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {\left (r -2\right ) r \left (-1+r \right )}{\left (r +8\right ) \left (r +6\right ) \left (r +7\right )} \] Which for the root \(r = 0\) becomes \[ a_{4}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+4}{r^{2}+6 r +5}\) | \(\frac {4}{5}\) |
| \(a_{2}\) | \(\frac {r^{3}-7 r +6}{r^{3}+12 r^{2}+41 r +30}\) | \(\frac {1}{5}\) |
| \(a_{3}\) | \(-\frac {\left (4+r \right ) r \left (-1+r \right ) \left (r -2\right )}{\left (r +7\right ) \left (r +6\right ) \left (r +5\right ) \left (r +1\right )}\) | \(0\) |
| \(a_{4}\) | \(\frac {\left (r -2\right ) r \left (-1+r \right )}{\left (r +8\right ) \left (r +6\right ) \left (r +7\right )}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {r \left (-1+r \right ) \left (r^{2}-4\right )}{\left (r^{2}+14 r +45\right ) \left (r +7\right ) \left (r +8\right )} \] Which for the root \(r = 0\) becomes \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+4}{r^{2}+6 r +5}\) | \(\frac {4}{5}\) |
| \(a_{2}\) | \(\frac {r^{3}-7 r +6}{r^{3}+12 r^{2}+41 r +30}\) | \(\frac {1}{5}\) |
| \(a_{3}\) | \(-\frac {\left (4+r \right ) r \left (-1+r \right ) \left (r -2\right )}{\left (r +7\right ) \left (r +6\right ) \left (r +5\right ) \left (r +1\right )}\) | \(0\) |
| \(a_{4}\) | \(\frac {\left (r -2\right ) r \left (-1+r \right )}{\left (r +8\right ) \left (r +6\right ) \left (r +7\right )}\) | \(0\) |
| \(a_{5}\) | \(-\frac {r \left (-1+r \right ) \left (r^{2}-4\right )}{\left (r^{2}+14 r +45\right ) \left (r +7\right ) \left (r +8\right )}\) | \(0\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {\left (r +3\right ) r \left (-1+r \right ) \left (r^{2}-4\right )}{\left (r +8\right ) \left (r^{2}+14 r +45\right ) \left (r^{2}+16 r +60\right )} \] Which for the root \(r = 0\) becomes \[ a_{6}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+4}{r^{2}+6 r +5}\) | \(\frac {4}{5}\) |
| \(a_{2}\) | \(\frac {r^{3}-7 r +6}{r^{3}+12 r^{2}+41 r +30}\) | \(\frac {1}{5}\) |
| \(a_{3}\) | \(-\frac {\left (4+r \right ) r \left (-1+r \right ) \left (r -2\right )}{\left (r +7\right ) \left (r +6\right ) \left (r +5\right ) \left (r +1\right )}\) | \(0\) |
| \(a_{4}\) | \(\frac {\left (r -2\right ) r \left (-1+r \right )}{\left (r +8\right ) \left (r +6\right ) \left (r +7\right )}\) | \(0\) |
| \(a_{5}\) | \(-\frac {r \left (-1+r \right ) \left (r^{2}-4\right )}{\left (r^{2}+14 r +45\right ) \left (r +7\right ) \left (r +8\right )}\) | \(0\) |
| \(a_{6}\) | \(\frac {\left (r +3\right ) r \left (-1+r \right ) \left (r^{2}-4\right )}{\left (r +8\right ) \left (r^{2}+14 r +45\right ) \left (r^{2}+16 r +60\right )}\) | \(0\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {-r^{6}-6 r^{5}-r^{4}+36 r^{3}+20 r^{2}-48 r}{r^{6}+48 r^{5}+946 r^{4}+9792 r^{3}+56113 r^{2}+168720 r +207900} \] Which for the root \(r = 0\) becomes \[ a_{7}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+4}{r^{2}+6 r +5}\) | \(\frac {4}{5}\) |
| \(a_{2}\) | \(\frac {r^{3}-7 r +6}{r^{3}+12 r^{2}+41 r +30}\) | \(\frac {1}{5}\) |
| \(a_{3}\) | \(-\frac {\left (4+r \right ) r \left (-1+r \right ) \left (r -2\right )}{\left (r +7\right ) \left (r +6\right ) \left (r +5\right ) \left (r +1\right )}\) | \(0\) |
| \(a_{4}\) | \(\frac {\left (r -2\right ) r \left (-1+r \right )}{\left (r +8\right ) \left (r +6\right ) \left (r +7\right )}\) | \(0\) |
| \(a_{5}\) | \(-\frac {r \left (-1+r \right ) \left (r^{2}-4\right )}{\left (r^{2}+14 r +45\right ) \left (r +7\right ) \left (r +8\right )}\) | \(0\) |
| \(a_{6}\) | \(\frac {\left (r +3\right ) r \left (-1+r \right ) \left (r^{2}-4\right )}{\left (r +8\right ) \left (r^{2}+14 r +45\right ) \left (r^{2}+16 r +60\right )}\) | \(0\) |
| \(a_{7}\) | \(\frac {-r^{6}-6 r^{5}-r^{4}+36 r^{3}+20 r^{2}-48 r}{r^{6}+48 r^{5}+946 r^{4}+9792 r^{3}+56113 r^{2}+168720 r +207900}\) | \(0\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= 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 \\ &= 1+\frac {4 x}{5}+\frac {x^{2}}{5}+O\left (x^{8}\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 {\left (r -2\right ) r \left (-1+r \right )}{\left (r +8\right ) \left (r +6\right ) \left (r +7\right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {\left (r -2\right ) r \left (-1+r \right )}{\left (r +8\right ) \left (r +6\right ) \left (r +7\right )}&= \lim _{r\rightarrow -4}\frac {\left (r -2\right ) r \left (-1+r \right )}{\left (r +8\right ) \left (r +6\right ) \left (r +7\right )}\\ &= -5 \end {align*}
The limit is \(-5\). 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 -4} \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} b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+b_{n} \left (n +r \right ) \left (n +r -1\right )+b_{n -1} \left (n +r -1\right )+5 \left (n +r \right ) b_{n}-4 b_{n -1} = 0 \end{equation} Which for for the root \(r = -4\) becomes \begin{equation} \tag{4A} b_{n -1} \left (n -5\right ) \left (n -6\right )+b_{n} \left (n -4\right ) \left (n -5\right )+b_{n -1} \left (n -5\right )+5 \left (n -4\right ) b_{n}-4 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = -\frac {b_{n -1} \left (n^{2}+2 n r +r^{2}-2 n -2 r -3\right )}{n^{2}+2 n r +r^{2}+4 n +4 r}\tag {5} \] Which for the root \(r = -4\) becomes \[ b_{n} = -\frac {b_{n -1} \left (n^{2}-10 n +21\right )}{n^{2}-4 n}\tag {6} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -4\) 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 \[ b_{1}=-\frac {r^{2}-4}{r^{2}+6 r +5} \] Which for the root \(r = -4\) becomes \[ b_{1}=4 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}+4}{r^{2}+6 r +5}\) | \(4\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {r^{3}-7 r +6}{\left (r +6\right ) \left (r^{2}+6 r +5\right )} \] Which for the root \(r = -4\) becomes \[ b_{2}=5 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}+4}{r^{2}+6 r +5}\) | \(4\) |
| \(b_{2}\) | \(\frac {r^{3}-7 r +6}{r^{3}+12 r^{2}+41 r +30}\) | \(5\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {\left (4+r \right ) r \left (r^{2}-3 r +2\right )}{\left (r +7\right ) \left (r +6\right ) \left (r^{2}+6 r +5\right )} \] Which for the root \(r = -4\) becomes \[ b_{3}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}+4}{r^{2}+6 r +5}\) | \(4\) |
| \(b_{2}\) | \(\frac {r^{3}-7 r +6}{r^{3}+12 r^{2}+41 r +30}\) | \(5\) |
| \(b_{3}\) | \(-\frac {\left (4+r \right ) r \left (-1+r \right ) \left (r -2\right )}{\left (r +7\right ) \left (r +6\right ) \left (r +5\right ) \left (r +1\right )}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {r \left (r^{2}-3 r +2\right )}{\left (r +8\right ) \left (r +6\right ) \left (r +7\right )} \] Which for the root \(r = -4\) becomes \[ b_{4}=-5 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}+4}{r^{2}+6 r +5}\) | \(4\) |
| \(b_{2}\) | \(\frac {r^{3}-7 r +6}{r^{3}+12 r^{2}+41 r +30}\) | \(5\) |
| \(b_{3}\) | \(-\frac {\left (4+r \right ) r \left (-1+r \right ) \left (r -2\right )}{\left (r +7\right ) \left (r +6\right ) \left (r +5\right ) \left (r +1\right )}\) | \(0\) |
| \(b_{4}\) | \(\frac {r \left (r^{2}-3 r +2\right )}{\left (r +8\right ) \left (r +6\right ) \left (r +7\right )}\) | \(-5\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {\left (r +2\right ) r \left (r^{2}-3 r +2\right )}{\left (r^{2}+14 r +45\right ) \left (r +7\right ) \left (r +8\right )} \] Which for the root \(r = -4\) becomes \[ b_{5}=-4 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}+4}{r^{2}+6 r +5}\) | \(4\) |
| \(b_{2}\) | \(\frac {r^{3}-7 r +6}{r^{3}+12 r^{2}+41 r +30}\) | \(5\) |
| \(b_{3}\) | \(-\frac {\left (4+r \right ) r \left (-1+r \right ) \left (r -2\right )}{\left (r +7\right ) \left (r +6\right ) \left (r +5\right ) \left (r +1\right )}\) | \(0\) |
| \(b_{4}\) | \(\frac {r \left (r^{2}-3 r +2\right )}{\left (r +8\right ) \left (r +6\right ) \left (r +7\right )}\) | \(-5\) |
| \(b_{5}\) | \(-\frac {r \left (-1+r \right ) \left (r -2\right ) \left (r +2\right )}{\left (9+r \right ) \left (r +5\right ) \left (r +7\right ) \left (r +8\right )}\) | \(-4\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=\frac {\left (r +3\right ) \left (r +2\right ) r \left (r^{2}-3 r +2\right )}{\left (r +8\right ) \left (r^{2}+14 r +45\right ) \left (r^{2}+16 r +60\right )} \] Which for the root \(r = -4\) becomes \[ b_{6}=-1 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}+4}{r^{2}+6 r +5}\) | \(4\) |
| \(b_{2}\) | \(\frac {r^{3}-7 r +6}{r^{3}+12 r^{2}+41 r +30}\) | \(5\) |
| \(b_{3}\) | \(-\frac {\left (4+r \right ) r \left (-1+r \right ) \left (r -2\right )}{\left (r +7\right ) \left (r +6\right ) \left (r +5\right ) \left (r +1\right )}\) | \(0\) |
| \(b_{4}\) | \(\frac {r \left (r^{2}-3 r +2\right )}{\left (r +8\right ) \left (r +6\right ) \left (r +7\right )}\) | \(-5\) |
| \(b_{5}\) | \(-\frac {r \left (-1+r \right ) \left (r -2\right ) \left (r +2\right )}{\left (9+r \right ) \left (r +5\right ) \left (r +7\right ) \left (r +8\right )}\) | \(-4\) |
| \(b_{6}\) | \(\frac {\left (r +3\right ) r \left (-1+r \right ) \left (r -2\right ) \left (r +2\right )}{\left (10+r \right ) \left (r +6\right ) \left (r +8\right ) \left (r +5\right ) \left (9+r \right )}\) | \(-1\) |
For \(n = 7\), using the above recursive equation gives \[ b_{7}=-\frac {\left (4+r \right ) \left (r +3\right ) \left (r +2\right ) r \left (r^{2}-3 r +2\right )}{\left (r^{2}+16 r +60\right ) \left (r^{2}+14 r +45\right ) \left (r^{2}+18 r +77\right )} \] Which for the root \(r = -4\) becomes \[ b_{7}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}+4}{r^{2}+6 r +5}\) | \(4\) |
| \(b_{2}\) | \(\frac {r^{3}-7 r +6}{r^{3}+12 r^{2}+41 r +30}\) | \(5\) |
| \(b_{3}\) | \(-\frac {\left (4+r \right ) r \left (-1+r \right ) \left (r -2\right )}{\left (r +7\right ) \left (r +6\right ) \left (r +5\right ) \left (r +1\right )}\) | \(0\) |
| \(b_{4}\) | \(\frac {r \left (r^{2}-3 r +2\right )}{\left (r +8\right ) \left (r +6\right ) \left (r +7\right )}\) | \(-5\) |
| \(b_{5}\) | \(-\frac {r \left (-1+r \right ) \left (r -2\right ) \left (r +2\right )}{\left (9+r \right ) \left (r +5\right ) \left (r +7\right ) \left (r +8\right )}\) | \(-4\) |
| \(b_{6}\) | \(\frac {\left (r +3\right ) r \left (-1+r \right ) \left (r -2\right ) \left (r +2\right )}{\left (10+r \right ) \left (r +6\right ) \left (r +8\right ) \left (r +5\right ) \left (9+r \right )}\) | \(-1\) |
| \(b_{7}\) | \(-\frac {\left (4+r \right ) \left (r +3\right ) \left (r +2\right ) r \left (r^{2}-3 r +2\right )}{\left (r^{2}+16 r +60\right ) \left (r^{2}+14 r +45\right ) \left (r^{2}+18 r +77\right )}\) | \(0\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= 1 \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 ) \\ &= \frac {1+4 x +5 x^{2}-5 x^{4}-4 x^{5}-x^{6}+O\left (x^{8}\right )}{x^{4}} \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} \left (1+\frac {4 x}{5}+\frac {x^{2}}{5}+O\left (x^{8}\right )\right ) + \frac {c_{2} \left (1+4 x +5 x^{2}-5 x^{4}-4 x^{5}-x^{6}+O\left (x^{8}\right )\right )}{x^{4}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (1+\frac {4 x}{5}+\frac {x^{2}}{5}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1+4 x +5 x^{2}-5 x^{4}-4 x^{5}-x^{6}+O\left (x^{8}\right )\right )}{x^{4}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (1+\frac {4 x}{5}+\frac {x^{2}}{5}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1+4 x +5 x^{2}-5 x^{4}-4 x^{5}-x^{6}+O\left (x^{8}\right )\right )}{x^{4}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1+\frac {4 x}{5}+\frac {x^{2}}{5}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1+4 x +5 x^{2}-5 x^{4}-4 x^{5}-x^{6}+O\left (x^{8}\right )\right )}{x^{4}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (1+x \right ) y^{\prime \prime }+\left (x +5\right ) y^{\prime }-4 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 {4 y}{x \left (1+x \right )}-\frac {\left (x +5\right ) y^{\prime }}{x \left (1+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 +5\right ) y^{\prime }}{x \left (1+x \right )}-\frac {4 y}{x \left (1+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 +5}{x \left (1+x \right )}, P_{3}\left (x \right )=-\frac {4}{x \left (1+x \right )}\right ] \\ {} & \circ & \left (1+x \right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (1+x \right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=-4 \\ {} & \circ & \left (1+x \right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (1+x \right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=0 \\ {} & \circ & x =-1\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}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x \left (1+x \right ) y^{\prime \prime }+\left (x +5\right ) y^{\prime }-4 y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{2}-u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (u +4\right ) \left (\frac {d}{d u}y \left (u \right )\right )-4 y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} r \left (-5+r \right ) u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-a_{k +1} \left (k +1+r \right ) \left (k -4+r \right )+a_{k} \left (k +r +2\right ) \left (k +r -2\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (-5+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 5\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -a_{k +1} \left (k +1+r \right ) \left (k -4+r \right )+a_{k} \left (k +r +2\right ) \left (k +r -2\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +r +2\right ) \left (k +r -2\right )}{\left (k +1+r \right ) \left (k -4+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =2 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +2\right ) \left (k -2\right )}{\left (k +1\right ) \left (k -4\right )} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=a_{0} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =1 \\ {} & {} & a_{2}=\frac {a_{1}}{2} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{2}=\frac {a_{0}}{2} \\ \bullet & {} & \textrm {Terminating series solution of the ODE for}\hspace {3pt} r =0\hspace {3pt}\textrm {. Use reduction of order to find the second linearly independent solution}\hspace {3pt} \\ {} & {} & y \left (u \right )=a_{0}\cdot \left (1+u +\frac {1}{2} u^{2}\right ) \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =1+x \\ {} & {} & \left [y=a_{0} \left (\frac {5}{2}+2 x +\frac {1}{2} x^{2}\right )\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =5 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +7\right ) \left (k +3\right )}{\left (k +6\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =5 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +5}, a_{k +1}=\frac {a_{k} \left (k +7\right ) \left (k +3\right )}{\left (k +6\right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =1+x \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (1+x \right )^{k +5}, a_{k +1}=\frac {a_{k} \left (k +7\right ) \left (k +3\right )}{\left (k +6\right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=a_{0} \left (\frac {5}{2}+2 x +\frac {1}{2} x^{2}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (1+x \right )^{k +5}\right ), b_{k +1}=\frac {b_{k} \left (k +7\right ) \left (k +3\right )}{\left (k +6\right ) \left (k +1\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) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 38
Order:=8; dsolve(x*(1+x)*diff(y(x),x$2)+(x+5)*diff(y(x),x)-4*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} \left (1+\frac {4}{5} x +\frac {1}{5} x^{2}+\operatorname {O}\left (x^{8}\right )\right )+\frac {c_{2} \left (-144-576 x -720 x^{2}+720 x^{4}+576 x^{5}+144 x^{6}+\operatorname {O}\left (x^{8}\right )\right )}{x^{4}} \]
✓ Solution by Mathematica
Time used: 0.13 (sec). Leaf size: 47
AsymptoticDSolveValue[x*(1+x)*y''[x]+(x+5)*y'[x]-4*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_2 \left (\frac {x^2}{5}+\frac {4 x}{5}+1\right )+c_1 \left (\frac {1}{x^4}+\frac {4}{x^3}-x^2+\frac {5}{x^2}-4 x-5\right ) \]