Internal problem ID [1337]
Internal file name [OUTPUT/1338_Sunday_June_05_2022_02_10_58_AM_50909819/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. 7.5 THE METHOD OF
FROBENIUS I. Exercises 7.5. Page 358
Problem number: 48.
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, _exact, _linear, _homogeneous]]
\[ \boxed {x^{2} \left (x^{2}+8\right ) y^{\prime \prime }+7 x \left (x^{2}+2\right ) y^{\prime }-\left (-9 x^{2}+2\right ) y=0} \] 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. \[ \left (x^{4}+8 x^{2}\right ) y^{\prime \prime }+\left (7 x^{3}+14 x \right ) y^{\prime }+\left (9 x^{2}-2\right ) y = 0 \] 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 {7 x^{2}+14}{x \left (x^{2}+8\right )}\\ q(x) &= \frac {9 x^{2}-2}{x^{2} \left (x^{2}+8\right )}\\ \end {align*}
| \(p(x)=\frac {7 x^{2}+14}{x \left (x^{2}+8\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -2 i \sqrt {2}\) | \(\text {``regular''}\) |
| \(x = 2 i \sqrt {2}\) | \(\text {``regular''}\) |
| \(q(x)=\frac {9 x^{2}-2}{x^{2} \left (x^{2}+8\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -2 i \sqrt {2}\) | \(\text {``regular''}\) |
| \(x = 2 i \sqrt {2}\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, -2 i \sqrt {2}, 2 i \sqrt {2}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (x^{2}+8\right ) y^{\prime \prime }+\left (7 x^{3}+14 x \right ) y^{\prime }+\left (9 x^{2}-2\right ) y = 0 \] 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} x^{2} \left (x^{2}+8\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 (7 x^{3}+14 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (9 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 }}x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}8 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}7 x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}14 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 x^{n +r +2} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-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 }}x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}7 x^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}7 a_{n -2} \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}9 x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}9 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 =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}8 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}7 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}14 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}9 a_{n -2} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 8 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+14 x^{n +r} a_{n} \left (n +r \right )-2 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 8 x^{r} a_{0} r \left (-1+r \right )+14 x^{r} a_{0} r -2 a_{0} x^{r} = 0 \] Or \[ \left (8 x^{r} r \left (-1+r \right )+14 x^{r} r -2 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (8 r^{2}+6 r -2\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 8 r^{2}+6 r -2 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{4}}\\ r_2 &= -1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (8 r^{2}+6 r -2\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{4}}, -1\right ]\).
Since \(r_1 - r_2 = {\frac {5}{4}}\) 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 +\frac {1}{4}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -1} \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 \[ a_{1} = 0 \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+8 a_{n} \left (n +r \right ) \left (n +r -1\right )+7 a_{n -2} \left (n +r -2\right )+14 a_{n} \left (n +r \right )+9 a_{n -2}-2 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {\left (1+n +r \right ) a_{n -2}}{2 \left (4 n +4 r -1\right )}\tag {4} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ a_{n} = -\frac {\left (5+4 n \right ) a_{n -2}}{32 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{4}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {-3-r}{14+8 r} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ a_{2}=-{\frac {13}{64}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {-3-r}{14+8 r}\) | \(-{\frac {13}{64}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {-3-r}{14+8 r}\) | \(-{\frac {13}{64}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r^{2}+8 r +15}{64 r^{2}+352 r +420} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ a_{4}={\frac {273}{8192}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {-3-r}{14+8 r}\) | \(-{\frac {13}{64}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {r^{2}+8 r +15}{64 r^{2}+352 r +420}\) | \(\frac {273}{8192}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {-3-r}{14+8 r}\) | \(-{\frac {13}{64}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {r^{2}+8 r +15}{64 r^{2}+352 r +420}\) | \(\frac {273}{8192}\) |
| \(a_{5}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {1}{4}} \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^{\frac {1}{4}} \left (1-\frac {13 x^{2}}{64}+\frac {273 x^{4}}{8192}+O\left (x^{6}\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 \[ b_{1} = 0 \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} b_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+8 b_{n} \left (n +r \right ) \left (n +r -1\right )+7 b_{n -2} \left (n +r -2\right )+14 b_{n} \left (n +r \right )+9 b_{n -2}-2 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {\left (1+n +r \right ) b_{n -2}}{2 \left (4 n +4 r -1\right )}\tag {4} \] Which for the root \(r = -1\) becomes \[ b_{n} = -\frac {n b_{n -2}}{8 n -10}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -1\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {-3-r}{14+8 r} \] Which for the root \(r = -1\) becomes \[ b_{2}=-{\frac {1}{3}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(\frac {-3-r}{14+8 r}\) | \(-{\frac {1}{3}}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(\frac {-3-r}{14+8 r}\) | \(-{\frac {1}{3}}\) |
| \(b_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {r^{2}+8 r +15}{64 r^{2}+352 r +420} \] Which for the root \(r = -1\) becomes \[ b_{4}={\frac {2}{33}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(\frac {-3-r}{14+8 r}\) | \(-{\frac {1}{3}}\) |
| \(b_{3}\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {r^{2}+8 r +15}{64 r^{2}+352 r +420}\) | \(\frac {2}{33}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(\frac {-3-r}{14+8 r}\) | \(-{\frac {1}{3}}\) |
| \(b_{3}\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {r^{2}+8 r +15}{64 r^{2}+352 r +420}\) | \(\frac {2}{33}\) |
| \(b_{5}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{\frac {1}{4}} \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 {x^{2}}{3}+\frac {2 x^{4}}{33}+O\left (x^{6}\right )}{x} \end {align*}
Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} x^{\frac {1}{4}} \left (1-\frac {13 x^{2}}{64}+\frac {273 x^{4}}{8192}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1-\frac {x^{2}}{3}+\frac {2 x^{4}}{33}+O\left (x^{6}\right )\right )}{x} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {1}{4}} \left (1-\frac {13 x^{2}}{64}+\frac {273 x^{4}}{8192}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {x^{2}}{3}+\frac {2 x^{4}}{33}+O\left (x^{6}\right )\right )}{x} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {1}{4}} \left (1-\frac {13 x^{2}}{64}+\frac {273 x^{4}}{8192}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {x^{2}}{3}+\frac {2 x^{4}}{33}+O\left (x^{6}\right )\right )}{x} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {1}{4}} \left (1-\frac {13 x^{2}}{64}+\frac {273 x^{4}}{8192}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {x^{2}}{3}+\frac {2 x^{4}}{33}+O\left (x^{6}\right )\right )}{x} \] Verified OK.
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 35
Order:=6; dsolve(x^2*(8+x^2)*diff(y(x),x$2)+7*x*(2+x^2)*diff(y(x),x)-(2-9*x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{2} x^{\frac {5}{4}} \left (1-\frac {13}{64} x^{2}+\frac {273}{8192} x^{4}+\operatorname {O}\left (x^{6}\right )\right )+c_{1} \left (1-\frac {1}{3} x^{2}+\frac {2}{33} x^{4}+\operatorname {O}\left (x^{6}\right )\right )}{x} \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 50
AsymptoticDSolveValue[x^2*(8+x^2)*y''[x]+7*x*(2+x^2)*y'[x]-(2-9*x^2)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \sqrt [4]{x} \left (\frac {273 x^4}{8192}-\frac {13 x^2}{64}+1\right )+\frac {c_2 \left (\frac {2 x^4}{33}-\frac {x^2}{3}+1\right )}{x} \]