Internal problem ID [5599]
Internal file name [OUTPUT/4847_Sunday_June_05_2022_03_07_54_PM_87154698/index.tex]
Book: A FIRST COURSE IN DIFFERENTIAL EQUATIONS with Modeling Applications.
Dennis G. Zill. 9th edition. Brooks/Cole. CA, USA.
Section: Chapter 6. SERIES SOLUTIONS OF LINEAR EQUATIONS. Exercises. 6.3.1 page
250
Problem number: 9.
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]]
\[ \boxed {x^{2} y^{\prime \prime }+y^{\prime } x +\left (25 x^{2}-\frac {4}{9}\right ) y=0} \] 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. \[ x^{2} y^{\prime \prime }+y^{\prime } x +\left (25 x^{2}-\frac {4}{9}\right ) y = 0 \] 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 {1}{x}\\ q(x) &= \frac {225 x^{2}-4}{9 x^{2}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} y^{\prime \prime }+y^{\prime } x +\left (25 x^{2}-\frac {4}{9}\right ) y = 0 \] 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 (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x +\left (25 x^{2}-\frac {4}{9}\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} 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 }}25 x^{n +r +2} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {4 a_{n} x^{n +r}}{9}\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 }}25 x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}25 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 }}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} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}25 a_{n -2} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {4 a_{n} x^{n +r}}{9}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right )-\frac {4 a_{n} x^{n +r}}{9} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r -\frac {4 a_{0} x^{r}}{9} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+x^{r} r -\frac {4 x^{r}}{9}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \frac {\left (9 r^{2}-4\right ) x^{r}}{9} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}-\frac {4}{9} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {2}{3}}\\ r_2 &= -{\frac {2}{3}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \frac {\left (9 r^{2}-4\right ) x^{r}}{9} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {2}{3}}, -{\frac {2}{3}}\right ]\).
Since \(r_1 - r_2 = {\frac {4}{3}}\) 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 {2}{3}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {2}{3}} \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} \left (n +r \right ) \left (n +r -1\right )+a_{n} \left (n +r \right )+25 a_{n -2}-\frac {4 a_{n}}{9} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {225 a_{n -2}}{9 n^{2}+18 n r +9 r^{2}-4}\tag {4} \] Which for the root \(r = {\frac {2}{3}}\) becomes \[ a_{n} = -\frac {75 a_{n -2}}{n \left (3 n +4\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {2}{3}}\) 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 \[ a_{2}=-\frac {225}{9 r^{2}+36 r +32} \] Which for the root \(r = {\frac {2}{3}}\) becomes \[ a_{2}=-{\frac {15}{4}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {225}{9 r^{2}+36 r +32}\) | \(-{\frac {15}{4}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {225}{9 r^{2}+36 r +32}\) | \(-{\frac {15}{4}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {50625}{\left (9 r^{2}+36 r +32\right ) \left (9 r^{2}+72 r +140\right )} \] Which for the root \(r = {\frac {2}{3}}\) becomes \[ a_{4}={\frac {1125}{256}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {225}{9 r^{2}+36 r +32}\) | \(-{\frac {15}{4}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {50625}{\left (9 r^{2}+36 r +32\right ) \left (9 r^{2}+72 r +140\right )}\) | \(\frac {1125}{256}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {225}{9 r^{2}+36 r +32}\) | \(-{\frac {15}{4}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {50625}{\left (9 r^{2}+36 r +32\right ) \left (9 r^{2}+72 r +140\right )}\) | \(\frac {1125}{256}\) |
| \(a_{5}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {2}{3}} \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 {2}{3}} \left (1-\frac {15 x^{2}}{4}+\frac {1125 x^{4}}{256}+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} \left (n +r \right ) \left (n +r -1\right )+b_{n} \left (n +r \right )+25 b_{n -2}-\frac {4 b_{n}}{9} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {225 b_{n -2}}{9 n^{2}+18 n r +9 r^{2}-4}\tag {4} \] Which for the root \(r = -{\frac {2}{3}}\) becomes \[ b_{n} = -\frac {75 b_{n -2}}{n \left (3 n -4\right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -{\frac {2}{3}}\) 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 \[ b_{2}=-\frac {225}{9 r^{2}+36 r +32} \] Which for the root \(r = -{\frac {2}{3}}\) becomes \[ b_{2}=-{\frac {75}{4}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {225}{9 r^{2}+36 r +32}\) | \(-{\frac {75}{4}}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {225}{9 r^{2}+36 r +32}\) | \(-{\frac {75}{4}}\) |
| \(b_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {50625}{\left (9 r^{2}+36 r +32\right ) \left (9 r^{2}+72 r +140\right )} \] Which for the root \(r = -{\frac {2}{3}}\) becomes \[ b_{4}={\frac {5625}{128}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {225}{9 r^{2}+36 r +32}\) | \(-{\frac {75}{4}}\) |
| \(b_{3}\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {50625}{\left (9 r^{2}+36 r +32\right ) \left (9 r^{2}+72 r +140\right )}\) | \(\frac {5625}{128}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {225}{9 r^{2}+36 r +32}\) | \(-{\frac {75}{4}}\) |
| \(b_{3}\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {50625}{\left (9 r^{2}+36 r +32\right ) \left (9 r^{2}+72 r +140\right )}\) | \(\frac {5625}{128}\) |
| \(b_{5}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{\frac {2}{3}} \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 {75 x^{2}}{4}+\frac {5625 x^{4}}{128}+O\left (x^{6}\right )}{x^{\frac {2}{3}}} \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 {2}{3}} \left (1-\frac {15 x^{2}}{4}+\frac {1125 x^{4}}{256}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1-\frac {75 x^{2}}{4}+\frac {5625 x^{4}}{128}+O\left (x^{6}\right )\right )}{x^{\frac {2}{3}}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {2}{3}} \left (1-\frac {15 x^{2}}{4}+\frac {1125 x^{4}}{256}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {75 x^{2}}{4}+\frac {5625 x^{4}}{128}+O\left (x^{6}\right )\right )}{x^{\frac {2}{3}}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {2}{3}} \left (1-\frac {15 x^{2}}{4}+\frac {1125 x^{4}}{256}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {75 x^{2}}{4}+\frac {5625 x^{4}}{128}+O\left (x^{6}\right )\right )}{x^{\frac {2}{3}}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {2}{3}} \left (1-\frac {15 x^{2}}{4}+\frac {1125 x^{4}}{256}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {75 x^{2}}{4}+\frac {5625 x^{4}}{128}+O\left (x^{6}\right )\right )}{x^{\frac {2}{3}}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime }+y^{\prime } x +\left (25 x^{2}-\frac {4}{9}\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 (225 x^{2}-4\right ) y}{9 x^{2}}-\frac {y^{\prime }}{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} \\ {} & {} & y^{\prime \prime }+\frac {y^{\prime }}{x}+\frac {\left (225 x^{2}-4\right ) y}{9 x^{2}}=0 \\ \bullet & {} & \textrm {Simplify ODE}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime }+25 y x^{2}+y^{\prime } x -\frac {4 y}{9}=0 \\ \bullet & {} & \textrm {Make a change of variables}\hspace {3pt} \\ {} & {} & t =5 x \\ \bullet & {} & \textrm {Compute}\hspace {3pt} y^{\prime } \\ {} & {} & y^{\prime }=5 \frac {d}{d t}y \left (t \right ) \\ \bullet & {} & \textrm {Compute second derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=25 \frac {d^{2}}{d t^{2}}y \left (t \right ) \\ \bullet & {} & \textrm {Apply change of variables to the ODE}\hspace {3pt} \\ {} & {} & t^{2} \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )+y \left (t \right ) t^{2}+\left (\frac {d}{d t}y \left (t \right )\right ) t -\frac {4 y \left (t \right )}{9}=0 \\ \bullet & {} & \textrm {ODE is now of the Bessel form}\hspace {3pt} \\ \bullet & {} & \textrm {Solution to Bessel ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=c_{1} \mathit {BesselJ}\left (\frac {2}{3}, t\right )+c_{2} \mathit {BesselY}\left (\frac {2}{3}, t\right ) \\ \bullet & {} & \textrm {Make the change from}\hspace {3pt} t \hspace {3pt}\textrm {back to}\hspace {3pt} x \\ {} & {} & y=c_{1} \mathit {BesselJ}\left (\frac {2}{3}, 5 x \right )+c_{2} \mathit {BesselY}\left (\frac {2}{3}, 5 x \right ) \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel <- Bessel successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 35
Order:=6; dsolve(x^2*diff(y(x),x$2)+x*diff(y(x),x)+(25*x^2-4/9)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{2} x^{\frac {4}{3}} \left (1-\frac {15}{4} x^{2}+\frac {1125}{256} x^{4}+\operatorname {O}\left (x^{6}\right )\right )+c_{1} \left (1-\frac {75}{4} x^{2}+\frac {5625}{128} x^{4}+\operatorname {O}\left (x^{6}\right )\right )}{x^{\frac {2}{3}}} \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 52
AsymptoticDSolveValue[x^2*y''[x]+x*y'[x]+(25*x^2-4/9)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 x^{2/3} \left (\frac {1125 x^4}{256}-\frac {15 x^2}{4}+1\right )+\frac {c_2 \left (\frac {5625 x^4}{128}-\frac {75 x^2}{4}+1\right )}{x^{2/3}} \]