Internal problem ID [2381]
Internal file name [OUTPUT/2381_Tuesday_February_27_2024_08_36_35_AM_31564535/index.tex]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 41, page 195
Problem number: 2.
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]]
\[ \boxed {3 x \left (2+3 x \right ) y^{\prime \prime }-4 y^{\prime }+4 y=0} \] 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. \[ \left (9 x^{2}+6 x \right ) y^{\prime \prime }+4 y-4 y^{\prime } = 0 \] 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 {4}{3 x \left (2+3 x \right )}\\ q(x) &= \frac {4}{3 x \left (2+3 x \right )}\\ \end {align*}
| \(p(x)=-\frac {4}{3 x \left (2+3 x \right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -{\frac {2}{3}}\) | \(\text {``regular''}\) |
| \(q(x)=\frac {4}{3 x \left (2+3 x \right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -{\frac {2}{3}}\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, -{\frac {2}{3}}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 3 x \left (2+3 x \right ) y^{\prime \prime }-4 y^{\prime }+4 y = 0 \] 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} 3 x \left (2+3 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 )-4 \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 }}9 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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 }}9 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}9 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} x^{n +r -1} \\ \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 }}9 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 }}6 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}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 \[ 6 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )-4 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ 6 x^{-1+r} a_{0} r \left (-1+r \right )-4 r a_{0} x^{-1+r} = 0 \] Or \[ \left (6 x^{-1+r} r \left (-1+r \right )-4 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (6 r^{2}-10 r \right ) x^{-1+r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 6 r^{2}-10 r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {5}{3}}\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (6 r^{2}-10 r \right ) x^{-1+r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {5}{3}}, 0\right ]\).
Since \(r_1 - r_2 = {\frac {5}{3}}\) 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 {5}{3}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n} \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 \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 9 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+6 a_{n} \left (n +r \right ) \left (n +r -1\right )-4 a_{n} \left (n +r \right )+4 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (9 n^{2}+18 n r +9 r^{2}-27 n -27 r +22\right )}{2 \left (3 n^{2}+6 n r +3 r^{2}-5 n -5 r \right )}\tag {4} \] Which for the root \(r = {\frac {5}{3}}\) becomes \[ a_{n} = -\frac {a_{n -1} \left (9 n^{2}+3 n +2\right )}{6 n^{2}+10 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {5}{3}}\) 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 \[ a_{1}=\frac {-9 r^{2}+9 r -4}{6 r^{2}+2 r -4} \] Which for the root \(r = {\frac {5}{3}}\) becomes \[ a_{1}=-{\frac {7}{8}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-9 r^{2}+9 r -4}{6 r^{2}+2 r -4}\) | \(-{\frac {7}{8}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {81 r^{4}-9 r^{2}+16}{36 r^{4}+96 r^{3}+28 r^{2}-48 r -16} \] Which for the root \(r = {\frac {5}{3}}\) becomes \[ a_{2}={\frac {7}{8}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-9 r^{2}+9 r -4}{6 r^{2}+2 r -4}\) | \(-{\frac {7}{8}}\) |
| \(a_{2}\) | \(\frac {81 r^{4}-9 r^{2}+16}{36 r^{4}+96 r^{3}+28 r^{2}-48 r -16}\) | \(\frac {7}{8}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-729 r^{6}-2187 r^{5}-1701 r^{4}+243 r^{3}+54 r^{2}-432 r -352}{216 r^{6}+1512 r^{5}+3528 r^{4}+2744 r^{3}-672 r^{2}-1568 r -384} \] Which for the root \(r = {\frac {5}{3}}\) becomes \[ a_{3}=-{\frac {23}{24}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-9 r^{2}+9 r -4}{6 r^{2}+2 r -4}\) | \(-{\frac {7}{8}}\) |
| \(a_{2}\) | \(\frac {81 r^{4}-9 r^{2}+16}{36 r^{4}+96 r^{3}+28 r^{2}-48 r -16}\) | \(\frac {7}{8}\) |
| \(a_{3}\) | \(\frac {-729 r^{6}-2187 r^{5}-1701 r^{4}+243 r^{3}+54 r^{2}-432 r -352}{216 r^{6}+1512 r^{5}+3528 r^{4}+2744 r^{3}-672 r^{2}-1568 r -384}\) | \(-{\frac {23}{24}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {\left (9 r^{2}+27 r +22\right ) \left (9 r^{2}+45 r +58\right ) \left (9 r^{2}-9 r +4\right ) \left (9 r^{2}+9 r +4\right )}{16 \left (27 r^{6}+189 r^{5}+441 r^{4}+343 r^{3}-84 r^{2}-196 r -48\right ) \left (3 r^{2}+19 r +28\right )} \] Which for the root \(r = {\frac {5}{3}}\) becomes \[ a_{4}={\frac {1817}{1632}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-9 r^{2}+9 r -4}{6 r^{2}+2 r -4}\) | \(-{\frac {7}{8}}\) |
| \(a_{2}\) | \(\frac {81 r^{4}-9 r^{2}+16}{36 r^{4}+96 r^{3}+28 r^{2}-48 r -16}\) | \(\frac {7}{8}\) |
| \(a_{3}\) | \(\frac {-729 r^{6}-2187 r^{5}-1701 r^{4}+243 r^{3}+54 r^{2}-432 r -352}{216 r^{6}+1512 r^{5}+3528 r^{4}+2744 r^{3}-672 r^{2}-1568 r -384}\) | \(-{\frac {23}{24}}\) |
| \(a_{4}\) | \(\frac {\left (9 r^{2}+27 r +22\right ) \left (9 r^{2}+45 r +58\right ) \left (9 r^{2}-9 r +4\right ) \left (9 r^{2}+9 r +4\right )}{16 \left (27 r^{6}+189 r^{5}+441 r^{4}+343 r^{3}-84 r^{2}-196 r -48\right ) \left (3 r^{2}+19 r +28\right )}\) | \(\frac {1817}{1632}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {\left (9 r^{2}+27 r +22\right ) \left (9 r^{2}+45 r +58\right ) \left (9 r^{2}-9 r +4\right ) \left (9 r^{2}+9 r +4\right ) \left (9 r^{2}+63 r +112\right )}{32 \left (27 r^{6}+189 r^{5}+441 r^{4}+343 r^{3}-84 r^{2}-196 r -48\right ) \left (3 r^{2}+19 r +28\right ) \left (3 r^{2}+25 r +50\right )} \] Which for the root \(r = {\frac {5}{3}}\) becomes \[ a_{5}=-{\frac {219857}{163200}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-9 r^{2}+9 r -4}{6 r^{2}+2 r -4}\) | \(-{\frac {7}{8}}\) |
| \(a_{2}\) | \(\frac {81 r^{4}-9 r^{2}+16}{36 r^{4}+96 r^{3}+28 r^{2}-48 r -16}\) | \(\frac {7}{8}\) |
| \(a_{3}\) | \(\frac {-729 r^{6}-2187 r^{5}-1701 r^{4}+243 r^{3}+54 r^{2}-432 r -352}{216 r^{6}+1512 r^{5}+3528 r^{4}+2744 r^{3}-672 r^{2}-1568 r -384}\) | \(-{\frac {23}{24}}\) |
| \(a_{4}\) | \(\frac {\left (9 r^{2}+27 r +22\right ) \left (9 r^{2}+45 r +58\right ) \left (9 r^{2}-9 r +4\right ) \left (9 r^{2}+9 r +4\right )}{16 \left (27 r^{6}+189 r^{5}+441 r^{4}+343 r^{3}-84 r^{2}-196 r -48\right ) \left (3 r^{2}+19 r +28\right )}\) | \(\frac {1817}{1632}\) |
| \(a_{5}\) | \(-\frac {\left (9 r^{2}+27 r +22\right ) \left (9 r^{2}+45 r +58\right ) \left (9 r^{2}-9 r +4\right ) \left (9 r^{2}+9 r +4\right ) \left (9 r^{2}+63 r +112\right )}{32 \left (27 r^{6}+189 r^{5}+441 r^{4}+343 r^{3}-84 r^{2}-196 r -48\right ) \left (3 r^{2}+19 r +28\right ) \left (3 r^{2}+25 r +50\right )}\) | \(-{\frac {219857}{163200}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {5}{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 {5}{3}} \left (1-\frac {7 x}{8}+\frac {7 x^{2}}{8}-\frac {23 x^{3}}{24}+\frac {1817 x^{4}}{1632}-\frac {219857 x^{5}}{163200}+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 \(b_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 9 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+6 b_{n} \left (n +r \right ) \left (n +r -1\right )-4 \left (n +r \right ) b_{n}+4 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {b_{n -1} \left (9 n^{2}+18 n r +9 r^{2}-27 n -27 r +22\right )}{2 \left (3 n^{2}+6 n r +3 r^{2}-5 n -5 r \right )}\tag {4} \] Which for the root \(r = 0\) becomes \[ b_{n} = -\frac {9 \left (n^{2}-3 n +\frac {22}{9}\right ) b_{n -1}}{6 n^{2}-10 n}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = 0\) 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 \[ b_{1}=\frac {-9 r^{2}+9 r -4}{6 r^{2}+2 r -4} \] Which for the root \(r = 0\) becomes \[ b_{1}=1 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-9 r^{2}+9 r -4}{6 r^{2}+2 r -4}\) | \(1\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {81 r^{4}-9 r^{2}+16}{36 r^{4}+96 r^{3}+28 r^{2}-48 r -16} \] Which for the root \(r = 0\) becomes \[ b_{2}=-1 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-9 r^{2}+9 r -4}{6 r^{2}+2 r -4}\) | \(1\) |
| \(b_{2}\) | \(\frac {81 r^{4}-9 r^{2}+16}{36 r^{4}+96 r^{3}+28 r^{2}-48 r -16}\) | \(-1\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {-729 r^{6}-2187 r^{5}-1701 r^{4}+243 r^{3}+54 r^{2}-432 r -352}{216 r^{6}+1512 r^{5}+3528 r^{4}+2744 r^{3}-672 r^{2}-1568 r -384} \] Which for the root \(r = 0\) becomes \[ b_{3}={\frac {11}{12}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-9 r^{2}+9 r -4}{6 r^{2}+2 r -4}\) | \(1\) |
| \(b_{2}\) | \(\frac {81 r^{4}-9 r^{2}+16}{36 r^{4}+96 r^{3}+28 r^{2}-48 r -16}\) | \(-1\) |
| \(b_{3}\) | \(\frac {-729 r^{6}-2187 r^{5}-1701 r^{4}+243 r^{3}+54 r^{2}-432 r -352}{216 r^{6}+1512 r^{5}+3528 r^{4}+2744 r^{3}-672 r^{2}-1568 r -384}\) | \(\frac {11}{12}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {\left (9 r^{2}+27 r +22\right ) \left (9 r^{2}+45 r +58\right ) \left (9 r^{2}-9 r +4\right ) \left (9 r^{2}+9 r +4\right )}{16 \left (27 r^{6}+189 r^{5}+441 r^{4}+343 r^{3}-84 r^{2}-196 r -48\right ) \left (3 r^{2}+19 r +28\right )} \] Which for the root \(r = 0\) becomes \[ b_{4}=-{\frac {319}{336}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-9 r^{2}+9 r -4}{6 r^{2}+2 r -4}\) | \(1\) |
| \(b_{2}\) | \(\frac {81 r^{4}-9 r^{2}+16}{36 r^{4}+96 r^{3}+28 r^{2}-48 r -16}\) | \(-1\) |
| \(b_{3}\) | \(\frac {-729 r^{6}-2187 r^{5}-1701 r^{4}+243 r^{3}+54 r^{2}-432 r -352}{216 r^{6}+1512 r^{5}+3528 r^{4}+2744 r^{3}-672 r^{2}-1568 r -384}\) | \(\frac {11}{12}\) |
| \(b_{4}\) | \(\frac {\left (9 r^{2}+27 r +22\right ) \left (9 r^{2}+45 r +58\right ) \left (9 r^{2}-9 r +4\right ) \left (9 r^{2}+9 r +4\right )}{16 \left (27 r^{6}+189 r^{5}+441 r^{4}+343 r^{3}-84 r^{2}-196 r -48\right ) \left (3 r^{2}+19 r +28\right )}\) | \(-{\frac {319}{336}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {\left (9 r^{2}+27 r +22\right ) \left (9 r^{2}+45 r +58\right ) \left (9 r^{2}-9 r +4\right ) \left (9 r^{2}+9 r +4\right ) \left (9 r^{2}+63 r +112\right )}{32 \left (27 r^{6}+189 r^{5}+441 r^{4}+343 r^{3}-84 r^{2}-196 r -48\right ) \left (3 r^{2}+19 r +28\right ) \left (3 r^{2}+25 r +50\right )} \] Which for the root \(r = 0\) becomes \[ b_{5}={\frac {319}{300}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-9 r^{2}+9 r -4}{6 r^{2}+2 r -4}\) | \(1\) |
| \(b_{2}\) | \(\frac {81 r^{4}-9 r^{2}+16}{36 r^{4}+96 r^{3}+28 r^{2}-48 r -16}\) | \(-1\) |
| \(b_{3}\) | \(\frac {-729 r^{6}-2187 r^{5}-1701 r^{4}+243 r^{3}+54 r^{2}-432 r -352}{216 r^{6}+1512 r^{5}+3528 r^{4}+2744 r^{3}-672 r^{2}-1568 r -384}\) | \(\frac {11}{12}\) |
| \(b_{4}\) | \(\frac {\left (9 r^{2}+27 r +22\right ) \left (9 r^{2}+45 r +58\right ) \left (9 r^{2}-9 r +4\right ) \left (9 r^{2}+9 r +4\right )}{16 \left (27 r^{6}+189 r^{5}+441 r^{4}+343 r^{3}-84 r^{2}-196 r -48\right ) \left (3 r^{2}+19 r +28\right )}\) | \(-{\frac {319}{336}}\) |
| \(b_{5}\) | \(-\frac {\left (9 r^{2}+27 r +22\right ) \left (9 r^{2}+45 r +58\right ) \left (9 r^{2}-9 r +4\right ) \left (9 r^{2}+9 r +4\right ) \left (9 r^{2}+63 r +112\right )}{32 \left (27 r^{6}+189 r^{5}+441 r^{4}+343 r^{3}-84 r^{2}-196 r -48\right ) \left (3 r^{2}+19 r +28\right ) \left (3 r^{2}+25 r +50\right )}\) | \(\frac {319}{300}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= 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 \\ &= 1+x -x^{2}+\frac {11 x^{3}}{12}-\frac {319 x^{4}}{336}+\frac {319 x^{5}}{300}+O\left (x^{6}\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^{\frac {5}{3}} \left (1-\frac {7 x}{8}+\frac {7 x^{2}}{8}-\frac {23 x^{3}}{24}+\frac {1817 x^{4}}{1632}-\frac {219857 x^{5}}{163200}+O\left (x^{6}\right )\right ) + c_{2} \left (1+x -x^{2}+\frac {11 x^{3}}{12}-\frac {319 x^{4}}{336}+\frac {319 x^{5}}{300}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {5}{3}} \left (1-\frac {7 x}{8}+\frac {7 x^{2}}{8}-\frac {23 x^{3}}{24}+\frac {1817 x^{4}}{1632}-\frac {219857 x^{5}}{163200}+O\left (x^{6}\right )\right )+c_{2} \left (1+x -x^{2}+\frac {11 x^{3}}{12}-\frac {319 x^{4}}{336}+\frac {319 x^{5}}{300}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {5}{3}} \left (1-\frac {7 x}{8}+\frac {7 x^{2}}{8}-\frac {23 x^{3}}{24}+\frac {1817 x^{4}}{1632}-\frac {219857 x^{5}}{163200}+O\left (x^{6}\right )\right )+c_{2} \left (1+x -x^{2}+\frac {11 x^{3}}{12}-\frac {319 x^{4}}{336}+\frac {319 x^{5}}{300}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {5}{3}} \left (1-\frac {7 x}{8}+\frac {7 x^{2}}{8}-\frac {23 x^{3}}{24}+\frac {1817 x^{4}}{1632}-\frac {219857 x^{5}}{163200}+O\left (x^{6}\right )\right )+c_{2} \left (1+x -x^{2}+\frac {11 x^{3}}{12}-\frac {319 x^{4}}{336}+\frac {319 x^{5}}{300}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 3 x \left (2+3 x \right ) y^{\prime \prime }-4 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}{3 x \left (2+3 x \right )}+\frac {4 y^{\prime }}{3 x \left (2+3 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 {4 y^{\prime }}{3 x \left (2+3 x \right )}+\frac {4 y}{3 x \left (2+3 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 {4}{3 x \left (2+3 x \right )}, P_{3}\left (x \right )=\frac {4}{3 x \left (2+3 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}}}=-\frac {2}{3} \\ {} & \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}}}=0 \\ {} & \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} \\ {} & {} & 3 x \left (2+3 x \right ) y^{\prime \prime }-4 y^{\prime }+4 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} y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+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 =1..2 \\ {} & {} & 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} \\ {} & {} & 2 a_{0} r \left (-5+3 r \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (2 a_{k +1} \left (k +1+r \right ) \left (3 k -2+3 r \right )+a_{k} \left (9 k^{2}+18 k r +9 r^{2}-9 k -9 r +4\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 2 r \left (-5+3 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {5}{3}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 6 \left (k -\frac {2}{3}+r \right ) \left (k +1+r \right ) a_{k +1}+9 \left (k^{2}+\left (2 r -1\right ) k +r^{2}-r +\frac {4}{9}\right ) a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {\left (9 k^{2}+18 k r +9 r^{2}-9 k -9 r +4\right ) a_{k}}{2 \left (3 k -2+3 r \right ) \left (k +1+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {\left (9 k^{2}-9 k +4\right ) a_{k}}{2 \left (3 k -2\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +1}=-\frac {\left (9 k^{2}-9 k +4\right ) a_{k}}{2 \left (3 k -2\right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {5}{3} \\ {} & {} & a_{k +1}=-\frac {\left (9 k^{2}+21 k +14\right ) a_{k}}{2 \left (3 k +3\right ) \left (k +\frac {8}{3}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {5}{3} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {5}{3}}, a_{k +1}=-\frac {\left (9 k^{2}+21 k +14\right ) a_{k}}{2 \left (3 k +3\right ) \left (k +\frac {8}{3}\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {5}{3}}\right ), a_{k +1}=-\frac {\left (9 k^{2}-9 k +4\right ) a_{k}}{2 \left (3 k -2\right ) \left (k +1\right )}, b_{k +1}=-\frac {\left (9 k^{2}+21 k +14\right ) b_{k}}{2 \left (3 k +3\right ) \left (k +\frac {8}{3}\right )}\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 -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach <- heuristic approach successful <- hypergeometric successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 44
Order:=6; dsolve(3*x*(2+3*x)*diff(y(x),x$2)-4*diff(y(x),x)+4*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{\frac {5}{3}} \left (1-\frac {7}{8} x +\frac {7}{8} x^{2}-\frac {23}{24} x^{3}+\frac {1817}{1632} x^{4}-\frac {219857}{163200} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (1+x -x^{2}+\frac {11}{12} x^{3}-\frac {319}{336} x^{4}+\frac {319}{300} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 79
AsymptoticDSolveValue[3*x*(2+3*x)*y''[x]-4*y'[x]+4*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \left (\frac {319 x^5}{300}-\frac {319 x^4}{336}+\frac {11 x^3}{12}-x^2+x+1\right )+c_1 \left (-\frac {219857 x^5}{163200}+\frac {1817 x^4}{1632}-\frac {23 x^3}{24}+\frac {7 x^2}{8}-\frac {7 x}{8}+1\right ) x^{5/3} \]