Internal problem ID [6487]
Internal file name [OUTPUT/5735_Sunday_June_05_2022_03_51_37_PM_14322178/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 4. Power Series Solutions and Special Functions. Problems for review and
discovert. (A) Drill Exercises . Page 194
Problem number: 2(d).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Complex roots"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {4 x^{2} y^{\prime \prime }+4 y^{\prime } x^{2}+2 y=0} \] 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. \[ 4 x^{2} y^{\prime \prime }+4 y^{\prime } x^{2}+2 y = 0 \] 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) &= 1\\ q(x) &= \frac {1}{2 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 \[ 4 x^{2} y^{\prime \prime }+4 y^{\prime } x^{2}+2 y = 0 \] 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 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) x^{2}+4 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x^{2}+2 \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 }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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 }}4 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} \left (n +r -1\right ) 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 }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+2 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 4 x^{r} a_{0} r \left (-1+r \right )+2 a_{0} x^{r} = 0 \] Or \[ \left (4 x^{r} r \left (-1+r \right )+2 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (4 r^{2}-4 r +2\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 4 r^{2}-4 r +2 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= \frac {1}{2}+\frac {i}{2}\\ r_2 &= \frac {1}{2}-\frac {i}{2} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (4 r^{2}-4 r +2\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [\frac {1}{2}+\frac {i}{2}, \frac {1}{2}-\frac {i}{2}\right ]\).
Since the roots are complex conjugates, then two linearly independent solutions can be constructed using \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}{2}+\frac {i}{2}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {1}{2}-\frac {i}{2}} \end {align*}
\(y_{1}\left (x \right )\) is found first. 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} 4 a_{n} \left (n +r \right ) \left (n +r -1\right )+4 a_{n -1} \left (n +r -1\right )+2 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {2 a_{n -1} \left (n +r -1\right )}{2 n^{2}+4 n r +2 r^{2}-2 n -2 r +1}\tag {4} \] Which for the root \(r = \frac {1}{2}+\frac {i}{2}\) becomes \[ a_{n} = -\frac {a_{n -1} \left (2 n -1+i\right )}{2 n \left (n +i\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = \frac {1}{2}+\frac {i}{2}\) 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 \[ a_{1}=-\frac {2 r}{2 r^{2}+2 r +1} \] Which for the root \(r = \frac {1}{2}+\frac {i}{2}\) becomes \[ a_{1}=-{\frac {1}{2}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2 r}{2 r^{2}+2 r +1}\) | \(-{\frac {1}{2}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {4 r \left (1+r \right )}{4 r^{4}+16 r^{3}+24 r^{2}+16 r +5} \] Which for the root \(r = \frac {1}{2}+\frac {i}{2}\) becomes \[ a_{2}=\frac {7}{40}-\frac {i}{40} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2 r}{2 r^{2}+2 r +1}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {4 r \left (1+r \right )}{4 r^{4}+16 r^{3}+24 r^{2}+16 r +5}\) | \(\frac {7}{40}-\frac {i}{40}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {8 r \left (1+r \right ) \left (2+r \right )}{8 r^{6}+72 r^{5}+260 r^{4}+480 r^{3}+482 r^{2}+258 r +65} \] Which for the root \(r = \frac {1}{2}+\frac {i}{2}\) becomes \[ a_{3}=-\frac {11}{240}+\frac {i}{80} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2 r}{2 r^{2}+2 r +1}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {4 r \left (1+r \right )}{4 r^{4}+16 r^{3}+24 r^{2}+16 r +5}\) | \(\frac {7}{40}-\frac {i}{40}\) |
| \(a_{3}\) | \(-\frac {8 r \left (1+r \right ) \left (2+r \right )}{8 r^{6}+72 r^{5}+260 r^{4}+480 r^{3}+482 r^{2}+258 r +65}\) | \(-\frac {11}{240}+\frac {i}{80}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {16 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{16 r^{8}+256 r^{7}+1728 r^{6}+6400 r^{5}+14184 r^{4}+19264 r^{3}+15792 r^{2}+7360 r +1625} \] Which for the root \(r = \frac {1}{2}+\frac {i}{2}\) becomes \[ a_{4}=\frac {31}{3264}-\frac {i}{272} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2 r}{2 r^{2}+2 r +1}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {4 r \left (1+r \right )}{4 r^{4}+16 r^{3}+24 r^{2}+16 r +5}\) | \(\frac {7}{40}-\frac {i}{40}\) |
| \(a_{3}\) | \(-\frac {8 r \left (1+r \right ) \left (2+r \right )}{8 r^{6}+72 r^{5}+260 r^{4}+480 r^{3}+482 r^{2}+258 r +65}\) | \(-\frac {11}{240}+\frac {i}{80}\) |
| \(a_{4}\) | \(\frac {16 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{16 r^{8}+256 r^{7}+1728 r^{6}+6400 r^{5}+14184 r^{4}+19264 r^{3}+15792 r^{2}+7360 r +1625}\) | \(\frac {31}{3264}-\frac {i}{272}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (r^{2}+9 r +\frac {41}{2}\right ) \left (r^{2}+5 r +\frac {13}{2}\right ) \left (r^{2}+r +\frac {1}{2}\right ) \left (r^{2}+7 r +\frac {25}{2}\right ) \left (r^{2}+3 r +\frac {5}{2}\right )} \] Which for the root \(r = \frac {1}{2}+\frac {i}{2}\) becomes \[ a_{5}=-\frac {53}{32640}+\frac {13 i}{16320} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2 r}{2 r^{2}+2 r +1}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {4 r \left (1+r \right )}{4 r^{4}+16 r^{3}+24 r^{2}+16 r +5}\) | \(\frac {7}{40}-\frac {i}{40}\) |
| \(a_{3}\) | \(-\frac {8 r \left (1+r \right ) \left (2+r \right )}{8 r^{6}+72 r^{5}+260 r^{4}+480 r^{3}+482 r^{2}+258 r +65}\) | \(-\frac {11}{240}+\frac {i}{80}\) |
| \(a_{4}\) | \(\frac {16 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{16 r^{8}+256 r^{7}+1728 r^{6}+6400 r^{5}+14184 r^{4}+19264 r^{3}+15792 r^{2}+7360 r +1625}\) | \(\frac {31}{3264}-\frac {i}{272}\) |
| \(a_{5}\) | \(-\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (r^{2}+9 r +\frac {41}{2}\right ) \left (r^{2}+5 r +\frac {13}{2}\right ) \left (r^{2}+r +\frac {1}{2}\right ) \left (r^{2}+7 r +\frac {25}{2}\right ) \left (r^{2}+3 r +\frac {5}{2}\right )}\) | \(-\frac {53}{32640}+\frac {13 i}{16320}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {64 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}{\left (2 r^{2}+18 r +41\right ) \left (2 r^{2}+10 r +13\right ) \left (2 r^{2}+2 r +1\right ) \left (2 r^{2}+14 r +25\right ) \left (2 r^{2}+6 r +5\right ) \left (2 r^{2}+22 r +61\right )} \] Which for the root \(r = \frac {1}{2}+\frac {i}{2}\) becomes \[ a_{6}=\frac {3421}{14492160}-\frac {223 i}{1610240} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2 r}{2 r^{2}+2 r +1}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {4 r \left (1+r \right )}{4 r^{4}+16 r^{3}+24 r^{2}+16 r +5}\) | \(\frac {7}{40}-\frac {i}{40}\) |
| \(a_{3}\) | \(-\frac {8 r \left (1+r \right ) \left (2+r \right )}{8 r^{6}+72 r^{5}+260 r^{4}+480 r^{3}+482 r^{2}+258 r +65}\) | \(-\frac {11}{240}+\frac {i}{80}\) |
| \(a_{4}\) | \(\frac {16 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{16 r^{8}+256 r^{7}+1728 r^{6}+6400 r^{5}+14184 r^{4}+19264 r^{3}+15792 r^{2}+7360 r +1625}\) | \(\frac {31}{3264}-\frac {i}{272}\) |
| \(a_{5}\) | \(-\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (r^{2}+9 r +\frac {41}{2}\right ) \left (r^{2}+5 r +\frac {13}{2}\right ) \left (r^{2}+r +\frac {1}{2}\right ) \left (r^{2}+7 r +\frac {25}{2}\right ) \left (r^{2}+3 r +\frac {5}{2}\right )}\) | \(-\frac {53}{32640}+\frac {13 i}{16320}\) |
| \(a_{6}\) | \(\frac {64 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}{\left (2 r^{2}+18 r +41\right ) \left (2 r^{2}+10 r +13\right ) \left (2 r^{2}+2 r +1\right ) \left (2 r^{2}+14 r +25\right ) \left (2 r^{2}+6 r +5\right ) \left (2 r^{2}+22 r +61\right )}\) | \(\frac {3421}{14492160}-\frac {223 i}{1610240}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=-\frac {128 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right ) \left (6+r \right )}{\left (2 r^{2}+18 r +41\right ) \left (2 r^{2}+10 r +13\right ) \left (2 r^{2}+2 r +1\right ) \left (2 r^{2}+14 r +25\right ) \left (2 r^{2}+6 r +5\right ) \left (2 r^{2}+22 r +61\right ) \left (2 r^{2}+26 r +85\right )} \] Which for the root \(r = \frac {1}{2}+\frac {i}{2}\) becomes \[ a_{7}=-\frac {30269}{1014451200}+\frac {977 i}{48307200} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2 r}{2 r^{2}+2 r +1}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {4 r \left (1+r \right )}{4 r^{4}+16 r^{3}+24 r^{2}+16 r +5}\) | \(\frac {7}{40}-\frac {i}{40}\) |
| \(a_{3}\) | \(-\frac {8 r \left (1+r \right ) \left (2+r \right )}{8 r^{6}+72 r^{5}+260 r^{4}+480 r^{3}+482 r^{2}+258 r +65}\) | \(-\frac {11}{240}+\frac {i}{80}\) |
| \(a_{4}\) | \(\frac {16 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{16 r^{8}+256 r^{7}+1728 r^{6}+6400 r^{5}+14184 r^{4}+19264 r^{3}+15792 r^{2}+7360 r +1625}\) | \(\frac {31}{3264}-\frac {i}{272}\) |
| \(a_{5}\) | \(-\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (r^{2}+9 r +\frac {41}{2}\right ) \left (r^{2}+5 r +\frac {13}{2}\right ) \left (r^{2}+r +\frac {1}{2}\right ) \left (r^{2}+7 r +\frac {25}{2}\right ) \left (r^{2}+3 r +\frac {5}{2}\right )}\) | \(-\frac {53}{32640}+\frac {13 i}{16320}\) |
| \(a_{6}\) | \(\frac {64 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}{\left (2 r^{2}+18 r +41\right ) \left (2 r^{2}+10 r +13\right ) \left (2 r^{2}+2 r +1\right ) \left (2 r^{2}+14 r +25\right ) \left (2 r^{2}+6 r +5\right ) \left (2 r^{2}+22 r +61\right )}\) | \(\frac {3421}{14492160}-\frac {223 i}{1610240}\) |
| \(a_{7}\) | \(-\frac {128 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right ) \left (6+r \right )}{\left (2 r^{2}+18 r +41\right ) \left (2 r^{2}+10 r +13\right ) \left (2 r^{2}+2 r +1\right ) \left (2 r^{2}+14 r +25\right ) \left (2 r^{2}+6 r +5\right ) \left (2 r^{2}+22 r +61\right ) \left (2 r^{2}+26 r +85\right )}\) | \(-\frac {30269}{1014451200}+\frac {977 i}{48307200}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= x^{\frac {1}{2}+\frac {i}{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}+a_{7} x^{7}+a_{8} x^{8}\dots \right ) \\ &= x^{\frac {1}{2}+\frac {i}{2}} \left (1-\frac {x}{2}+\left (\frac {7}{40}-\frac {i}{40}\right ) x^{2}+\left (-\frac {11}{240}+\frac {i}{80}\right ) x^{3}+\left (\frac {31}{3264}-\frac {i}{272}\right ) x^{4}+\left (-\frac {53}{32640}+\frac {13 i}{16320}\right ) x^{5}+\left (\frac {3421}{14492160}-\frac {223 i}{1610240}\right ) x^{6}+\left (-\frac {30269}{1014451200}+\frac {977 i}{48307200}\right ) x^{7}+O\left (x^{8}\right )\right ) \\ \end{align*} The second solution \(y_{2}\left (x \right )\) is found by taking the complex conjugate of \(y_{1}\left (x \right )\) which gives \[ y_{2}\left (x \right )= x^{\frac {1}{2}-\frac {i}{2}} \left (1-\frac {x}{2}+\left (\frac {7}{40}+\frac {i}{40}\right ) x^{2}+\left (-\frac {11}{240}-\frac {i}{80}\right ) x^{3}+\left (\frac {31}{3264}+\frac {i}{272}\right ) x^{4}+\left (-\frac {53}{32640}-\frac {13 i}{16320}\right ) x^{5}+\left (\frac {3421}{14492160}+\frac {223 i}{1610240}\right ) x^{6}+\left (-\frac {30269}{1014451200}-\frac {977 i}{48307200}\right ) x^{7}+O\left (x^{8}\right )\right ) \] 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}{2}+\frac {i}{2}} \left (1-\frac {x}{2}+\left (\frac {7}{40}-\frac {i}{40}\right ) x^{2}+\left (-\frac {11}{240}+\frac {i}{80}\right ) x^{3}+\left (\frac {31}{3264}-\frac {i}{272}\right ) x^{4}+\left (-\frac {53}{32640}+\frac {13 i}{16320}\right ) x^{5}+\left (\frac {3421}{14492160}-\frac {223 i}{1610240}\right ) x^{6}+\left (-\frac {30269}{1014451200}+\frac {977 i}{48307200}\right ) x^{7}+O\left (x^{8}\right )\right ) + c_{2} x^{\frac {1}{2}-\frac {i}{2}} \left (1-\frac {x}{2}+\left (\frac {7}{40}+\frac {i}{40}\right ) x^{2}+\left (-\frac {11}{240}-\frac {i}{80}\right ) x^{3}+\left (\frac {31}{3264}+\frac {i}{272}\right ) x^{4}+\left (-\frac {53}{32640}-\frac {13 i}{16320}\right ) x^{5}+\left (\frac {3421}{14492160}+\frac {223 i}{1610240}\right ) x^{6}+\left (-\frac {30269}{1014451200}-\frac {977 i}{48307200}\right ) x^{7}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {1}{2}+\frac {i}{2}} \left (1-\frac {x}{2}+\left (\frac {7}{40}-\frac {i}{40}\right ) x^{2}+\left (-\frac {11}{240}+\frac {i}{80}\right ) x^{3}+\left (\frac {31}{3264}-\frac {i}{272}\right ) x^{4}+\left (-\frac {53}{32640}+\frac {13 i}{16320}\right ) x^{5}+\left (\frac {3421}{14492160}-\frac {223 i}{1610240}\right ) x^{6}+\left (-\frac {30269}{1014451200}+\frac {977 i}{48307200}\right ) x^{7}+O\left (x^{8}\right )\right )+c_{2} x^{\frac {1}{2}-\frac {i}{2}} \left (1-\frac {x}{2}+\left (\frac {7}{40}+\frac {i}{40}\right ) x^{2}+\left (-\frac {11}{240}-\frac {i}{80}\right ) x^{3}+\left (\frac {31}{3264}+\frac {i}{272}\right ) x^{4}+\left (-\frac {53}{32640}-\frac {13 i}{16320}\right ) x^{5}+\left (\frac {3421}{14492160}+\frac {223 i}{1610240}\right ) x^{6}+\left (-\frac {30269}{1014451200}-\frac {977 i}{48307200}\right ) x^{7}+O\left (x^{8}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {1}{2}+\frac {i}{2}} \left (1-\frac {x}{2}+\left (\frac {7}{40}-\frac {i}{40}\right ) x^{2}+\left (-\frac {11}{240}+\frac {i}{80}\right ) x^{3}+\left (\frac {31}{3264}-\frac {i}{272}\right ) x^{4}+\left (-\frac {53}{32640}+\frac {13 i}{16320}\right ) x^{5}+\left (\frac {3421}{14492160}-\frac {223 i}{1610240}\right ) x^{6}+\left (-\frac {30269}{1014451200}+\frac {977 i}{48307200}\right ) x^{7}+O\left (x^{8}\right )\right )+c_{2} x^{\frac {1}{2}-\frac {i}{2}} \left (1-\frac {x}{2}+\left (\frac {7}{40}+\frac {i}{40}\right ) x^{2}+\left (-\frac {11}{240}-\frac {i}{80}\right ) x^{3}+\left (\frac {31}{3264}+\frac {i}{272}\right ) x^{4}+\left (-\frac {53}{32640}-\frac {13 i}{16320}\right ) x^{5}+\left (\frac {3421}{14492160}+\frac {223 i}{1610240}\right ) x^{6}+\left (-\frac {30269}{1014451200}-\frac {977 i}{48307200}\right ) x^{7}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {1}{2}+\frac {i}{2}} \left (1-\frac {x}{2}+\left (\frac {7}{40}-\frac {i}{40}\right ) x^{2}+\left (-\frac {11}{240}+\frac {i}{80}\right ) x^{3}+\left (\frac {31}{3264}-\frac {i}{272}\right ) x^{4}+\left (-\frac {53}{32640}+\frac {13 i}{16320}\right ) x^{5}+\left (\frac {3421}{14492160}-\frac {223 i}{1610240}\right ) x^{6}+\left (-\frac {30269}{1014451200}+\frac {977 i}{48307200}\right ) x^{7}+O\left (x^{8}\right )\right )+c_{2} x^{\frac {1}{2}-\frac {i}{2}} \left (1-\frac {x}{2}+\left (\frac {7}{40}+\frac {i}{40}\right ) x^{2}+\left (-\frac {11}{240}-\frac {i}{80}\right ) x^{3}+\left (\frac {31}{3264}+\frac {i}{272}\right ) x^{4}+\left (-\frac {53}{32640}-\frac {13 i}{16320}\right ) x^{5}+\left (\frac {3421}{14492160}+\frac {223 i}{1610240}\right ) x^{6}+\left (-\frac {30269}{1014451200}-\frac {977 i}{48307200}\right ) x^{7}+O\left (x^{8}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 \left (\frac {d}{d x}y^{\prime }\right ) x^{2}+4 y^{\prime } x^{2}+2 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-y^{\prime }-\frac {y}{2 x^{2}} \\ \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} \\ {} & {} & \frac {d}{d x}y^{\prime }+y^{\prime }+\frac {y}{2 x^{2}}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=1, P_{3}\left (x \right )=\frac {1}{2 x^{2}}\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}}}=0 \\ {} & \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 {1}{2} \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 2 \left (\frac {d}{d x}y^{\prime }\right ) x^{2}+2 y^{\prime } x^{2}+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^{2}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot 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 \\ {} & {} & x^{2}\cdot 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^{2}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k -1+r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (2 r^{2}-2 r +1\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (2 k^{2}+4 k r +2 r^{2}-2 k -2 r +1\right )+2 a_{k -1} \left (k -1+r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 2 r^{2}-2 r +1=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\frac {1}{2}-\frac {\mathrm {I}}{2}, \frac {1}{2}+\frac {\mathrm {I}}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (2 k^{2}+\left (4 r -2\right ) k +2 r^{2}-2 r +1\right ) a_{k}+2 a_{k -1} \left (k -1+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (2 \left (k +1\right )^{2}+\left (4 r -2\right ) \left (k +1\right )+2 r^{2}-2 r +1\right ) a_{k +1}+2 a_{k} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {2 a_{k} \left (k +r \right )}{2 k^{2}+4 k r +2 r^{2}+2 k +2 r +1} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2}-\frac {\mathrm {I}}{2} \\ {} & {} & a_{k +1}=-\frac {2 a_{k} \left (k +\frac {1}{2}-\frac {\mathrm {I}}{2}\right )}{2 k^{2}+\left (2-2 \,\mathrm {I}\right ) k +2-2 \,\mathrm {I}+2 k} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2}-\frac {\mathrm {I}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}-\frac {\mathrm {I}}{2}}, a_{k +1}=-\frac {2 a_{k} \left (k +\frac {1}{2}-\frac {\mathrm {I}}{2}\right )}{2 k^{2}+\left (2-2 \,\mathrm {I}\right ) k +2-2 \,\mathrm {I}+2 k}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2}+\frac {\mathrm {I}}{2} \\ {} & {} & a_{k +1}=-\frac {2 a_{k} \left (k +\frac {1}{2}+\frac {\mathrm {I}}{2}\right )}{2 k^{2}+\left (2+2 \,\mathrm {I}\right ) k +2+2 \,\mathrm {I}+2 k} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2}+\frac {\mathrm {I}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}+\frac {\mathrm {I}}{2}}, a_{k +1}=-\frac {2 a_{k} \left (k +\frac {1}{2}+\frac {\mathrm {I}}{2}\right )}{2 k^{2}+\left (2+2 \,\mathrm {I}\right ) k +2+2 \,\mathrm {I}+2 k}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}-\frac {\mathrm {I}}{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{2}+\frac {\mathrm {I}}{2}}\right ), a_{k +1}=-\frac {2 a_{k} \left (k +\frac {1}{2}-\frac {\mathrm {I}}{2}\right )}{2 k^{2}+\left (2-2 \,\mathrm {I}\right ) k +2-2 \,\mathrm {I}+2 k}, b_{k +1}=-\frac {2 b_{k} \left (k +\frac {1}{2}+\frac {\mathrm {I}}{2}\right )}{2 k^{2}+\left (2+2 \,\mathrm {I}\right ) k +2+2 \,\mathrm {I}+2 k}\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: 0.032 (sec). Leaf size: 83
Order:=8; dsolve(4*x^2*diff(y(x),x$2)+4*x^2*diff(y(x),x)+2*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{\frac {1}{2}-\frac {i}{2}} \left (1-\frac {1}{2} x +\left (\frac {7}{40}+\frac {i}{40}\right ) x^{2}+\left (-\frac {11}{240}-\frac {i}{80}\right ) x^{3}+\left (\frac {31}{3264}+\frac {i}{272}\right ) x^{4}+\left (-\frac {53}{32640}-\frac {13 i}{16320}\right ) x^{5}+\left (\frac {3421}{14492160}+\frac {223 i}{1610240}\right ) x^{6}+\left (-\frac {30269}{1014451200}-\frac {977 i}{48307200}\right ) x^{7}+\operatorname {O}\left (x^{8}\right )\right )+c_{2} x^{\frac {1}{2}+\frac {i}{2}} \left (1-\frac {1}{2} x +\left (\frac {7}{40}-\frac {i}{40}\right ) x^{2}+\left (-\frac {11}{240}+\frac {i}{80}\right ) x^{3}+\left (\frac {31}{3264}-\frac {i}{272}\right ) x^{4}+\left (-\frac {53}{32640}+\frac {13 i}{16320}\right ) x^{5}+\left (\frac {3421}{14492160}-\frac {223 i}{1610240}\right ) x^{6}+\left (-\frac {30269}{1014451200}+\frac {977 i}{48307200}\right ) x^{7}+\operatorname {O}\left (x^{8}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.023 (sec). Leaf size: 226
AsymptoticDSolveValue[4*x^2*y''[x]+4*x^2*y'[x]+2*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \left (\left (\frac {3421}{14492160}-\frac {223 i}{1610240}\right ) x^{\frac {13}{2}+\frac {i}{2}}-\left (\frac {53}{32640}-\frac {13 i}{16320}\right ) x^{\frac {11}{2}+\frac {i}{2}}+\left (\frac {31}{3264}-\frac {i}{272}\right ) x^{\frac {9}{2}+\frac {i}{2}}-\left (\frac {11}{240}-\frac {i}{80}\right ) x^{\frac {7}{2}+\frac {i}{2}}+\left (\frac {7}{40}-\frac {i}{40}\right ) x^{\frac {5}{2}+\frac {i}{2}}-\frac {1}{2} x^{\frac {3}{2}+\frac {i}{2}}+x^{\frac {1}{2}+\frac {i}{2}}\right )+c_2 \left (\left (\frac {3421}{14492160}+\frac {223 i}{1610240}\right ) x^{\frac {13}{2}-\frac {i}{2}}-\left (\frac {53}{32640}+\frac {13 i}{16320}\right ) x^{\frac {11}{2}-\frac {i}{2}}+\left (\frac {31}{3264}+\frac {i}{272}\right ) x^{\frac {9}{2}-\frac {i}{2}}-\left (\frac {11}{240}+\frac {i}{80}\right ) x^{\frac {7}{2}-\frac {i}{2}}+\left (\frac {7}{40}+\frac {i}{40}\right ) x^{\frac {5}{2}-\frac {i}{2}}-\frac {1}{2} x^{\frac {3}{2}-\frac {i}{2}}+x^{\frac {1}{2}-\frac {i}{2}}\right ) \]