Internal problem ID [6471]
Internal file name [OUTPUT/5719_Sunday_June_05_2022_03_48_43_PM_20153980/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. Section 4.6. Gauss’s
Hypergeometric Equation. Page 187
Problem number: 2(b).
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]]
\[ \boxed {\left (2 x^{2}+2 x \right ) y^{\prime \prime }+\left (5 x +1\right ) y^{\prime }+y=0} \] 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. \[ \left (2 x^{2}+2 x \right ) y^{\prime \prime }+\left (5 x +1\right ) y^{\prime }+y = 0 \] 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 {5 x +1}{2 x \left (1+x \right )}\\ q(x) &= \frac {1}{2 x \left (1+x \right )}\\ \end {align*}
| \(p(x)=\frac {5 x +1}{2 x \left (1+x \right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=\frac {1}{2 x \left (1+x \right )}\) | |
| 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 \[ 2 x \left (1+x \right ) y^{\prime \prime }+\left (5 x +1\right ) y^{\prime }+y = 0 \] 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} 2 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 (5 x +1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\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 }}2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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 }}2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}5 x^{n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}5 a_{n -1} \left (n +r -1\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}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 }}2 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 }}2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}5 a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+\left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ 2 x^{-1+r} a_{0} r \left (-1+r \right )+r a_{0} x^{-1+r} = 0 \] Or \[ \left (2 x^{-1+r} r \left (-1+r \right )+r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (2 r -1\right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 2 r^{2}-r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{2}}\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (2 r -1\right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{2}}, 0\right ]\).
Since \(r_1 - r_2 = {\frac {1}{2}}\) 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}{2}}\\ 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} 2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+2 a_{n} \left (n +r \right ) \left (n +r -1\right )+5 a_{n -1} \left (n +r -1\right )+a_{n} \left (n +r \right )+a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -a_{n -1}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{n} = -a_{n -1}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{2}}\) 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 \[ a_{1}=-1 \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{1}=-1 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-1\) | \(-1\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=1 \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{2}=1 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-1\) | \(-1\) |
| \(a_{2}\) | \(1\) | \(1\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-1 \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{3}=-1 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-1\) | \(-1\) |
| \(a_{2}\) | \(1\) | \(1\) |
| \(a_{3}\) | \(-1\) | \(-1\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=1 \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{4}=1 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-1\) | \(-1\) |
| \(a_{2}\) | \(1\) | \(1\) |
| \(a_{3}\) | \(-1\) | \(-1\) |
| \(a_{4}\) | \(1\) | \(1\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-1 \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{5}=-1 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-1\) | \(-1\) |
| \(a_{2}\) | \(1\) | \(1\) |
| \(a_{3}\) | \(-1\) | \(-1\) |
| \(a_{4}\) | \(1\) | \(1\) |
| \(a_{5}\) | \(-1\) | \(-1\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=1 \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{6}=1 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-1\) | \(-1\) |
| \(a_{2}\) | \(1\) | \(1\) |
| \(a_{3}\) | \(-1\) | \(-1\) |
| \(a_{4}\) | \(1\) | \(1\) |
| \(a_{5}\) | \(-1\) | \(-1\) |
| \(a_{6}\) | \(1\) | \(1\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=-1 \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{7}=-1 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-1\) | \(-1\) |
| \(a_{2}\) | \(1\) | \(1\) |
| \(a_{3}\) | \(-1\) | \(-1\) |
| \(a_{4}\) | \(1\) | \(1\) |
| \(a_{5}\) | \(-1\) | \(-1\) |
| \(a_{6}\) | \(1\) | \(1\) |
| \(a_{7}\) | \(-1\) | \(-1\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= \sqrt {x} \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 ) \\ &= \sqrt {x}\, \left (1-x +x^{2}-x^{3}+x^{4}-x^{5}+x^{6}-x^{7}+O\left (x^{8}\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} 2 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+2 b_{n} \left (n +r \right ) \left (n +r -1\right )+5 b_{n -1} \left (n +r -1\right )+\left (n +r \right ) b_{n}+b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -b_{n -1}\tag {4} \] Which for the root \(r = 0\) becomes \[ b_{n} = -b_{n -1}\tag {5} \] 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 \[ b_{1}=-1 \] Which for the root \(r = 0\) becomes \[ b_{1}=-1 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-1\) | \(-1\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=1 \] Which for the root \(r = 0\) becomes \[ b_{2}=1 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-1\) | \(-1\) |
| \(b_{2}\) | \(1\) | \(1\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-1 \] Which for the root \(r = 0\) becomes \[ b_{3}=-1 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-1\) | \(-1\) |
| \(b_{2}\) | \(1\) | \(1\) |
| \(b_{3}\) | \(-1\) | \(-1\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=1 \] Which for the root \(r = 0\) becomes \[ b_{4}=1 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-1\) | \(-1\) |
| \(b_{2}\) | \(1\) | \(1\) |
| \(b_{3}\) | \(-1\) | \(-1\) |
| \(b_{4}\) | \(1\) | \(1\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-1 \] Which for the root \(r = 0\) becomes \[ b_{5}=-1 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-1\) | \(-1\) |
| \(b_{2}\) | \(1\) | \(1\) |
| \(b_{3}\) | \(-1\) | \(-1\) |
| \(b_{4}\) | \(1\) | \(1\) |
| \(b_{5}\) | \(-1\) | \(-1\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=1 \] Which for the root \(r = 0\) becomes \[ b_{6}=1 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-1\) | \(-1\) |
| \(b_{2}\) | \(1\) | \(1\) |
| \(b_{3}\) | \(-1\) | \(-1\) |
| \(b_{4}\) | \(1\) | \(1\) |
| \(b_{5}\) | \(-1\) | \(-1\) |
| \(b_{6}\) | \(1\) | \(1\) |
For \(n = 7\), using the above recursive equation gives \[ b_{7}=-1 \] Which for the root \(r = 0\) becomes \[ b_{7}=-1 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-1\) | \(-1\) |
| \(b_{2}\) | \(1\) | \(1\) |
| \(b_{3}\) | \(-1\) | \(-1\) |
| \(b_{4}\) | \(1\) | \(1\) |
| \(b_{5}\) | \(-1\) | \(-1\) |
| \(b_{6}\) | \(1\) | \(1\) |
| \(b_{7}\) | \(-1\) | \(-1\) |
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}+b_{7} x^{7}+b_{8} x^{8}\dots \\ &= 1-x +x^{2}-x^{3}+x^{4}-x^{5}+x^{6}-x^{7}+O\left (x^{8}\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} \sqrt {x}\, \left (1-x +x^{2}-x^{3}+x^{4}-x^{5}+x^{6}-x^{7}+O\left (x^{8}\right )\right ) + c_{2} \left (1-x +x^{2}-x^{3}+x^{4}-x^{5}+x^{6}-x^{7}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \sqrt {x}\, \left (1-x +x^{2}-x^{3}+x^{4}-x^{5}+x^{6}-x^{7}+O\left (x^{8}\right )\right )+c_{2} \left (1-x +x^{2}-x^{3}+x^{4}-x^{5}+x^{6}-x^{7}+O\left (x^{8}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \sqrt {x}\, \left (1-x +x^{2}-x^{3}+x^{4}-x^{5}+x^{6}-x^{7}+O\left (x^{8}\right )\right )+c_{2} \left (1-x +x^{2}-x^{3}+x^{4}-x^{5}+x^{6}-x^{7}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \sqrt {x}\, \left (1-x +x^{2}-x^{3}+x^{4}-x^{5}+x^{6}-x^{7}+O\left (x^{8}\right )\right )+c_{2} \left (1-x +x^{2}-x^{3}+x^{4}-x^{5}+x^{6}-x^{7}+O\left (x^{8}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x \left (1+x \right ) y^{\prime \prime }+\left (5 x +1\right ) y^{\prime }+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 {y}{2 x \left (1+x \right )}-\frac {\left (5 x +1\right ) y^{\prime }}{2 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 (5 x +1\right ) y^{\prime }}{2 x \left (1+x \right )}+\frac {y}{2 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 {5 x +1}{2 x \left (1+x \right )}, P_{3}\left (x \right )=\frac {1}{2 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}}}=2 \\ {} & \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} \\ {} & {} & 2 x \left (1+x \right ) y^{\prime \prime }+\left (5 x +1\right ) y^{\prime }+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 (2 u^{2}-2 u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (5 u -4\right ) \left (\frac {d}{d u}y \left (u \right )\right )+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} \\ {} & {} & -2 a_{0} r \left (1+r \right ) u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-2 a_{k +1} \left (k +r +1\right ) \left (k +2+r \right )+a_{k} \left (k +r +1\right ) \left (2 k +2 r +1\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -2 r \left (1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, 0\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 2 \left (k +r +1\right ) \left (\left (-k -r -2\right ) a_{k +1}+a_{k} \left (k +r +\frac {1}{2}\right )\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (2 k +2 r +1\right )}{2 \left (k +2+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (2 k -1\right )}{2 \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-1 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -1}, a_{k +1}=\frac {a_{k} \left (2 k -1\right )}{2 \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 -1}, a_{k +1}=\frac {a_{k} \left (2 k -1\right )}{2 \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (2 k +1\right )}{2 \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +1}=\frac {a_{k} \left (2 k +1\right )}{2 \left (k +2\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}, a_{k +1}=\frac {a_{k} \left (2 k +1\right )}{2 \left (k +2\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (1+x \right )^{k -1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (1+x \right )^{k}\right ), a_{k +1}=\frac {a_{k} \left (2 k -1\right )}{2 \left (k +1\right )}, b_{k +1}=\frac {b_{k} \left (2 k +1\right )}{2 \left (k +2\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)] <- linear_1 successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 52
Order:=8; dsolve((2*x^2+2*x)*diff(y(x),x$2)+(1+5*x)*diff(y(x),x)+y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \left (-x^{7}+x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x +1\right ) \left (c_{1} \sqrt {x}+c_{2} \right )+O\left (x^{8}\right ) \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 73
AsymptoticDSolveValue[(2*x^2+2*x)*y''[x]+(1+5*x)*y'[x]+y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \sqrt {x} \left (-x^7+x^6-x^5+x^4-x^3+x^2-x+1\right )+c_2 \left (-x^7+x^6-x^5+x^4-x^3+x^2-x+1\right ) \]