4.5 problem 5

4.5.1 Maple step by step solution

Internal problem ID [6921]
Internal file name [OUTPUT/6164_Tuesday_August_09_2022_05_23_52_AM_82600808/index.tex]

Book: Elementary differential equations. Rainville, Bedient, Bedient. Prentice Hall. NJ. 8th edition. 1997.
Section: CHAPTER 18. Power series solutions. 18.4 Indicial Equation with Difference of Roots Nonintegral. Exercises page 365
Problem number: 5.
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]]

\[ \boxed {2 x^{2} \left (1-x \right ) y^{\prime \prime }-x \left (1+7 x \right ) y^{\prime }+y=0} \] 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. \[ \left (-2 x^{3}+2 x^{2}\right ) y^{\prime \prime }+\left (-7 x^{2}-x \right ) y^{\prime }+y = 0 \] 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 {1+7 x}{2 x \left (-1+x \right )}\\ q(x) &= -\frac {1}{2 \left (-1+x \right ) x^{2}}\\ \end {align*}

Table 9: Table \(p(x),q(x)\) singularites.
\(p(x)=\frac {1+7 x}{2 x \left (-1+x \right )}\)
singularity type
\(x = 0\) \(\text {``regular''}\)
\(x = 1\) \(\text {``regular''}\)
\(q(x)=-\frac {1}{2 \left (-1+x \right ) x^{2}}\)
singularity type
\(x = 0\) \(\text {``regular''}\)
\(x = 1\) \(\text {``regular''}\)

Combining everything together gives the following summary of singularities for the ode as

Regular singular points : \([0, 1, \infty ]\)

Irregular singular points : \([]\)

Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ -2 y^{\prime \prime } x^{2} \left (-1+x \right )+\left (-7 x^{2}-x \right ) y^{\prime }+y = 0 \] 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 \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} \left (-1+x \right )+\left (-7 x^{2}-x \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} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{1+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} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-7 x^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\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 \(n +r\) 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 }}\left (-2 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-7 x^{1+n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-7 a_{n -1} \left (n +r -1\right ) x^{n +r}\right ) \\ \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} \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-7 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 2 x^{r} a_{0} r \left (-1+r \right )-x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (2 x^{r} r \left (-1+r \right )-x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (2 r^{2}-3 r +1\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 2 r^{2}-3 r +1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1\\ r_2 &= {\frac {1}{2}} \end {align*}

Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (2 r^{2}-3 r +1\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [1, {\frac {1}{2}}\right ]\).

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^{1+n}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {1}{2}} \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 )-7 a_{n -1} \left (n +r -1\right )-a_{n} \left (n +r \right )+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {\left (2 n +2 r +3\right ) a_{n -1}}{2 n +2 r -1}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {\left (2 n +5\right ) a_{n -1}}{2 n +1}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 1\) 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 \[ a_{1}=\frac {5+2 r}{1+2 r} \] Which for the root \(r = 1\) becomes \[ a_{1}={\frac {7}{3}} \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(\frac {5+2 r}{1+2 r}\) \(\frac {7}{3}\)

For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {4 r^{2}+24 r +35}{4 r^{2}+8 r +3} \] Which for the root \(r = 1\) becomes \[ a_{2}={\frac {21}{5}} \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(\frac {5+2 r}{1+2 r}\) \(\frac {7}{3}\)
\(a_{2}\) \(\frac {4 r^{2}+24 r +35}{4 r^{2}+8 r +3}\) \(\frac {21}{5}\)

For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {4 r^{2}+32 r +63}{4 r^{2}+8 r +3} \] Which for the root \(r = 1\) becomes \[ a_{3}={\frac {33}{5}} \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(\frac {5+2 r}{1+2 r}\) \(\frac {7}{3}\)
\(a_{2}\) \(\frac {4 r^{2}+24 r +35}{4 r^{2}+8 r +3}\) \(\frac {21}{5}\)
\(a_{3}\) \(\frac {4 r^{2}+32 r +63}{4 r^{2}+8 r +3}\) \(\frac {33}{5}\)

For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {4 r^{2}+40 r +99}{4 r^{2}+8 r +3} \] Which for the root \(r = 1\) becomes \[ a_{4}={\frac {143}{15}} \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(\frac {5+2 r}{1+2 r}\) \(\frac {7}{3}\)
\(a_{2}\) \(\frac {4 r^{2}+24 r +35}{4 r^{2}+8 r +3}\) \(\frac {21}{5}\)
\(a_{3}\) \(\frac {4 r^{2}+32 r +63}{4 r^{2}+8 r +3}\) \(\frac {33}{5}\)
\(a_{4}\) \(\frac {4 r^{2}+40 r +99}{4 r^{2}+8 r +3}\) \(\frac {143}{15}\)

For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {4 r^{2}+48 r +143}{4 r^{2}+8 r +3} \] Which for the root \(r = 1\) becomes \[ a_{5}=13 \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(\frac {5+2 r}{1+2 r}\) \(\frac {7}{3}\)
\(a_{2}\) \(\frac {4 r^{2}+24 r +35}{4 r^{2}+8 r +3}\) \(\frac {21}{5}\)
\(a_{3}\) \(\frac {4 r^{2}+32 r +63}{4 r^{2}+8 r +3}\) \(\frac {33}{5}\)
\(a_{4}\) \(\frac {4 r^{2}+40 r +99}{4 r^{2}+8 r +3}\) \(\frac {143}{15}\)
\(a_{5}\) \(\frac {4 r^{2}+48 r +143}{4 r^{2}+8 r +3}\) \(13\)

For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {4 r^{2}+56 r +195}{4 r^{2}+8 r +3} \] Which for the root \(r = 1\) becomes \[ a_{6}=17 \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(\frac {5+2 r}{1+2 r}\) \(\frac {7}{3}\)
\(a_{2}\) \(\frac {4 r^{2}+24 r +35}{4 r^{2}+8 r +3}\) \(\frac {21}{5}\)
\(a_{3}\) \(\frac {4 r^{2}+32 r +63}{4 r^{2}+8 r +3}\) \(\frac {33}{5}\)
\(a_{4}\) \(\frac {4 r^{2}+40 r +99}{4 r^{2}+8 r +3}\) \(\frac {143}{15}\)
\(a_{5}\) \(\frac {4 r^{2}+48 r +143}{4 r^{2}+8 r +3}\) \(13\)
\(a_{6}\) \(\frac {4 r^{2}+56 r +195}{4 r^{2}+8 r +3}\) \(17\)

For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {4 r^{2}+64 r +255}{4 r^{2}+8 r +3} \] Which for the root \(r = 1\) becomes \[ a_{7}={\frac {323}{15}} \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(\frac {5+2 r}{1+2 r}\) \(\frac {7}{3}\)
\(a_{2}\) \(\frac {4 r^{2}+24 r +35}{4 r^{2}+8 r +3}\) \(\frac {21}{5}\)
\(a_{3}\) \(\frac {4 r^{2}+32 r +63}{4 r^{2}+8 r +3}\) \(\frac {33}{5}\)
\(a_{4}\) \(\frac {4 r^{2}+40 r +99}{4 r^{2}+8 r +3}\) \(\frac {143}{15}\)
\(a_{5}\) \(\frac {4 r^{2}+48 r +143}{4 r^{2}+8 r +3}\) \(13\)
\(a_{6}\) \(\frac {4 r^{2}+56 r +195}{4 r^{2}+8 r +3}\) \(17\)
\(a_{7}\) \(\frac {4 r^{2}+64 r +255}{4 r^{2}+8 r +3}\) \(\frac {323}{15}\)

Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= 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 ) \\ &= x \left (1+\frac {7 x}{3}+\frac {21 x^{2}}{5}+\frac {33 x^{3}}{5}+\frac {143 x^{4}}{15}+13 x^{5}+17 x^{6}+\frac {323 x^{7}}{15}+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 )-7 b_{n -1} \left (n +r -1\right )-b_{n} \left (n +r \right )+b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {\left (2 n +2 r +3\right ) b_{n -1}}{2 n +2 r -1}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{n} = \frac {\left (n +2\right ) b_{n -1}}{n}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = {\frac {1}{2}}\) 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 \[ b_{1}=\frac {5+2 r}{1+2 r} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{1}=3 \] And the table now becomes

\(n\) \(b_{n ,r}\) \(b_{n}\)
\(b_{0}\) \(1\) \(1\)
\(b_{1}\) \(\frac {5+2 r}{1+2 r}\) \(3\)

For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {4 r^{2}+24 r +35}{4 r^{2}+8 r +3} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{2}=6 \] And the table now becomes

\(n\) \(b_{n ,r}\) \(b_{n}\)
\(b_{0}\) \(1\) \(1\)
\(b_{1}\) \(\frac {5+2 r}{1+2 r}\) \(3\)
\(b_{2}\) \(\frac {4 r^{2}+24 r +35}{4 r^{2}+8 r +3}\) \(6\)

For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {4 r^{2}+32 r +63}{4 r^{2}+8 r +3} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{3}=10 \] And the table now becomes

\(n\) \(b_{n ,r}\) \(b_{n}\)
\(b_{0}\) \(1\) \(1\)
\(b_{1}\) \(\frac {5+2 r}{1+2 r}\) \(3\)
\(b_{2}\) \(\frac {4 r^{2}+24 r +35}{4 r^{2}+8 r +3}\) \(6\)
\(b_{3}\) \(\frac {4 r^{2}+32 r +63}{4 r^{2}+8 r +3}\) \(10\)

For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {4 r^{2}+40 r +99}{4 r^{2}+8 r +3} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{4}=15 \] And the table now becomes

\(n\) \(b_{n ,r}\) \(b_{n}\)
\(b_{0}\) \(1\) \(1\)
\(b_{1}\) \(\frac {5+2 r}{1+2 r}\) \(3\)
\(b_{2}\) \(\frac {4 r^{2}+24 r +35}{4 r^{2}+8 r +3}\) \(6\)
\(b_{3}\) \(\frac {4 r^{2}+32 r +63}{4 r^{2}+8 r +3}\) \(10\)
\(b_{4}\) \(\frac {4 r^{2}+40 r +99}{4 r^{2}+8 r +3}\) \(15\)

For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {4 r^{2}+48 r +143}{4 r^{2}+8 r +3} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{5}=21 \] And the table now becomes

\(n\) \(b_{n ,r}\) \(b_{n}\)
\(b_{0}\) \(1\) \(1\)
\(b_{1}\) \(\frac {5+2 r}{1+2 r}\) \(3\)
\(b_{2}\) \(\frac {4 r^{2}+24 r +35}{4 r^{2}+8 r +3}\) \(6\)
\(b_{3}\) \(\frac {4 r^{2}+32 r +63}{4 r^{2}+8 r +3}\) \(10\)
\(b_{4}\) \(\frac {4 r^{2}+40 r +99}{4 r^{2}+8 r +3}\) \(15\)
\(b_{5}\) \(\frac {4 r^{2}+48 r +143}{4 r^{2}+8 r +3}\) \(21\)

For \(n = 6\), using the above recursive equation gives \[ b_{6}=\frac {4 r^{2}+56 r +195}{4 r^{2}+8 r +3} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{6}=28 \] And the table now becomes

\(n\) \(b_{n ,r}\) \(b_{n}\)
\(b_{0}\) \(1\) \(1\)
\(b_{1}\) \(\frac {5+2 r}{1+2 r}\) \(3\)
\(b_{2}\) \(\frac {4 r^{2}+24 r +35}{4 r^{2}+8 r +3}\) \(6\)
\(b_{3}\) \(\frac {4 r^{2}+32 r +63}{4 r^{2}+8 r +3}\) \(10\)
\(b_{4}\) \(\frac {4 r^{2}+40 r +99}{4 r^{2}+8 r +3}\) \(15\)
\(b_{5}\) \(\frac {4 r^{2}+48 r +143}{4 r^{2}+8 r +3}\) \(21\)
\(b_{6}\) \(\frac {4 r^{2}+56 r +195}{4 r^{2}+8 r +3}\) \(28\)

For \(n = 7\), using the above recursive equation gives \[ b_{7}=\frac {4 r^{2}+64 r +255}{4 r^{2}+8 r +3} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{7}=36 \] And the table now becomes

\(n\) \(b_{n ,r}\) \(b_{n}\)
\(b_{0}\) \(1\) \(1\)
\(b_{1}\) \(\frac {5+2 r}{1+2 r}\) \(3\)
\(b_{2}\) \(\frac {4 r^{2}+24 r +35}{4 r^{2}+8 r +3}\) \(6\)
\(b_{3}\) \(\frac {4 r^{2}+32 r +63}{4 r^{2}+8 r +3}\) \(10\)
\(b_{4}\) \(\frac {4 r^{2}+40 r +99}{4 r^{2}+8 r +3}\) \(15\)
\(b_{5}\) \(\frac {4 r^{2}+48 r +143}{4 r^{2}+8 r +3}\) \(21\)
\(b_{6}\) \(\frac {4 r^{2}+56 r +195}{4 r^{2}+8 r +3}\) \(28\)
\(b_{7}\) \(\frac {4 r^{2}+64 r +255}{4 r^{2}+8 r +3}\) \(36\)

Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x \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}+b_{7} x^{7}+b_{8} x^{8}\dots \right ) \\ &= \sqrt {x}\, \left (1+3 x +6 x^{2}+10 x^{3}+15 x^{4}+21 x^{5}+28 x^{6}+36 x^{7}+O\left (x^{8}\right )\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 \left (1+\frac {7 x}{3}+\frac {21 x^{2}}{5}+\frac {33 x^{3}}{5}+\frac {143 x^{4}}{15}+13 x^{5}+17 x^{6}+\frac {323 x^{7}}{15}+O\left (x^{8}\right )\right ) + c_{2} \sqrt {x}\, \left (1+3 x +6 x^{2}+10 x^{3}+15 x^{4}+21 x^{5}+28 x^{6}+36 x^{7}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x \left (1+\frac {7 x}{3}+\frac {21 x^{2}}{5}+\frac {33 x^{3}}{5}+\frac {143 x^{4}}{15}+13 x^{5}+17 x^{6}+\frac {323 x^{7}}{15}+O\left (x^{8}\right )\right )+c_{2} \sqrt {x}\, \left (1+3 x +6 x^{2}+10 x^{3}+15 x^{4}+21 x^{5}+28 x^{6}+36 x^{7}+O\left (x^{8}\right )\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x \left (1+\frac {7 x}{3}+\frac {21 x^{2}}{5}+\frac {33 x^{3}}{5}+\frac {143 x^{4}}{15}+13 x^{5}+17 x^{6}+\frac {323 x^{7}}{15}+O\left (x^{8}\right )\right )+c_{2} \sqrt {x}\, \left (1+3 x +6 x^{2}+10 x^{3}+15 x^{4}+21 x^{5}+28 x^{6}+36 x^{7}+O\left (x^{8}\right )\right ) \\ \end{align*}

Verification of solutions

\[ y = c_{1} x \left (1+\frac {7 x}{3}+\frac {21 x^{2}}{5}+\frac {33 x^{3}}{5}+\frac {143 x^{4}}{15}+13 x^{5}+17 x^{6}+\frac {323 x^{7}}{15}+O\left (x^{8}\right )\right )+c_{2} \sqrt {x}\, \left (1+3 x +6 x^{2}+10 x^{3}+15 x^{4}+21 x^{5}+28 x^{6}+36 x^{7}+O\left (x^{8}\right )\right ) \] Verified OK.

4.5.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -2 \left (\frac {d}{d x}y^{\prime }\right ) x^{2} \left (-1+x \right )+\left (-7 x^{2}-x \right ) y^{\prime }+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 }=\frac {y}{2 x^{2} \left (-1+x \right )}-\frac {\left (1+7 x \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} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {\left (1+7 x \right ) y^{\prime }}{2 x \left (-1+x \right )}-\frac {y}{2 x^{2} \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 {1+7 x}{2 x \left (-1+x \right )}, P_{3}\left (x \right )=-\frac {1}{2 \left (-1+x \right ) 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}}}=-\frac {1}{2} \\ {} & \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}\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} \left (-1+x \right )+x \left (1+7 x \right ) y^{\prime }-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^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..3 \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\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 \left (\frac {d}{d x}y^{\prime }\right )=\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} \\ {} & {} & -a_{0} \left (-1+2 r \right ) \left (-1+r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (-a_{k} \left (2 k +2 r -1\right ) \left (k +r -1\right )+a_{k -1} \left (k +r -1\right ) \left (2 k +3+2 r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -\left (-1+2 r \right ) \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{1, \frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -2 \left (\left (-k -r -\frac {3}{2}\right ) a_{k -1}+a_{k} \left (k +r -\frac {1}{2}\right )\right ) \left (k +r -1\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & -2 \left (\left (-k -\frac {5}{2}-r \right ) a_{k}+a_{k +1} \left (k +\frac {1}{2}+r \right )\right ) \left (k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {\left (2 k +2 r +5\right ) a_{k}}{2 k +1+2 r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +1}=\frac {\left (2 k +7\right ) a_{k}}{2 k +3} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}, a_{k +1}=\frac {\left (2 k +7\right ) a_{k}}{2 k +3}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +1}=\frac {\left (2 k +6\right ) a_{k}}{2 k +2} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}}, a_{k +1}=\frac {\left (2 k +6\right ) a_{k}}{2 k +2}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{2}}\right ), a_{k +1}=\frac {\left (2 k +7\right ) a_{k}}{2 k +3}, b_{k +1}=\frac {\left (2 k +6\right ) b_{k}}{2 k +2}\right ] \end {array} \]

Maple trace Kovacic algorithm successful

`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 
   A Liouvillian solution exists 
   Reducible group (found an exponential solution) 
   Reducible group (found another exponential solution) 
<- Kovacics algorithm successful`
 

Solution by Maple

Time used: 0.031 (sec). Leaf size: 53

Order:=8; 
dsolve(2*x^2*(1-x)*diff(y(x),x$2)-x*(1+7*x)*diff(y(x),x)+y(x)=0,y(x),type='series',x=0);
 

\[ y \left (x \right ) = c_{1} \sqrt {x}\, \left (1+3 x +6 x^{2}+10 x^{3}+15 x^{4}+21 x^{5}+28 x^{6}+36 x^{7}+\operatorname {O}\left (x^{8}\right )\right )+c_{2} x \left (1+\frac {7}{3} x +\frac {21}{5} x^{2}+\frac {33}{5} x^{3}+\frac {143}{15} x^{4}+13 x^{5}+17 x^{6}+\frac {323}{15} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) \]

Solution by Mathematica

Time used: 0.007 (sec). Leaf size: 96

AsymptoticDSolveValue[2*x^2*(1-x)*y''[x]-x*(1+7*x)*y'[x]+y[x]==0,y[x],{x,0,7}]
 

\[ y(x)\to c_1 x \left (\frac {323 x^7}{15}+17 x^6+13 x^5+\frac {143 x^4}{15}+\frac {33 x^3}{5}+\frac {21 x^2}{5}+\frac {7 x}{3}+1\right )+c_2 \sqrt {x} \left (36 x^7+28 x^6+21 x^5+15 x^4+10 x^3+6 x^2+3 x+1\right ) \]