problem 626

Internal problem ID [15394]
Internal ﬁle name [OUTPUT/15395_Wednesday_May_08_2024_03_57_50_PM_74824681/index.tex]

Book: A book of problems in ordinary diﬀerential equations. M.L. KRASNOV, A.L. KISELYOV, G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 15.4 Nonhomogeneous linear equations with constant coeﬃcients. The Euler equations. Exercises page 143
Problem number: 626.
ODE order: 3.
ODE degree: 1.

\[ \boxed {\left (x +1\right )^{2} y^{\prime \prime \prime }-12 y^{\prime }=0} \] Since \(y\) is missing from the ode then we can use the substitution \(y^{\prime } = v \left (x \right )\) to reduce the order by one. The ODE becomes \begin {align*} -12 v \left (x \right )+\left (x^{2}+2 x +1\right ) v^{\prime \prime }\left (x \right ) = 0 \end {align*}

Writing the ode as \begin {align*} v^{\prime \prime }\left (x \right ) \left (x +1\right )^{2}-12 v \left (x \right ) &= 0 \tag {1} \\ A v^{\prime \prime }\left (x \right ) + B v^{\prime }\left (x \right ) + C v \left (x \right ) &= 0 \tag {2} \end {align*}

Comparing (1) and (2) shows that \begin {align*} A &= \left (x +1\right )^{2} \\ B &= 0\tag {3} \\ C &= -12 \end {align*}

Applying the Liouville transformation on the dependent variable gives \begin {align*} z(x) &= v \left (x \right ) e^{\int \frac {B}{2 A} \,dx} \end {align*}

Where \(r\) is given by \begin {align*} r &= \frac {s}{t}\tag {5} \\ &= \frac {2 A B' - 2 B A' + B^2 - 4 A C}{4 A^2} \end {align*}

Substituting the values of \(A,B,C\) from (3) in the above and simplifying gives \begin {align*} r &= \frac {12}{\left (x +1\right )^{2}}\tag {6} \end {align*}

Comparing the above to (5) shows that \begin {align*} s &= 12\\ t &= \left (x +1\right )^{2} \end {align*}

Therefore eq. (4) becomes \begin {align*} z''(x) &= \left ( \frac {12}{\left (x +1\right )^{2}}\right ) z(x)\tag {7} \end {align*}

Equation (7) is now solved. After ﬁnding \(z(x)\) then \(v \left (x \right )\) is found using the inverse transformation \begin {align*} v \left (x \right ) &= z \left (x \right ) e^{-\int \frac {B}{2 A} \,dx} \end {align*}

The ﬁrst step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases depending on the order of poles of \(r\) and the order of \(r\) at \(\infty \). The following table summarizes these cases.


Case	Allowed pole order for \(r\)	Allowed value for \(\mathcal {O}(\infty )\)

1	\(\left \{ 0,1,2,4,6,8,\cdots \right \} \)	\(\left \{ \cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \right \} \)

2	Need to have at least one pole that is either order \(2\) or odd order greater than \(2\). Any other pole order is allowed as long as the above condition is satisﬁed. Hence the following set of pole orders are all allowed. \(\{1,2\}\),\(\{1,3\}\),\(\{2\}\),\(\{3\}\),\(\{3,4\}\),\(\{1,2,5\}\).	no condition

3	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)

Table 314: Necessary conditions for each Kovacic case

The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\). Therefore \begin {align*} O\left (\infty \right ) &= \text {deg}(t) - \text {deg}(s) \\ &= 2 - 0 \\ &= 2 \end {align*}

The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=\left (x +1\right )^{2}\). There is a pole at \(x=-1\) of order \(2\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(2\) then the necessary conditions for case one are met. Since there is a pole of order \(2\) then necessary conditions for case two are met. Since pole order is not larger than \(2\) and the order at \(\infty \) is \(2\) then the necessary conditions for case three are met. Therefore \begin {align*} L &= [1, 2, 4, 6, 12] \end {align*}

Looking at poles of order 2. The partial fractions decomposition of \(r\) is \[ r = \frac {12}{\left (x +1\right )^{2}} \] For the pole at \(x=-1\) let \(b\) be the coeﬃcient of \(\frac {1}{ \left (x +1\right )^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b=12\). Hence \begin {alignat*} {2} [\sqrt r]_c &= 0 \\ \alpha _c^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= 4\\ \alpha _c^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= -3 \end {alignat*}

Since the order of \(r\) at \(\infty \) is 2 then \([\sqrt r]_\infty =0\). Let \(b\) be the coeﬃcient of \(\frac {1}{x^{2}}\) in the Laurent series expansion of \(r\) at \(\infty \). which can be found by dividing the leading coeﬃcient of \(s\) by the leading coeﬃcient of \(t\) from \begin {alignat*} {2} r &= \frac {s}{t} &&= \frac {12}{\left (x +1\right )^{2}} \end {alignat*}

Since the \(\text {gcd}(s,t)=1\). This gives \(b=12\). Hence \begin {alignat*} {2} [\sqrt r]_\infty &= 0 \\ \alpha _{\infty }^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= 4\\ \alpha _{\infty }^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= -3 \end {alignat*}

The following table summarizes the ﬁndings so far for poles and for the order of \(r\) at \(\infty \) where \(r\) is \[ r=\frac {12}{\left (x +1\right )^{2}} \]

Now that the all \([\sqrt r]_c\) and its associated \(\alpha _c^{\pm }\) have been determined for all the poles in the set \(\Gamma \) and \([\sqrt r]_\infty \) and its associated \(\alpha _\infty ^{\pm }\) have also been found, the next step is to determine possible non negative integer \(d\) from these using \begin {align*} d &= \alpha _\infty ^{s(\infty )} - \sum _{c \in \Gamma } \alpha _c^{s(c)} \end {align*}

Where \(s(c)\) is either \(+\) or \(-\) and \(s(\infty )\) is the sign of \(\alpha _\infty ^{\pm }\). This is done by trial over all set of families \(s=(s(c))_{c \in \Gamma \cup {\infty }}\) until such \(d\) is found to work in ﬁnding candidate \(\omega \). Trying \(\alpha _\infty ^{-} = -3\) then \begin {align*} d &= \alpha _\infty ^{-} - \left ( \alpha _{c_1}^{-} \right ) \\ &= -3 - \left ( -3 \right ) \\ &= 0 \end {align*}

Since \(d\) an integer and \(d \geq 0\) then it can be used to ﬁnd \(\omega \) using \begin {align*} \omega &= \sum _{c \in \Gamma } \left ( s(c) [\sqrt r]_c + \frac {\alpha _c^{s(c)}}{x-c} \right ) + s(\infty ) [\sqrt r]_\infty \end {align*}

The above gives \begin {align*} \omega &= \left ( (-)[\sqrt r]_{c_1} + \frac { \alpha _{c_1}^{-} }{x- c_1}\right ) + (-) [\sqrt r]_\infty \\ &= -\frac {3}{x +1} + (-) \left ( 0 \right ) \\ &= -\frac {3}{x +1}\\ &= -\frac {3}{x +1} \end {align*}

Now that \(\omega \) is determined, the next step is ﬁnd a corresponding minimal polynomial \(p(x)\) of degree \(d=0\) to solve the ode. The polynomial \(p(x)\) needs to satisfy the equation \begin {align*} p'' + 2 \omega p' + \left ( \omega ' +\omega ^2 -r\right ) p = 0 \tag {1A} \end {align*}

Substituting the above in eq. (1A) gives \begin {align*} \left (0\right ) + 2 \left (-\frac {3}{x +1}\right ) \left (0\right ) + \left ( \left (\frac {3}{\left (x +1\right )^{2}}\right ) + \left (-\frac {3}{x +1}\right )^2 - \left (\frac {12}{\left (x +1\right )^{2}}\right ) \right ) &= 0\\ 0 = 0 \end {align*}

The equation is satisﬁed since both sides are zero. Therefore the ﬁrst solution to the ode \(z'' = r z\) is \begin {align*} z_1(x) &= p e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int -\frac {3}{x +1}d x}\\ &= \frac {1}{\left (x +1\right )^{3}} \end {align*}

The ﬁrst solution to the original ode in \(v \left (x \right )\) is found from \[ v_1 = z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dx} \]

Since \(B=0\) then the above reduces to \begin{align*} v_1 &= z_1 \\ &= \frac {1}{\left (x +1\right )^{3}} \\ \end{align*} Which simpliﬁes to \[ v_1 = \frac {1}{\left (x +1\right )^{3}} \] The second solution \(v_2\) to the original ode is found using reduction of order \[ v_2 = v_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{v_1^2} \,dx \] Since \(B=0\) then the above becomes \begin{align*} v_2 &= v_1 \int \frac {1}{v_1^2} \,dx \\ &= \frac {1}{\left (x +1\right )^{3}}\int \frac {1}{\frac {1}{\left (x +1\right )^{6}}} \,dx \\ &= \frac {1}{\left (x +1\right )^{3}}\left (\frac {\left (x +1\right )^{7}}{7}\right ) \\ \end{align*} Therefore the solution is

\begin{align*} v \left (x \right ) &= c_{1} v_1 + c_{2} v_2 \\ &= c_{1} \left (\frac {1}{\left (x +1\right )^{3}}\right ) + c_{2} \left (\frac {1}{\left (x +1\right )^{3}}\left (\frac {\left (x +1\right )^{7}}{7}\right )\right ) \\ \end{align*} But since \(y^{\prime } = v \left (x \right )\) then we now need to solve the ode \(y^{\prime } = \frac {c_{1}}{\left (x +1\right )^{3}}+\frac {c_{2} \left (x +1\right )^{4}}{7}\). Integrating both sides gives \begin {align*} y &= \int { \frac {c_{2} x^{7}+7 c_{2} x^{6}+21 c_{2} x^{5}+35 c_{2} x^{4}+35 c_{2} x^{3}+21 c_{2} x^{2}+7 x c_{2} +7 c_{1} +c_{2}}{7 \left (x +1\right )^{3}}\,\mathop {\mathrm {d}x}}\\ &= \frac {c_{2} x^{5}}{35}+\frac {c_{2} x^{4}}{7}+\frac {2 c_{2} x^{3}}{7}+\frac {2 c_{2} x^{2}}{7}+\frac {x c_{2}}{7}-\frac {c_{1}}{2 \left (x +1\right )^{2}}+c_{3} \end {align*}

This is higher order nonhomogeneous ODE. Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ODE And \(y_p\) is a particular solution to the nonhomogeneous ODE. \(y_h\) is the solution to \[ -12 y^{\prime }+\left (x^{2}+2 x +1\right ) y^{\prime \prime \prime } = 0 \] Let the particular solution be \[ y_p = U_1 y_1+U_2 y_2+U_3 y_3 \] Where \(y_i\) are the basis solutions found above for the homogeneous solution \(y_h\) and \(U_i(x)\) are functions to be determined as follows \[ U_i = (-1)^{n-i} \int { \frac {F(x) W_i(x) }{a W(x)} \, dx} \] Where \(W(x)\) is the Wronskian and \(W_i(x)\) is the Wronskian that results after deleting the last row and the \(i\)-th column of the determinant and \(n\) is the order of the ODE or equivalently, the number of basis solutions, and \(a\) is the coeﬃcient of the leading derivative in the ODE, and \(F(x)\) is the RHS of the ODE. Therefore, the ﬁrst step is to ﬁnd the Wronskian \(W \left (x \right )\). This is given by \begin {equation*} W(x) = \begin {vmatrix} y_1&y_2&y_3\\ y_1'&y_2'&y_3'\\ y_1''&y_2''&y_3''\\ \end {vmatrix} \end {equation*} Substituting the fundamental set of solutions \(y_i\) found above in the Wronskian gives \begin {align*} W &= \left [\begin {array}{ccc} 1 & \frac {1}{\left (x +1\right )^{2}} & \frac {x \left (x^{4}+5 x^{3}+10 x^{2}+10 x +5\right )}{35} \\ 0 & -\frac {2}{\left (x +1\right )^{3}} & \frac {\left (x +1\right )^{4}}{7} \\ 0 & \frac {6}{\left (x +1\right )^{4}} & \frac {4 \left (x +1\right )^{3}}{7} \end {array}\right ] \\ |W| &= -2 \end {align*}

Now we determine \(W_i\) for each \(U_i\). \begin {align*} W_1(x) &= \det \,\left [\begin {array}{cc} \frac {1}{\left (x +1\right )^{2}} & \frac {x \left (x^{4}+5 x^{3}+10 x^{2}+10 x +5\right )}{35} \\ -\frac {2}{\left (x +1\right )^{3}} & \frac {\left (x +1\right )^{4}}{7} \end {array}\right ] \\ &= \frac {7 x^{5}+35 x^{4}+70 x^{3}+70 x^{2}+35 x +5}{35 \left (x +1\right )^{3}} \end {align*}

\begin {align*} W_2(x) &= \det \,\left [\begin {array}{cc} 1 & \frac {x \left (x^{4}+5 x^{3}+10 x^{2}+10 x +5\right )}{35} \\ 0 & \frac {\left (x +1\right )^{4}}{7} \end {array}\right ] \\ &= \frac {\left (x +1\right )^{4}}{7} \end {align*}

\begin {align*} W_3(x) &= \det \,\left [\begin {array}{cc} 1 & \frac {1}{\left (x +1\right )^{2}} \\ 0 & -\frac {2}{\left (x +1\right )^{3}} \end {array}\right ] \\ &= -\frac {2}{\left (x +1\right )^{3}} \end {align*}

Now we are ready to evaluate each \(U_i(x)\). \begin {align*} U_1 &= (-1)^{3-1} \int { \frac {F(x) W_1(x) }{a W(x)} \, dx}\\ &= (-1)^{2} \int { \frac { \left (-\left (x +1\right )^{2} y^{\prime \prime \prime }+\left (x^{2}+2 x +1\right ) y^{\prime \prime \prime }\right ) \left (\frac {7 x^{5}+35 x^{4}+70 x^{3}+70 x^{2}+35 x +5}{35 \left (x +1\right )^{3}}\right )}{\left (x^{2}+2 x +1\right ) \left (-2\right )} \, dx} \\ &= \int { \frac {\frac {\left (-\left (x +1\right )^{2} y^{\prime \prime \prime }+\left (x^{2}+2 x +1\right ) y^{\prime \prime \prime }\right ) \left (7 x^{5}+35 x^{4}+70 x^{3}+70 x^{2}+35 x +5\right )}{35 \left (x +1\right )^{3}}}{-2 x^{2}-4 x -2} \, dx}\\ &= \int {\left (0\right ) \, dx} \\ &= 0 \end {align*}

\begin {align*} U_2 &= (-1)^{3-2} \int { \frac {F(x) W_2(x) }{a W(x)} \, dx}\\ &= (-1)^{1} \int { \frac { \left (-\left (x +1\right )^{2} y^{\prime \prime \prime }+\left (x^{2}+2 x +1\right ) y^{\prime \prime \prime }\right ) \left (\frac {\left (x +1\right )^{4}}{7}\right )}{\left (x^{2}+2 x +1\right ) \left (-2\right )} \, dx} \\ &= - \int { \frac {\frac {\left (-\left (x +1\right )^{2} y^{\prime \prime \prime }+\left (x^{2}+2 x +1\right ) y^{\prime \prime \prime }\right ) \left (x +1\right )^{4}}{7}}{-2 x^{2}-4 x -2} \, dx}\\ &= - \int {\left (0\right ) \, dx} \\ &= 0 \end {align*}

\begin {align*} U_3 &= (-1)^{3-3} \int { \frac {F(x) W_3(x) }{a W(x)} \, dx}\\ &= (-1)^{0} \int { \frac { \left (-\left (x +1\right )^{2} y^{\prime \prime \prime }+\left (x^{2}+2 x +1\right ) y^{\prime \prime \prime }\right ) \left (-\frac {2}{\left (x +1\right )^{3}}\right )}{\left (x^{2}+2 x +1\right ) \left (-2\right )} \, dx} \\ &= \int { \frac {-\frac {2 \left (-\left (x +1\right )^{2} y^{\prime \prime \prime }+\left (x^{2}+2 x +1\right ) y^{\prime \prime \prime }\right )}{\left (x +1\right )^{3}}}{-2 x^{2}-4 x -2} \, dx}\\ &= \int {\left (0\right ) \, dx} \\ &= 0 \end {align*}

Now that all the \(U_i\) functions have been determined, the particular solution is found from \[ y_p = U_1 y_1+U_2 y_2+U_3 y_3 \] Hence \begin {equation*} \begin {split} y_p &= \left (0\right ) \\ &+\left (0\right ) \left (\frac {1}{\left (x +1\right )^{2}}\right ) \\ &+\left (0\right ) \left (\frac {1}{35} x^{5}+\frac {1}{7} x^{4}+\frac {2}{7} x^{3}+\frac {2}{7} x^{2}+\frac {1}{7} x\right ) \end {split} \end {equation*} Therefore the particular solution is \[ y_p = 0 \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (y &= \frac {c_{2} x^{5}}{35}+\frac {c_{2} x^{4}}{7}+\frac {2 c_{2} x^{3}}{7}+\frac {2 c_{2} x^{2}}{7}+\frac {x c_{2}}{7}-\frac {c_{1}}{2 \left (x +1\right )^{2}}+c_{3}\right ) + \left (0\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{2} x^{5}}{35}+\frac {c_{2} x^{4}}{7}+\frac {2 c_{2} x^{3}}{7}+\frac {2 c_{2} x^{2}}{7}+\frac {x c_{2}}{7}-\frac {c_{1}}{2 \left (x +1\right )^{2}}+c_{3} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \frac {c_{2} x^{5}}{35}+\frac {c_{2} x^{4}}{7}+\frac {2 c_{2} x^{3}}{7}+\frac {2 c_{2} x^{2}}{7}+\frac {x c_{2}}{7}-\frac {c_{1}}{2 \left (x +1\right )^{2}}+c_{3} \] Veriﬁed OK.

19.9.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x +1\right )^{2} y^{\prime \prime \prime }-12 y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \prime \prime } \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=-1\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=0, P_{3}\left (x \right )=-\frac {12}{\left (x +1\right )^{2}}, P_{4}\left (x \right )=0\right ] \\ {} & \circ & \left (x +1\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=0 \\ {} & \circ & \left (x +1\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=-12 \\ {} & \circ & \left (x +1\right )^{3}\cdot P_{4}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )^{3}\cdot P_{4}\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}=-1\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=-1 \\ \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 \\ {} & {} & u^{2} \left (\frac {d^{3}}{d u^{3}}y \left (u \right )\right )-12 \frac {d}{d u}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 DE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} \frac {d}{d u}y \left (u \right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d}{d u}y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \frac {d}{d u}y \left (u \right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{2}\cdot \left (\frac {d^{3}}{d u^{3}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & u^{2}\cdot \left (\frac {d^{3}}{d u^{3}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) \left (k +r -2\right ) u^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & u^{2}\cdot \left (\frac {d^{3}}{d u^{3}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) \left (k +r -1\right ) u^{k +r} \\ & {} & \textrm {Rewrite DE with series expansions}\hspace {3pt} \\ {} & {} & \moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +3+r \right ) \left (k -4+r \right ) u^{k +r}=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1\right ) \left (k +3\right ) \left (k -4\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=0 \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=0 \\ \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}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k}, a_{k +1}=0\right ] \end {array} \]

Maple trace

`Methods for third order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
checking if the LODE is of Euler type 
<- LODE of Euler type successful`

\[ y \left (x \right ) = c_{1} +\frac {c_{2}}{\left (1+x \right )^{2}}+c_{3} \left (1+x \right )^{5} \]


pole \(c\) location	pole order	\([\sqrt r]_c\)	\(\alpha _c^{+}\)	\(\alpha _c^{-}\)

\(-1\)	\(2\)	\(0\)	\(4\)	\(-3\)


Order of \(r\) at \(\infty \)	\([\sqrt r]_\infty \)	\(\alpha _\infty ^{+}\)	\(\alpha _\infty ^{-}\)

\(2\)	\(0\)	\(4\)	\(-3\)

19.9 problem 626

19.9.1 Maple step by step solution