Internal problem ID [10999]
Internal file name [OUTPUT/10256_Sunday_December_31_2023_11_33_06_AM_29844681/index.tex]
Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev.
Second edition
Section: Chapter 2, Second-Order Differential Equations. section 2.1.2-5 Equation of form
\((a x^2+b x+c) y''+f(x)y'+g(x)y=0\)
Problem number: 176.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_non_constant_coeff_transformation_on_B"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {\left (a \,x^{2}+b x +c \right ) y^{\prime \prime }+\left (k x +d \right ) y^{\prime }-y k=0} \]
Given an ode of the form \begin {align*} A y^{\prime \prime } + B y^{\prime } + C y &= F(x) \end {align*}
This method reduces the order ode the ODE by one by applying the transformation \begin {align*} y&= B v \end {align*}
This results in \begin {align*} y' &=B' v+ v' B \\ y'' &=B'' v+ B' v' +v'' B + v' B' \\ &=v'' B+2 v'+ B'+B'' v \end {align*}
And now the original ode becomes \begin {align*} A\left ( v'' B+2v'B'+ B'' v\right )+B\left ( B'v+ v' B\right ) +CBv & =0\\ ABv'' +\left ( 2AB'+B^{2}\right ) v'+\left (AB''+BB'+CB\right ) v & =0 \tag {1} \end {align*}
If the term \(AB''+BB'+CB\) is zero, then this method works and can be used to solve \[ ABv''+\left ( 2AB' +B^{2}\right ) v'=0 \] By Using \(u=v'\) which reduces the order of the above ode to one. The new ode is \[ ABu'+\left ( 2AB'+B^{2}\right ) u=0 \] The above ode is first order ode which is solved for \(u\). Now a new ode \(v'=u\) is solved for \(v\) as first order ode. Then the final solution is obtain from \(y=Bv\).
This method works only if the term \(AB''+BB'+CB\) is zero. The given ODE shows that \begin {align*} A &= a \,x^{2}+b x +c\\ B &= k x +d\\ C &= -k\\ F &= 0 \end {align*}
The above shows that for this ode \begin {align*} AB''+BB'+CB &= \left (a \,x^{2}+b x +c\right ) \left (0\right ) + \left (k x +d\right ) \left (k\right ) + \left (-k\right ) \left (k x +d\right ) \\ &=0 \end {align*}
Hence the ode in \(v\) given in (1) now simplifies to \begin {align*} \left (a \,x^{2}+b x +c \right ) \left (k x +d \right ) v'' +\left ( 2 k \left (a \,x^{2}+b x +c \right )+\left (k x +d \right )^{2}\right ) v' & =0 \end {align*}
Now by applying \(v'=u\) the above becomes \begin {align*} \left (a \,x^{2}+b x +c \right ) \left (k x +d \right ) u^{\prime }\left (x \right )+2 \left (k \left (a +\frac {k}{2}\right ) x^{2}+k \left (b +d \right ) x +c k +\frac {d^{2}}{2}\right ) u \left (x \right ) = 0 \end {align*}
Which is now solved for \(u\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {u \left (2 a k \,x^{2}+k^{2} x^{2}+2 b k x +2 d k x +2 c k +d^{2}\right )}{\left (a \,x^{2}+b x +c \right ) \left (k x +d \right )} \end {align*}
Where \(f(x)=-\frac {2 a k \,x^{2}+k^{2} x^{2}+2 b k x +2 d k x +2 c k +d^{2}}{\left (a \,x^{2}+b x +c \right ) \left (k x +d \right )}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {2 a k \,x^{2}+k^{2} x^{2}+2 b k x +2 d k x +2 c k +d^{2}}{\left (a \,x^{2}+b x +c \right ) \left (k x +d \right )} \,d x\\ \int { \frac {1}{u} \,du} &= \int {-\frac {2 a k \,x^{2}+k^{2} x^{2}+2 b k x +2 d k x +2 c k +d^{2}}{\left (a \,x^{2}+b x +c \right ) \left (k x +d \right )} \,d x}\\ \ln \left (u \right )&=-2 \ln \left (k x +d \right )-\frac {k \ln \left (a \,x^{2}+b x +c \right )}{2 a}-\frac {2 \left (d -\frac {k b}{2 a}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}+c_{1}\\ u&={\mathrm e}^{-2 \ln \left (k x +d \right )-\frac {k \ln \left (a \,x^{2}+b x +c \right )}{2 a}-\frac {2 \left (d -\frac {k b}{2 a}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}+c_{1}}\\ &=c_{1} {\mathrm e}^{-2 \ln \left (k x +d \right )-\frac {k \ln \left (a \,x^{2}+b x +c \right )}{2 a}-\frac {2 \left (d -\frac {k b}{2 a}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}} \end {align*}
The ode for \(v\) now becomes \begin {align*} v' &= u\\ &=c_{1} {\mathrm e}^{-2 \ln \left (k x +d \right )-\frac {k \ln \left (a \,x^{2}+b x +c \right )}{2 a}-\frac {2 \left (d -\frac {k b}{2 a}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}} \end {align*}
Which is now solved for \(v\). Integrating both sides gives \begin {align*} v \left (x \right ) = \int c_{1} {\mathrm e}^{-2 \ln \left (k x +d \right )-\frac {k \ln \left (a \,x^{2}+b x +c \right )}{2 a}-\frac {2 \left (d -\frac {k b}{2 a}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}d x +c_{2} \end {align*}
Therefore the solution is \begin {align*} y(x) &= B v\\ &= \left (k x +d\right ) \left (\int c_{1} {\mathrm e}^{-2 \ln \left (k x +d \right )-\frac {k \ln \left (a \,x^{2}+b x +c \right )}{2 a}-\frac {2 \left (d -\frac {k b}{2 a}\right ) \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}d x +c_{2}\right ) \\ &= \left (k x +d \right ) \left (c_{1} \left (\int \frac {\left (a \,x^{2}+b x +c \right )^{-\frac {k}{2 a}} {\mathrm e}^{-\frac {2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right ) \left (a d -\frac {b k}{2}\right )}{\sqrt {4 a c -b^{2}}\, a}}}{\left (k x +d \right )^{2}}d x \right )+c_{2} \right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (k x +d \right ) \left (c_{1} \left (\int \frac {\left (a \,x^{2}+b x +c \right )^{-\frac {k}{2 a}} {\mathrm e}^{-\frac {2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right ) \left (a d -\frac {b k}{2}\right )}{\sqrt {4 a c -b^{2}}\, a}}}{\left (k x +d \right )^{2}}d x \right )+c_{2} \right ) \\ \end{align*}
Verification of solutions
\[ y = \left (k x +d \right ) \left (c_{1} \left (\int \frac {\left (a \,x^{2}+b x +c \right )^{-\frac {k}{2 a}} {\mathrm e}^{-\frac {2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right ) \left (a d -\frac {b k}{2}\right )}{\sqrt {4 a c -b^{2}}\, a}}}{\left (k x +d \right )^{2}}d x \right )+c_{2} \right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (a \,x^{2}+b x +c \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (k x +d \right ) y^{\prime }-y k =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 {k y}{a \,x^{2}+b x +c}-\frac {\left (k x +d \right ) y^{\prime }}{a \,x^{2}+b x +c} \\ \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 (k x +d \right ) y^{\prime }}{a \,x^{2}+b x +c}-\frac {k y}{a \,x^{2}+b x +c}=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 {k x +d}{a \,x^{2}+b x +c}, P_{3}\left (x \right )=-\frac {k}{a \,x^{2}+b x +c}\right ] \\ {} & \circ & \left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a} \\ {} & {} & \left (\left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}}}}=0 \\ {} & \circ & {\left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )}^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a} \\ {} & {} & \left ({\left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )}^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}}}}=0 \\ {} & \circ & x =\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\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}=\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a} \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (a \,x^{2}+b x +c \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (k x +d \right ) y^{\prime }-y k =0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u +\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (a \,u^{2}+u \sqrt {-4 a c +b^{2}}\right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (k u -\frac {k b}{2 a}+\frac {k \sqrt {-4 a c +b^{2}}}{2 a}+d \right ) \left (\frac {d}{d u}y \left (u \right )\right )-k 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}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}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}{d u}\frac {d}{d u}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} \\ {} & {} & \frac {a_{0} r \left (2 \sqrt {-4 a c +b^{2}}\, a r -2 \sqrt {-4 a c +b^{2}}\, a +\sqrt {-4 a c +b^{2}}\, k +2 a d -b k \right ) u^{r -1}}{2 a}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (\frac {a_{k +1} \left (k +1+r \right ) \left (2 \sqrt {-4 a c +b^{2}}\, a \left (k +1\right )+2 \sqrt {-4 a c +b^{2}}\, a r -2 \sqrt {-4 a c +b^{2}}\, a +\sqrt {-4 a c +b^{2}}\, k +2 a d -b k \right )}{2 a}+a_{k} \left (k +r -1\right ) \left (a k +a r +k \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \frac {r \left (2 \sqrt {-4 a c +b^{2}}\, a r -2 \sqrt {-4 a c +b^{2}}\, a +\sqrt {-4 a c +b^{2}}\, k +2 a d -b k \right )}{2 a}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k}{2 \sqrt {-4 a c +b^{2}}\, a}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \frac {2 \left (\left (k +r \right ) a +\frac {k}{2}\right ) \left (k +1+r \right ) a_{k +1} \sqrt {-4 a c +b^{2}}+2 a_{k} \left (k +r \right ) \left (k +r -1\right ) a^{2}+\left (2 d \left (k +1+r \right ) a_{k +1}+2 k a_{k} \left (k +r -1\right )\right ) a -b k a_{k +1} \left (k +1+r \right )}{2 a}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {2 a a_{k} \left (a \,k^{2}+2 a k r +a \,r^{2}-a k -a r +k k +k r -k \right )}{2 \sqrt {-4 a c +b^{2}}\, a \,k^{2}+4 \sqrt {-4 a c +b^{2}}\, a k r +2 \sqrt {-4 a c +b^{2}}\, a \,r^{2}+2 \sqrt {-4 a c +b^{2}}\, a k +2 \sqrt {-4 a c +b^{2}}\, a r +\sqrt {-4 a c +b^{2}}\, k k +\sqrt {-4 a c +b^{2}}\, k r +2 a d k +2 a d r -b k k -b k r +\sqrt {-4 a c +b^{2}}\, k +2 a d -b k} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =1 \\ {} & {} & a_{k +1}=-\frac {2 a a_{k} \left (a \,k^{2}-a k +k k -k \right )}{2 \sqrt {-4 a c +b^{2}}\, a \,k^{2}+2 \sqrt {-4 a c +b^{2}}\, a k +\sqrt {-4 a c +b^{2}}\, k k +2 a d k -b k k +\sqrt {-4 a c +b^{2}}\, k +2 a d -b k} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=\frac {2 a a_{0} k}{\sqrt {-4 a c +b^{2}}\, k +2 a d -b k} \\ \bullet & {} & \textrm {Terminating series solution of the ODE for}\hspace {3pt} r =0\hspace {3pt}\textrm {. Use reduction of order to find the second linearly independent solution}\hspace {3pt} \\ {} & {} & y \left (u \right )=a_{0}\cdot \left (1+\frac {2 a k u}{\sqrt {-4 a c +b^{2}}\, k +2 a d -b k}\right ) \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a} \\ {} & {} & \left [y=\frac {2 a_{0} \left (k x +d \right ) a}{\sqrt {-4 a c +b^{2}}\, k +2 a d -b k}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k}{2 \sqrt {-4 a c +b^{2}}\, a} \\ {} & {} & a_{k +1}=-\frac {2 a a_{k} \left (a \,k^{2}+\frac {k \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )}{\sqrt {-4 a c +b^{2}}}+\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )^{2}}{4 a \left (-4 a c +b^{2}\right )}-a k -\frac {2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k}{2 \sqrt {-4 a c +b^{2}}}+k k +\frac {k \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )}{2 \sqrt {-4 a c +b^{2}}\, a}-k \right )}{2 \sqrt {-4 a c +b^{2}}\, a \,k^{2}+2 k \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )+\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )^{2}}{2 \sqrt {-4 a c +b^{2}}\, a}+2 \sqrt {-4 a c +b^{2}}\, a k +2 \sqrt {-4 a c +b^{2}}\, a +\sqrt {-4 a c +b^{2}}\, k k +\frac {k \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )}{2 a}+2 a d k +\frac {d \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )}{\sqrt {-4 a c +b^{2}}}-b k k -\frac {b k \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )}{2 \sqrt {-4 a c +b^{2}}\, a}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k}{2 \sqrt {-4 a c +b^{2}}\, a} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k}{2 \sqrt {-4 a c +b^{2}}\, a}}, a_{k +1}=-\frac {2 a a_{k} \left (a \,k^{2}+\frac {k \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )}{\sqrt {-4 a c +b^{2}}}+\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )^{2}}{4 a \left (-4 a c +b^{2}\right )}-a k -\frac {2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k}{2 \sqrt {-4 a c +b^{2}}}+k k +\frac {k \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )}{2 \sqrt {-4 a c +b^{2}}\, a}-k \right )}{2 \sqrt {-4 a c +b^{2}}\, a \,k^{2}+2 k \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )+\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )^{2}}{2 \sqrt {-4 a c +b^{2}}\, a}+2 \sqrt {-4 a c +b^{2}}\, a k +2 \sqrt {-4 a c +b^{2}}\, a +\sqrt {-4 a c +b^{2}}\, k k +\frac {k \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )}{2 a}+2 a d k +\frac {d \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )}{\sqrt {-4 a c +b^{2}}}-b k k -\frac {b k \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )}{2 \sqrt {-4 a c +b^{2}}\, a}}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} {\left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )}^{k +\frac {2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k}{2 \sqrt {-4 a c +b^{2}}\, a}}, a_{k +1}=-\frac {2 a a_{k} \left (a \,k^{2}+\frac {k \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )}{\sqrt {-4 a c +b^{2}}}+\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )^{2}}{4 a \left (-4 a c +b^{2}\right )}-a k -\frac {2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k}{2 \sqrt {-4 a c +b^{2}}}+k k +\frac {k \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )}{2 \sqrt {-4 a c +b^{2}}\, a}-k \right )}{2 \sqrt {-4 a c +b^{2}}\, a \,k^{2}+2 k \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )+\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )^{2}}{2 \sqrt {-4 a c +b^{2}}\, a}+2 \sqrt {-4 a c +b^{2}}\, a k +2 \sqrt {-4 a c +b^{2}}\, a +\sqrt {-4 a c +b^{2}}\, k k +\frac {k \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )}{2 a}+2 a d k +\frac {d \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )}{\sqrt {-4 a c +b^{2}}}-b k k -\frac {b k \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )}{2 \sqrt {-4 a c +b^{2}}\, a}}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\frac {2 e_{0} \left (k x +d \right ) a}{\sqrt {-4 a c +b^{2}}\, k +2 a d -b k}+\left (\moverset {\infty }{\munderset {m =0}{\sum }}f_{m} {\left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}\right )}^{m +\frac {2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k}{2 \sqrt {-4 a c +b^{2}}\, a}}\right ), f_{m +1}=-\frac {2 a f_{m} \left (m^{2} a +\frac {m \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )}{\sqrt {-4 a c +b^{2}}}+\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )^{2}}{4 a \left (-4 a c +b^{2}\right )}-m a -\frac {2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k}{2 \sqrt {-4 a c +b^{2}}}+k m +\frac {k \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )}{2 \sqrt {-4 a c +b^{2}}\, a}-k \right )}{2 \sqrt {-4 a c +b^{2}}\, a \,m^{2}+2 m \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )+\frac {\left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )^{2}}{2 \sqrt {-4 a c +b^{2}}\, a}+2 \sqrt {-4 a c +b^{2}}\, a m +2 \sqrt {-4 a c +b^{2}}\, a +\sqrt {-4 a c +b^{2}}\, k m +\frac {k \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )}{2 a}+2 a d m +\frac {d \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )}{\sqrt {-4 a c +b^{2}}}-b k m -\frac {b k \left (2 \sqrt {-4 a c +b^{2}}\, a -\sqrt {-4 a c +b^{2}}\, k -2 a d +b k \right )}{2 \sqrt {-4 a c +b^{2}}\, a}}\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) Group is reducible, not completely reducible Solution has integrals. Trying a special function solution free of integrals... -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach <- heuristic approach successful <- hypergeometric successful <- special function solution successful -> Trying to convert hypergeometric functions to elementary form... <- elementary form for at least one hypergeometric solution is achieved - returning with no uncomputed integrals <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.094 (sec). Leaf size: 315
dsolve((a*x^2+b*x+c)*diff(y(x),x$2)+(k*x+d)*diff(y(x),x)-k*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \left (k x +d \right )+c_{2} {\left (2 \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}\, x \,a^{2}+\sqrt {\frac {-4 a c +b^{2}}{a^{2}}}\, b a -4 a c +b^{2}\right )}^{\frac {a \left (a -\frac {k}{2}\right ) \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}+a d -\frac {k b}{2}}{\sqrt {\frac {-4 a c +b^{2}}{a^{2}}}\, a^{2}}} \operatorname {hypergeom}\left (\left [-\frac {k \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}\, a -2 a d +k b}{2 a^{2} \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}}, \frac {a \left (a +\frac {k}{2}\right ) \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}+a d -\frac {k b}{2}}{\sqrt {\frac {-4 a c +b^{2}}{a^{2}}}\, a^{2}}\right ], \left [\frac {4 a^{2} \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}-k \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}\, a +2 a d -k b}{2 a^{2} \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}}\right ], \frac {\left (-2 a^{2} x -a b \right ) \sqrt {\frac {-4 a c +b^{2}}{a^{2}}}+4 a c -b^{2}}{8 a c -2 b^{2}}\right ) \]
✓ Solution by Mathematica
Time used: 4.256 (sec). Leaf size: 107
DSolve[(a*x^2+b*x+c)*y''[x]+(k*x+d)*y'[x]-k*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {(d+k x) \left (c_2 \int _1^x\frac {\exp \left (\frac {(b k-2 a d) \arctan \left (\frac {b+2 a K[1]}{\sqrt {4 a c-b^2}}\right )}{a \sqrt {4 a c-b^2}}\right ) (c+K[1] (b+a K[1]))^{-\frac {k}{2 a}}}{(d+k K[1])^2}dK[1]+c_1\right )}{d} \]