Internal problem ID [9474]
Internal file name [OUTPUT/8414_Monday_June_06_2022_03_00_41_AM_78583540/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1145.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {\left (\operatorname {a2} x +\operatorname {b2} \right ) y^{\prime \prime }+\left (\operatorname {a1} x +\operatorname {b1} \right ) y^{\prime }+\left (\operatorname {a0} x +\operatorname {b0} \right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\mathit {a2} x +\mathit {b2} \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (\mathit {a1} x +\mathit {b1} \right ) y^{\prime }+\left (\mathit {a0} x +\mathit {b0} \right ) 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 {\left (\mathit {a0} x +\mathit {b0} \right ) y}{\mathit {a2} x +\mathit {b2}}-\frac {\left (\mathit {a1} x +\mathit {b1} \right ) y^{\prime }}{\mathit {a2} x +\mathit {b2}} \\ \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 (\mathit {a1} x +\mathit {b1} \right ) y^{\prime }}{\mathit {a2} x +\mathit {b2}}+\frac {\left (\mathit {a0} x +\mathit {b0} \right ) y}{\mathit {a2} x +\mathit {b2}}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=-\frac {\mathit {b2}}{\mathit {a2}}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {\mathit {a1} x +\mathit {b1}}{\mathit {a2} x +\mathit {b2}}, P_{3}\left (x \right )=\frac {\mathit {a0} x +\mathit {b0}}{\mathit {a2} x +\mathit {b2}}\right ] \\ {} & \circ & \left (x +\frac {\mathit {b2}}{\mathit {a2}}\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-\frac {\mathit {b2}}{\mathit {a2}} \\ {} & {} & \left (\left (x +\frac {\mathit {b2}}{\mathit {a2}}\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-\frac {\mathit {b2}}{\mathit {a2}}}}}=\frac {-\frac {\mathit {a1} \mathit {b2}}{\mathit {a2}}+\mathit {b1}}{\mathit {a2}} \\ {} & \circ & \left (x +\frac {\mathit {b2}}{\mathit {a2}}\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-\frac {\mathit {b2}}{\mathit {a2}} \\ {} & {} & \left (\left (x +\frac {\mathit {b2}}{\mathit {a2}}\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-\frac {\mathit {b2}}{\mathit {a2}}}}}=0 \\ {} & \circ & x =-\frac {\mathit {b2}}{\mathit {a2}}\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=-\frac {\mathit {b2}}{\mathit {a2}}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=-\frac {\mathit {b2}}{\mathit {a2}} \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (\mathit {a2} x +\mathit {b2} \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (\mathit {a1} x +\mathit {b1} \right ) y^{\prime }+\left (\mathit {a0} x +\mathit {b0} \right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -\frac {\mathit {b2}}{\mathit {a2}}\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \mathit {a2} u \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (\mathit {a1} u -\frac {\mathit {a1} \mathit {b2}}{\mathit {a2}}+\mathit {b1} \right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (\mathit {a0} u -\frac {\mathit {a0} \mathit {b2}}{\mathit {a2}}+\mathit {b0} \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 y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \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 \cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & u \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 -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & u \cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -\frac {a_{0} r \left (-\mathit {a2}^{2} r +\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1} \right ) u^{-1+r}}{\mathit {a2}}+\left (-\frac {a_{1} \left (1+r \right ) \left (-\mathit {a2}^{2} r +\mathit {a1} \mathit {b2} -\mathit {a2} \mathit {b1} \right )}{\mathit {a2}}-\frac {a_{0} \left (-\mathit {a1} \mathit {a2} r +\mathit {a0} \mathit {b2} -\mathit {a2} \mathit {b0} \right )}{\mathit {a2}}\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (-\frac {a_{k +1} \left (k +1+r \right ) \left (-\mathit {a2}^{2} \left (k +1\right )-\mathit {a2}^{2} r +\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1} \right )}{\mathit {a2}}-\frac {a_{k} \left (-\mathit {a1} \mathit {a2} k -\mathit {a1} \mathit {a2} r +\mathit {a0} \mathit {b2} -\mathit {a2} \mathit {b0} \right )}{\mathit {a2}}+a_{k -1} \mathit {a0} \right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -\frac {r \left (-\mathit {a2}^{2} r +\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1} \right )}{\mathit {a2}}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1}}{\mathit {a2}^{2}}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & -\frac {a_{1} \left (1+r \right ) \left (-\mathit {a2}^{2} r +\mathit {a1} \mathit {b2} -\mathit {a2} \mathit {b1} \right )}{\mathit {a2}}-\frac {a_{0} \left (-\mathit {a1} \mathit {a2} r +\mathit {a0} \mathit {b2} -\mathit {a2} \mathit {b0} \right )}{\mathit {a2}}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \frac {a_{k +1} \left (k +1+r \right ) \left (k +r \right ) \mathit {a2}^{2}+\left (\mathit {b1} \left (k +1+r \right ) a_{k +1}+k \mathit {a1} a_{k}+r \mathit {a1} a_{k}+a_{k -1} \mathit {a0} +a_{k} \mathit {b0} \right ) \mathit {a2} -\mathit {b2} \left (\mathit {a1} \left (k +1+r \right ) a_{k +1}+a_{k} \mathit {a0} \right )}{\mathit {a2}}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \frac {a_{k +2} \left (k +2+r \right ) \left (k +1+r \right ) \mathit {a2}^{2}+\left (\mathit {b1} \left (k +2+r \right ) a_{k +2}+\left (k +1\right ) \mathit {a1} a_{k +1}+r \mathit {a1} a_{k +1}+a_{k} \mathit {a0} +a_{k +1} \mathit {b0} \right ) \mathit {a2} -\mathit {b2} \left (\mathit {a1} \left (k +2+r \right ) a_{k +2}+a_{k +1} \mathit {a0} \right )}{\mathit {a2}}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {\mathit {a1} \mathit {a2} k a_{k +1}+\mathit {a1} \mathit {a2} r a_{k +1}+\mathit {a0} \mathit {a2} a_{k}-\mathit {a0} \mathit {b2} a_{k +1}+\mathit {a1} \mathit {a2} a_{k +1}+\mathit {a2} \mathit {b0} a_{k +1}}{\left (k +2+r \right ) \left (-\mathit {a2}^{2} k -\mathit {a2}^{2} r +\mathit {a1} \mathit {b2} -\mathit {a2}^{2}-\mathit {a2} \mathit {b1} \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {\mathit {a1} \mathit {a2} k a_{k +1}+\mathit {a0} \mathit {a2} a_{k}-\mathit {a0} \mathit {b2} a_{k +1}+\mathit {a1} \mathit {a2} a_{k +1}+\mathit {a2} \mathit {b0} a_{k +1}}{\left (k +2\right ) \left (-\mathit {a2}^{2} k +\mathit {a1} \mathit {b2} -\mathit {a2}^{2}-\mathit {a2} \mathit {b1} \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 +2}=\frac {\mathit {a1} \mathit {a2} k a_{k +1}+\mathit {a0} \mathit {a2} a_{k}-\mathit {a0} \mathit {b2} a_{k +1}+\mathit {a1} \mathit {a2} a_{k +1}+\mathit {a2} \mathit {b0} a_{k +1}}{\left (k +2\right ) \left (-\mathit {a2}^{2} k +\mathit {a1} \mathit {b2} -\mathit {a2}^{2}-\mathit {a2} \mathit {b1} \right )}, -\frac {a_{1} \left (\mathit {a1} \mathit {b2} -\mathit {a2} \mathit {b1} \right )}{\mathit {a2}}-\frac {a_{0} \left (\mathit {a0} \mathit {b2} -\mathit {a2} \mathit {b0} \right )}{\mathit {a2}}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +\frac {\mathit {b2}}{\mathit {a2}} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +\frac {\mathit {b2}}{\mathit {a2}}\right )^{k}, a_{k +2}=\frac {\mathit {a1} \mathit {a2} k a_{k +1}+\mathit {a0} \mathit {a2} a_{k}-\mathit {a0} \mathit {b2} a_{k +1}+\mathit {a1} \mathit {a2} a_{k +1}+\mathit {a2} \mathit {b0} a_{k +1}}{\left (k +2\right ) \left (-\mathit {a2}^{2} k +\mathit {a1} \mathit {b2} -\mathit {a2}^{2}-\mathit {a2} \mathit {b1} \right )}, -\frac {a_{1} \left (\mathit {a1} \mathit {b2} -\mathit {a2} \mathit {b1} \right )}{\mathit {a2}}-\frac {a_{0} \left (\mathit {a0} \mathit {b2} -\mathit {a2} \mathit {b0} \right )}{\mathit {a2}}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1}}{\mathit {a2}^{2}} \\ {} & {} & a_{k +2}=\frac {\mathit {a1} \mathit {a2} k a_{k +1}+\frac {\mathit {a1} \left (\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1} \right ) a_{k +1}}{\mathit {a2}}+\mathit {a0} \mathit {a2} a_{k}-\mathit {a0} \mathit {b2} a_{k +1}+\mathit {a1} \mathit {a2} a_{k +1}+\mathit {a2} \mathit {b0} a_{k +1}}{\left (k +2+\frac {\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1}}{\mathit {a2}^{2}}\right ) \left (-\mathit {a2}^{2} k -2 \mathit {a2}^{2}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1}}{\mathit {a2}^{2}} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1}}{\mathit {a2}^{2}}}, a_{k +2}=\frac {\mathit {a1} \mathit {a2} k a_{k +1}+\frac {\mathit {a1} \left (\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1} \right ) a_{k +1}}{\mathit {a2}}+\mathit {a0} \mathit {a2} a_{k}-\mathit {a0} \mathit {b2} a_{k +1}+\mathit {a1} \mathit {a2} a_{k +1}+\mathit {a2} \mathit {b0} a_{k +1}}{\left (k +2+\frac {\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1}}{\mathit {a2}^{2}}\right ) \left (-\mathit {a2}^{2} k -2 \mathit {a2}^{2}\right )}, a_{1} \left (1+\frac {\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1}}{\mathit {a2}^{2}}\right ) \mathit {a2} -\frac {a_{0} \left (-\frac {\mathit {a1} \left (\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1} \right )}{\mathit {a2}}+\mathit {a0} \mathit {b2} -\mathit {a2} \mathit {b0} \right )}{\mathit {a2}}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +\frac {\mathit {b2}}{\mathit {a2}} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +\frac {\mathit {b2}}{\mathit {a2}}\right )^{k +\frac {\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1}}{\mathit {a2}^{2}}}, a_{k +2}=\frac {\mathit {a1} \mathit {a2} k a_{k +1}+\frac {\mathit {a1} \left (\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1} \right ) a_{k +1}}{\mathit {a2}}+\mathit {a0} \mathit {a2} a_{k}-\mathit {a0} \mathit {b2} a_{k +1}+\mathit {a1} \mathit {a2} a_{k +1}+\mathit {a2} \mathit {b0} a_{k +1}}{\left (k +2+\frac {\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1}}{\mathit {a2}^{2}}\right ) \left (-\mathit {a2}^{2} k -2 \mathit {a2}^{2}\right )}, a_{1} \left (1+\frac {\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1}}{\mathit {a2}^{2}}\right ) \mathit {a2} -\frac {a_{0} \left (-\frac {\mathit {a1} \left (\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1} \right )}{\mathit {a2}}+\mathit {a0} \mathit {b2} -\mathit {a2} \mathit {b0} \right )}{\mathit {a2}}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +\frac {\mathit {b2}}{\mathit {a2}}\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +\frac {\mathit {b2}}{\mathit {a2}}\right )^{k +\frac {\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1}}{\mathit {a2}^{2}}}\right ), a_{k +2}=\frac {\mathit {a1} \mathit {a2} k a_{k +1}+\mathit {a0} \mathit {a2} a_{k}-\mathit {a0} \mathit {b2} a_{k +1}+\mathit {a1} \mathit {a2} a_{k +1}+\mathit {a2} \mathit {b0} a_{k +1}}{\left (k +2\right ) \left (-\mathit {a2}^{2} k +\mathit {a1} \mathit {b2} -\mathit {a2}^{2}-\mathit {a2} \mathit {b1} \right )}, -\frac {a_{1} \left (\mathit {a1} \mathit {b2} -\mathit {a2} \mathit {b1} \right )}{\mathit {a2}}-\frac {a_{0} \left (\mathit {a0} \mathit {b2} -\mathit {a2} \mathit {b0} \right )}{\mathit {a2}}=0, b_{k +2}=\frac {\mathit {a1} \mathit {a2} k b_{k +1}+\frac {\mathit {a1} \left (\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1} \right ) b_{k +1}}{\mathit {a2}}+\mathit {a0} \mathit {a2} b_{k}-\mathit {a0} \mathit {b2} b_{k +1}+\mathit {a1} \mathit {a2} b_{k +1}+\mathit {a2} \mathit {b0} b_{k +1}}{\left (k +2+\frac {\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1}}{\mathit {a2}^{2}}\right ) \left (-\mathit {a2}^{2} k -2 \mathit {a2}^{2}\right )}, b_{1} \left (1+\frac {\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1}}{\mathit {a2}^{2}}\right ) \mathit {a2} -\frac {b_{0} \left (-\frac {\mathit {a1} \left (\mathit {a1} \mathit {b2} +\mathit {a2}^{2}-\mathit {a2} \mathit {b1} \right )}{\mathit {a2}}+\mathit {a0} \mathit {b2} -\mathit {a2} \mathit {b0} \right )}{\mathit {a2}}=0\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 -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Kummer successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.078 (sec). Leaf size: 248
dsolve((a2*x+b2)*diff(diff(y(x),x),x)+(a1*x+b1)*diff(y(x),x)+(a0*x+b0)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{-\frac {\left (\sqrt {-4 \operatorname {a0} \operatorname {a2} +\operatorname {a1}^{2}}+\operatorname {a1} \right ) x}{2 \operatorname {a2}}} \left (\operatorname {a2} x +\operatorname {b2} \right )^{\frac {\operatorname {a1} \operatorname {b2} +\operatorname {a2}^{2}-\operatorname {a2} \operatorname {b1}}{\operatorname {a2}^{2}}} \left (\operatorname {KummerU}\left (\frac {\left (\operatorname {a1} \operatorname {b2} +2 \operatorname {a2}^{2}-\operatorname {a2} \operatorname {b1} \right ) \sqrt {-4 \operatorname {a0} \operatorname {a2} +\operatorname {a1}^{2}}-2 \operatorname {a2}^{2} \operatorname {b0} +\left (2 \operatorname {a0} \operatorname {b2} +\operatorname {a1} \operatorname {b1} \right ) \operatorname {a2} -\operatorname {a1}^{2} \operatorname {b2}}{2 \sqrt {-4 \operatorname {a0} \operatorname {a2} +\operatorname {a1}^{2}}\, \operatorname {a2}^{2}}, \frac {\operatorname {a1} \operatorname {b2} +2 \operatorname {a2}^{2}-\operatorname {a2} \operatorname {b1}}{\operatorname {a2}^{2}}, \frac {\sqrt {-4 \operatorname {a0} \operatorname {a2} +\operatorname {a1}^{2}}\, \left (\operatorname {a2} x +\operatorname {b2} \right )}{\operatorname {a2}^{2}}\right ) c_{2} +\operatorname {KummerM}\left (\frac {\left (\operatorname {a1} \operatorname {b2} +2 \operatorname {a2}^{2}-\operatorname {a2} \operatorname {b1} \right ) \sqrt {-4 \operatorname {a0} \operatorname {a2} +\operatorname {a1}^{2}}-2 \operatorname {a2}^{2} \operatorname {b0} +\left (2 \operatorname {a0} \operatorname {b2} +\operatorname {a1} \operatorname {b1} \right ) \operatorname {a2} -\operatorname {a1}^{2} \operatorname {b2}}{2 \sqrt {-4 \operatorname {a0} \operatorname {a2} +\operatorname {a1}^{2}}\, \operatorname {a2}^{2}}, \frac {\operatorname {a1} \operatorname {b2} +2 \operatorname {a2}^{2}-\operatorname {a2} \operatorname {b1}}{\operatorname {a2}^{2}}, \frac {\sqrt {-4 \operatorname {a0} \operatorname {a2} +\operatorname {a1}^{2}}\, \left (\operatorname {a2} x +\operatorname {b2} \right )}{\operatorname {a2}^{2}}\right ) c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.301 (sec). Leaf size: 301
DSolve[(b0 + a0*x)*y[x] + (b1 + a1*x)*y'[x] + (b2 + a2*x)*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{-\frac {x \left (\sqrt {\text {a1}^2-4 \text {a0} \text {a2}}+\text {a1}\right )}{2 \text {a2}}} (\text {a2} x+\text {b2})^{\frac {\text {a1} \text {b2}+\text {a2}^2-\text {a2} \text {b1}}{\text {a2}^2}} \left (c_1 \operatorname {HypergeometricU}\left (\frac {2 \left (\sqrt {\text {a1}^2-4 \text {a0} \text {a2}}-\text {b0}\right ) \text {a2}^2+\left (\text {a1} \text {b1}-\sqrt {\text {a1}^2-4 \text {a0} \text {a2}} \text {b1}+2 \text {a0} \text {b2}\right ) \text {a2}+\text {a1} \left (\sqrt {\text {a1}^2-4 \text {a0} \text {a2}}-\text {a1}\right ) \text {b2}}{2 \text {a2}^2 \sqrt {\text {a1}^2-4 \text {a0} \text {a2}}},-\frac {\text {b1}}{\text {a2}}+\frac {\text {a1} \text {b2}}{\text {a2}^2}+2,\frac {\sqrt {\text {a1}^2-4 \text {a0} \text {a2}} (\text {b2}+\text {a2} x)}{\text {a2}^2}\right )+c_2 L_{\frac {-2 \left (\sqrt {\text {a1}^2-4 \text {a0} \text {a2}}-\text {b0}\right ) \text {a2}^2+\left (-\text {a1} \text {b1}+\sqrt {\text {a1}^2-4 \text {a0} \text {a2}} \text {b1}-2 \text {a0} \text {b2}\right ) \text {a2}+\text {a1} \left (\text {a1}-\sqrt {\text {a1}^2-4 \text {a0} \text {a2}}\right ) \text {b2}}{2 \text {a2}^2 \sqrt {\text {a1}^2-4 \text {a0} \text {a2}}}}^{\frac {\text {a2}^2-\text {b1} \text {a2}+\text {a1} \text {b2}}{\text {a2}^2}}\left (\frac {\sqrt {\text {a1}^2-4 \text {a0} \text {a2}} (\text {b2}+\text {a2} x)}{\text {a2}^2}\right )\right ) \]