28.48 problem 108

28.48.1 Maple step by step solution

Internal problem ID [10931]
Internal file name [OUTPUT/10188_Sunday_December_31_2023_11_03_46_AM_19163397/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 2.1.2-3 Equation of form \((a x + b)y''+f(x)y'+g(x)y=0\)
Problem number: 108.
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.

\[ \boxed {\left (a_{2} x +b_{2} \right ) y^{\prime \prime }+\left (a_{1} x +b_{1} \right ) y^{\prime }+\left (a_{0} x +b_{0} \right ) y=0} \]

28.48.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (a_{2} x +b_{2} \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (a_{1} x +b_{1} \right ) y^{\prime }+\left (a_{0} x +b_{0} \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 (a_{0} x +b_{0} \right ) y}{a_{2} x +b_{2}}-\frac {\left (a_{1} x +b_{1} \right ) y^{\prime }}{a_{2} x +b_{2}} \\ \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 (a_{1} x +b_{1} \right ) y^{\prime }}{a_{2} x +b_{2}}+\frac {\left (a_{0} x +b_{0} \right ) y}{a_{2} x +b_{2}}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=-\frac {b_{2}}{a_{2}}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {a_{1} x +b_{1}}{a_{2} x +b_{2}}, P_{3}\left (x \right )=\frac {a_{0} x +b_{0}}{a_{2} x +b_{2}}\right ] \\ {} & \circ & \left (x +\frac {b_{2}}{a_{2}}\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-\frac {b_{2}}{a_{2}} \\ {} & {} & \left (\left (x +\frac {b_{2}}{a_{2}}\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-\frac {b_{2}}{a_{2}}}}}=\frac {-\frac {a_{1} b_{2}}{a_{2}}+b_{1}}{a_{2}} \\ {} & \circ & \left (x +\frac {b_{2}}{a_{2}}\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-\frac {b_{2}}{a_{2}} \\ {} & {} & \left (\left (x +\frac {b_{2}}{a_{2}}\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-\frac {b_{2}}{a_{2}}}}}=0 \\ {} & \circ & x =-\frac {b_{2}}{a_{2}}\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=-\frac {b_{2}}{a_{2}}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=-\frac {b_{2}}{a_{2}} \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (a_{2} x +b_{2} \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (a_{1} x +b_{1} \right ) y^{\prime }+\left (a_{0} x +b_{0} \right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -\frac {b_{2}}{a_{2}}\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & a_{2} u \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (a_{1} u -\frac {a_{1} b_{2}}{a_{2}}+b_{1} \right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (a_{0} u -\frac {a_{0} b_{2}}{a_{2}}+b_{0} \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 (-a_{2}^{2} r +a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1} \right ) u^{-1+r}}{a_{2}}+\left (-\frac {a_{1} \left (1+r \right ) \left (-a_{2}^{2} r +a_{1} b_{2} -a_{2} b_{1} \right )}{a_{2}}-\frac {a_{0} \left (-a_{2} a_{1} r +a_{0} b_{2} -a_{2} b_{0} \right )}{a_{2}}\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (-\frac {a_{k +1} \left (k +1+r \right ) \left (-a_{2}^{2} \left (k +1\right )-a_{2}^{2} r +a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1} \right )}{a_{2}}-\frac {a_{k} \left (-a_{2} a_{1} k -a_{2} a_{1} r +a_{0} b_{2} -a_{2} b_{0} \right )}{a_{2}}+a_{k -1} a_{0} \right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -\frac {r \left (-a_{2}^{2} r +a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1} \right )}{a_{2}}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & -\frac {a_{1} \left (1+r \right ) \left (-a_{2}^{2} r +a_{1} b_{2} -a_{2} b_{1} \right )}{a_{2}}-\frac {a_{0} \left (-a_{2} a_{1} r +a_{0} b_{2} -a_{2} b_{0} \right )}{a_{2}}=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 ) a_{2}^{2}+\left (b_{1} \left (k +1+r \right ) a_{k +1}+k a_{1} a_{k}+r a_{1} a_{k}+a_{k -1} a_{0} +a_{k} b_{0} \right ) a_{2} -\left (a_{1} \left (k +1+r \right ) a_{k +1}+a_{k} a_{0} \right ) b_{2}}{a_{2}}=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 ) a_{2}^{2}+\left (b_{1} \left (k +2+r \right ) a_{k +2}+\left (k +1\right ) a_{1} a_{k +1}+r a_{1} a_{k +1}+a_{k} a_{0} +a_{k +1} b_{0} \right ) a_{2} -\left (a_{1} \left (k +2+r \right ) a_{k +2}+a_{k +1} a_{0} \right ) b_{2}}{a_{2}}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {a_{1} a_{2} k a_{k +1}+a_{1} a_{2} r a_{k +1}+a_{0} a_{2} a_{k}-a_{0} b_{2} a_{k +1}+a_{1} a_{2} a_{k +1}+a_{2} b_{0} a_{k +1}}{\left (k +2+r \right ) \left (-a_{2}^{2} k -a_{2}^{2} r +a_{1} b_{2} -a_{2}^{2}-a_{2} b_{1} \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {a_{1} a_{2} k a_{k +1}+a_{0} a_{2} a_{k}-a_{0} b_{2} a_{k +1}+a_{1} a_{2} a_{k +1}+a_{2} b_{0} a_{k +1}}{\left (k +2\right ) \left (-a_{2}^{2} k +a_{1} b_{2} -a_{2}^{2}-a_{2} b_{1} \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 {a_{1} a_{2} k a_{k +1}+a_{0} a_{2} a_{k}-a_{0} b_{2} a_{k +1}+a_{1} a_{2} a_{k +1}+a_{2} b_{0} a_{k +1}}{\left (k +2\right ) \left (-a_{2}^{2} k +a_{1} b_{2} -a_{2}^{2}-a_{2} b_{1} \right )}, -\frac {a_{1} \left (a_{1} b_{2} -a_{2} b_{1} \right )}{a_{2}}-\frac {a_{0} \left (a_{0} b_{2} -a_{2} b_{0} \right )}{a_{2}}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +\frac {b_{2}}{a_{2}} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +\frac {b_{2}}{a_{2}}\right )^{k}, a_{k +2}=\frac {a_{1} a_{2} k a_{k +1}+a_{0} a_{2} a_{k}-a_{0} b_{2} a_{k +1}+a_{1} a_{2} a_{k +1}+a_{2} b_{0} a_{k +1}}{\left (k +2\right ) \left (-a_{2}^{2} k +a_{1} b_{2} -a_{2}^{2}-a_{2} b_{1} \right )}, -\frac {a_{1} \left (a_{1} b_{2} -a_{2} b_{1} \right )}{a_{2}}-\frac {a_{0} \left (a_{0} b_{2} -a_{2} b_{0} \right )}{a_{2}}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}} \\ {} & {} & a_{k +2}=\frac {a_{1} a_{2} k a_{k +1}+\frac {a_{1} \left (a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1} \right ) a_{k +1}}{a_{2}}+a_{0} a_{2} a_{k}-a_{0} b_{2} a_{k +1}+a_{1} a_{2} a_{k +1}+a_{2} b_{0} a_{k +1}}{\left (k +2+\frac {a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}\right ) \left (-a_{2}^{2} k -2 a_{2}^{2}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}}, a_{k +2}=\frac {a_{1} a_{2} k a_{k +1}+\frac {a_{1} \left (a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1} \right ) a_{k +1}}{a_{2}}+a_{0} a_{2} a_{k}-a_{0} b_{2} a_{k +1}+a_{1} a_{2} a_{k +1}+a_{2} b_{0} a_{k +1}}{\left (k +2+\frac {a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}\right ) \left (-a_{2}^{2} k -2 a_{2}^{2}\right )}, a_{1} \left (1+\frac {a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}\right ) a_{2} -\frac {a_{0} \left (-\frac {a_{1} \left (a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1} \right )}{a_{2}}+a_{0} b_{2} -a_{2} b_{0} \right )}{a_{2}}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +\frac {b_{2}}{a_{2}} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +\frac {b_{2}}{a_{2}}\right )^{k +\frac {a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}}, a_{k +2}=\frac {a_{1} a_{2} k a_{k +1}+\frac {a_{1} \left (a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1} \right ) a_{k +1}}{a_{2}}+a_{0} a_{2} a_{k}-a_{0} b_{2} a_{k +1}+a_{1} a_{2} a_{k +1}+a_{2} b_{0} a_{k +1}}{\left (k +2+\frac {a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}\right ) \left (-a_{2}^{2} k -2 a_{2}^{2}\right )}, a_{1} \left (1+\frac {a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}\right ) a_{2} -\frac {a_{0} \left (-\frac {a_{1} \left (a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1} \right )}{a_{2}}+a_{0} b_{2} -a_{2} b_{0} \right )}{a_{2}}=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 {b_{2}}{a_{2}}\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +\frac {b_{2}}{a_{2}}\right )^{k +\frac {a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}}\right ), a_{k +2}=\frac {a_{1} a_{2} k a_{k +1}+a_{0} a_{2} a_{k}-a_{0} b_{2} a_{k +1}+a_{1} a_{2} a_{k +1}+a_{2} b_{0} a_{k +1}}{\left (k +2\right ) \left (-a_{2}^{2} k +a_{1} b_{2} -a_{2}^{2}-a_{2} b_{1} \right )}, -\frac {a_{1} \left (a_{1} b_{2} -a_{2} b_{1} \right )}{a_{2}}-\frac {a_{0} \left (a_{0} b_{2} -a_{2} b_{0} \right )}{a_{2}}=0, b_{k +2}=\frac {a_{1} a_{2} k b_{k +1}+\frac {a_{1} \left (a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1} \right ) b_{k +1}}{a_{2}}+a_{0} a_{2} b_{k}-a_{0} b_{2} b_{k +1}+a_{1} a_{2} b_{k +1}+a_{2} b_{0} b_{k +1}}{\left (k +2+\frac {a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}\right ) \left (-a_{2}^{2} k -2 a_{2}^{2}\right )}, b_{1} \left (1+\frac {a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}\right ) a_{2} -\frac {b_{0} \left (-\frac {a_{1} \left (a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1} \right )}{a_{2}}+a_{0} b_{2} -a_{2} b_{0} \right )}{a_{2}}=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: 0.047 (sec). Leaf size: 248

dsolve((a__2*x+b__2)*diff(y(x),x$2)+(a__1*x+b__1)*diff(y(x),x)+(a__0*x+b__0)*y(x)=0,y(x), singsol=all)
 

\[ y \left (x \right ) = \left (a_{2} x +b_{2} \right )^{\frac {a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}} {\mathrm e}^{-\frac {\left (\sqrt {-4 a_{0} a_{2} +a_{1}^{2}}+a_{1} \right ) x}{2 a_{2}}} \left (\operatorname {KummerM}\left (\frac {\left (a_{1} b_{2} +2 a_{2}^{2}-a_{2} b_{1} \right ) \sqrt {-4 a_{0} a_{2} +a_{1}^{2}}-2 a_{2}^{2} b_{0} +\left (2 a_{0} b_{2} +b_{1} a_{1} \right ) a_{2} -a_{1}^{2} b_{2}}{2 \sqrt {-4 a_{0} a_{2} +a_{1}^{2}}\, a_{2}^{2}}, \frac {a_{1} b_{2} +2 a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}, \frac {\sqrt {-4 a_{0} a_{2} +a_{1}^{2}}\, \left (a_{2} x +b_{2} \right )}{a_{2}^{2}}\right ) c_{1} +\operatorname {KummerU}\left (\frac {\left (a_{1} b_{2} +2 a_{2}^{2}-a_{2} b_{1} \right ) \sqrt {-4 a_{0} a_{2} +a_{1}^{2}}-2 a_{2}^{2} b_{0} +\left (2 a_{0} b_{2} +b_{1} a_{1} \right ) a_{2} -a_{1}^{2} b_{2}}{2 \sqrt {-4 a_{0} a_{2} +a_{1}^{2}}\, a_{2}^{2}}, \frac {a_{1} b_{2} +2 a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}, \frac {\sqrt {-4 a_{0} a_{2} +a_{1}^{2}}\, \left (a_{2} x +b_{2} \right )}{a_{2}^{2}}\right ) c_{2} \right ) \]

Solution by Mathematica

Time used: 0.526 (sec). Leaf size: 301

DSolve[(a2*x+b2)*y''[x]+(a1*x+b1)*y'[x]+(a0*x+b0)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to e^{-\frac {x \left (\sqrt {\text {a1}^2-4 \text {a0} \text {a2}}+\text {a1}\right )}{2 \text {a2}}} (\text {a2} x+\text {b2})^{\frac {\text {a1} \text {b2}+\text {a2}^2-\text {a2} \text {b1}}{\text {a2}^2}} \left (c_1 \operatorname {HypergeometricU}\left (\frac {2 \left (\sqrt {\text {a1}^2-4 \text {a0} \text {a2}}-\text {b0}\right ) \text {a2}^2+\left (\text {a1} \text {b1}-\sqrt {\text {a1}^2-4 \text {a0} \text {a2}} \text {b1}+2 \text {a0} \text {b2}\right ) \text {a2}+\text {a1} \left (\sqrt {\text {a1}^2-4 \text {a0} \text {a2}}-\text {a1}\right ) \text {b2}}{2 \text {a2}^2 \sqrt {\text {a1}^2-4 \text {a0} \text {a2}}},-\frac {\text {b1}}{\text {a2}}+\frac {\text {a1} \text {b2}}{\text {a2}^2}+2,\frac {\sqrt {\text {a1}^2-4 \text {a0} \text {a2}} (\text {b2}+\text {a2} x)}{\text {a2}^2}\right )+c_2 L_{\frac {-2 \left (\sqrt {\text {a1}^2-4 \text {a0} \text {a2}}-\text {b0}\right ) \text {a2}^2+\left (-\text {a1} \text {b1}+\sqrt {\text {a1}^2-4 \text {a0} \text {a2}} \text {b1}-2 \text {a0} \text {b2}\right ) \text {a2}+\text {a1} \left (\text {a1}-\sqrt {\text {a1}^2-4 \text {a0} \text {a2}}\right ) \text {b2}}{2 \text {a2}^2 \sqrt {\text {a1}^2-4 \text {a0} \text {a2}}}}^{\frac {\text {a2}^2-\text {b1} \text {a2}+\text {a1} \text {b2}}{\text {a2}^2}}\left (\frac {\sqrt {\text {a1}^2-4 \text {a0} \text {a2}} (\text {b2}+\text {a2} x)}{\text {a2}^2}\right )\right ) \]