2.32.20 Problem 202

Internal problem ID [13862]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 2, Second-Order Differential Equations. section 2.1.2-6
Problem number : 202
Date solved : Sunday, January 18, 2026 at 09:26:28 PM
CAS classification : [[_2nd_order, _with_linear_symmetries], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]

2.32.20.1 second order change of variable on x method 2

0.809 (sec)

\begin{align*} 2 x \left (a \,x^{2}+b x +c \right ) y^{\prime \prime }+\left (a \,x^{2}-c \right ) y^{\prime }+y \lambda \,x^{2}&=0 \\ \end{align*}

Entering second order change of variable on \(x\) method 2 solverIn normal form the ode

\begin{align*} 2 x \left (a \,x^{2}+b x +c \right ) y^{\prime \prime }+\left (a \,x^{2}-c \right ) y^{\prime }+y \lambda \,x^{2} = 0\tag {1} \end{align*}

Becomes

\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end{align*}

Where

\begin{align*} p \left (x \right )&=\frac {a \,x^{2}-c}{2 x \left (a \,x^{2}+b x +c \right )}\\ q \left (x \right )&=\frac {\lambda x}{2 a \,x^{2}+2 b x +2 c} \end{align*}

Applying change of variables \(\tau = g \left (x \right )\) to (2) gives

\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end{align*}

Where \(\tau \) is the new independent variable, and

\begin{align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end{align*}

Let \(p_{1} = 0\). Eq (4) simplifies to

\begin{align*} \tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )&=0 \end{align*}

This ode is solved resulting in

\begin{align*} \tau &= \int {\mathrm e}^{-\int p \left (x \right )d x}d x\\ &= \int {\mathrm e}^{-\int \frac {a \,x^{2}-c}{2 x \left (a \,x^{2}+b x +c \right )}d x}d x\\ &= \int e^{-\frac {\ln \left (a \,x^{2}+b x +c \right )}{2}+\frac {\ln \left (x \right )}{2}} \,dx\\ &= \int \frac {\sqrt {x}}{\sqrt {a \,x^{2}+b x +c}}d x\\ &= \int \frac {\sqrt {x}}{\sqrt {a \,x^{2}+b x +c}}d x\tag {6} \end{align*}

Using (6) to evaluate \(q_{1}\) from (5) gives

\begin{align*} q_{1} \left (\tau \right ) &= \frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &= \frac {\frac {\lambda x}{2 a \,x^{2}+2 b x +2 c}}{\frac {x}{a \,x^{2}+b x +c}}\\ &= \frac {\lambda }{2}\tag {7} \end{align*}

Substituting the above in (3) and noting that now \(p_{1} = 0\) results in

\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+q_{1} y \left (\tau \right )&=0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+\frac {\lambda y \left (\tau \right )}{2}&=0 \end{align*}

The above ode is now solved for \(y \left (\tau \right )\).Entering second order linear constant coefficient ode solver

This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

\[ A y''(\tau ) + B y'(\tau ) + C y(\tau ) = 0 \]

Where in the above \(A=1, B=0, C=\frac {\lambda }{2}\). Let the solution be \(y \left (\tau \right )=e^{\lambda \tau }\). Substituting this into the ODE gives

\[ \lambda ^{2} {\mathrm e}^{\tau \lambda }+\frac {\lambda \,{\mathrm e}^{\tau \lambda }}{2} = 0 \tag {1} \]

Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda \tau }\) gives

\[ \lambda ^{2}+\frac {\lambda }{2} = 0 \tag {2} \]

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form. Using the quadratic formula the roots are

\[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \]

Substituting \(A=1, B=0, C=\frac {\lambda }{2}\) into the above gives

\begin{align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {0^2 - (4) \left (1\right )\left (\frac {\lambda }{2}\right )}\\ &= \pm \frac {\sqrt {-2 \lambda }}{2} \end{align*}

Hence

\begin{align*} \lambda _1 &= + \frac {\sqrt {-2 \lambda }}{2} \\ \lambda _2 &= - \frac {\sqrt {-2 \lambda }}{2} \\ \end{align*}

Which simplifies to

\begin{align*} \lambda _1 &= \frac {\sqrt {2}\, \sqrt {-\lambda }}{2} \\ \lambda _2 &= -\frac {\sqrt {2}\, \sqrt {-\lambda }}{2} \\ \end{align*}

Since the roots are distinct, the solution is

\begin{align*} y \left (\tau \right ) &= c_1 e^{\lambda _1 \tau } + c_2 e^{\lambda _2 \tau } \\ y \left (\tau \right ) &= c_1 e^{\left (\frac {\sqrt {2}\, \sqrt {-\lambda }}{2}\right )\tau } +c_2 e^{\left (-\frac {\sqrt {2}\, \sqrt {-\lambda }}{2}\right )\tau } \\ \end{align*}

Or

\[ y \left (\tau \right ) =c_1 \,{\mathrm e}^{\frac {\sqrt {2}\, \sqrt {-\lambda }\, \tau }{2}}+c_2 \,{\mathrm e}^{-\frac {\sqrt {2}\, \sqrt {-\lambda }\, \tau }{2}} \]

The above solution is now transformed back to \(y\) using (6) which results in

\[ y = c_1 \,{\mathrm e}^{\frac {\sqrt {2}\, \sqrt {-\lambda }\, \int \frac {\sqrt {x}}{\sqrt {a \,x^{2}+b x +c}}d x}{2}}+c_2 \,{\mathrm e}^{-\frac {\sqrt {2}\, \sqrt {-\lambda }\, \int \frac {\sqrt {x}}{\sqrt {a \,x^{2}+b x +c}}d x}{2}} \]

Summary of solutions found

\begin{align*} y &= c_1 \,{\mathrm e}^{\frac {\sqrt {2}\, \sqrt {-\lambda }\, \int \frac {\sqrt {x}}{\sqrt {a \,x^{2}+b x +c}}d x}{2}}+c_2 \,{\mathrm e}^{-\frac {\sqrt {2}\, \sqrt {-\lambda }\, \int \frac {\sqrt {x}}{\sqrt {a \,x^{2}+b x +c}}d x}{2}} \\ \end{align*}

2.32.20.2 ✓ Maple. Time used: 0.033 (sec). Leaf size: 65

ode:=2*x*(a*x^2+b*x+c)*diff(diff(y(x),x),x)+(a*x^2-c)*diff(y(x),x)+lambda*x^2*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ y = \left (c_1 \,{\mathrm e}^{i \sqrt {2}\, \sqrt {\lambda }\, \int \frac {\sqrt {x}}{\sqrt {a \,x^{2}+b x +c}}d x}+c_2 \right ) {\mathrm e}^{-\frac {i \sqrt {2}\, \sqrt {\lambda }\, \int \frac {\sqrt {x}}{\sqrt {a \,x^{2}+b x +c}}d x}{2}} \]

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)] 
   Solution is available but has integrals. Trying a simpler solution using Kov\ 
acics algorithm... 
   -> Trying a Liouvillian solution using Kovacics algorithm 
      A Liouvillian solution exists 
      Group is reducible or imprimitive 
      Solution has integrals. Trying a special function solution free of integr\ 
als... 
      -> Trying a solution in terms of special functions: 
         -> Bessel 
         -> elliptic 
         -> Legendre 
         -> Kummer 
            -> hyper3: Equivalence to 1F1 under a power @ Moebius 
         -> hypergeometric 
            -> heuristic approach 
            -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
         -> Mathieu 
            -> Equivalence to the rational form of Mathieu ODE under a power @\ 
 Moebius 
         -> Heun: Equivalence to the GHE or one of its 4 confluent cases under \ 
a power @ Moebius 
      No special function solution was found. 
   <- Kovacics algorithm successful 
   Solution via Kovacic is not simpler. Returning default solution 
   <- linear_1 successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x \left (a \,x^{2}+b x +c \right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (a \,x^{2}-c \right ) \left (\frac {d}{d x}y \left (x \right )\right )+\lambda \,x^{2} y \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=-\frac {\lambda x y \left (x \right )}{2 \left (a \,x^{2}+b x +c \right )}-\frac {\left (a \,x^{2}-c \right ) \left (\frac {d}{d x}y \left (x \right )\right )}{2 x \left (a \,x^{2}+b x +c \right )} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (x \right )\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^{2}}{d x^{2}}y \left (x \right )+\frac {\left (a \,x^{2}-c \right ) \left (\frac {d}{d x}y \left (x \right )\right )}{2 x \left (a \,x^{2}+b x +c \right )}+\frac {\lambda x y \left (x \right )}{2 \left (a \,x^{2}+b x +c \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 {a \,x^{2}-c}{2 x \left (a \,x^{2}+b x +c \right )}, P_{3}\left (x \right )=\frac {\lambda x}{2 \left (a \,x^{2}+b x +c \right )}\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}}}=0 \\ {} & \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 x \left (a \,x^{2}+b x +c \right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (a \,x^{2}-c \right ) \left (\frac {d}{d x}y \left (x \right )\right )+\lambda \,x^{2} y \left (x \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=\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^{2}\cdot y \left (x \right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -2 \\ {} & {} & x^{2}\cdot y \left (x \right )=\moverset {\infty }{\munderset {k =2}{\sum }}a_{k -2} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot \left (\frac {d}{d x}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y \left (x \right )\right )=\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 \left (\frac {d}{d x}y \left (x \right )\right )=\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^{2}}{d x^{2}}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & x^{m}\cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\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^{2}}{d x^{2}}y \left (x \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 ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & c a_{0} r \left (-3+2 r \right ) x^{-1+r}+\left (c a_{1} \left (1+r \right ) \left (-1+2 r \right )+2 a_{0} r \left (-1+r \right ) b \right ) x^{r}+\left (c a_{2} \left (2+r \right ) \left (1+2 r \right )+2 a_{1} \left (1+r \right ) r b +a a_{0} r \left (-1+2 r \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (c a_{k +1} \left (k +1+r \right ) \left (2 k +2 r -1\right )+2 a_{k} \left (k +r \right ) \left (k +r -1\right ) b +a a_{k -1} \left (k +r -1\right ) \left (2 k +2 r -3\right )+\lambda a_{k -2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & c r \left (-3+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {3}{2}\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [c a_{1} \left (1+r \right ) \left (-1+2 r \right )+2 a_{0} r \left (-1+r \right ) b =0, c a_{2} \left (2+r \right ) \left (1+2 r \right )+2 a_{1} \left (1+r \right ) r b +a a_{0} r \left (-1+2 r \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=-\frac {2 a_{0} r \left (-1+r \right ) b}{c \left (2 r^{2}+r -1\right )}, a_{2}=-\frac {a_{0} r \left (4 a c \,r^{2}-4 b^{2} r^{2}-4 a c r +4 b^{2} r +a c \right )}{c^{2} \left (4 r^{3}+8 r^{2}-r -2\right )}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & c a_{k +1} \left (k +1+r \right ) \left (2 k +2 r -1\right )+2 a_{k} \left (k +r \right ) \left (k +r -1\right ) b +a a_{k -1} \left (k +r -1\right ) \left (2 k +2 r -3\right )+\lambda a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & c a_{k +3} \left (k +3+r \right ) \left (2 k +3+2 r \right )+2 a_{k +2} \left (k +r +2\right ) \left (k +1+r \right ) b +a a_{k +1} \left (k +1+r \right ) \left (2 k +1+2 r \right )+\lambda a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=-\frac {2 a \,k^{2} a_{k +1}+4 a k r a_{k +1}+2 a \,r^{2} a_{k +1}+2 b \,k^{2} a_{k +2}+4 b k r a_{k +2}+2 b \,r^{2} a_{k +2}+3 a k a_{k +1}+3 a r a_{k +1}+6 b k a_{k +2}+6 b r a_{k +2}+a a_{k +1}+4 b a_{k +2}+\lambda a_{k}}{c \left (k +3+r \right ) \left (2 k +3+2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=-\frac {2 a \,k^{2} a_{k +1}+2 b \,k^{2} a_{k +2}+3 a k a_{k +1}+6 b k a_{k +2}+a a_{k +1}+4 b a_{k +2}+\lambda a_{k}}{c \left (k +3\right ) \left (2 k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +3}=-\frac {2 a \,k^{2} a_{k +1}+2 b \,k^{2} a_{k +2}+3 a k a_{k +1}+6 b k a_{k +2}+a a_{k +1}+4 b a_{k +2}+\lambda a_{k}}{c \left (k +3\right ) \left (2 k +3\right )}, a_{1}=0, a_{2}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {3}{2} \\ {} & {} & a_{k +3}=-\frac {2 a \,k^{2} a_{k +1}+2 b \,k^{2} a_{k +2}+9 a k a_{k +1}+12 b k a_{k +2}+10 a a_{k +1}+\frac {35}{2} b a_{k +2}+\lambda a_{k}}{c \left (k +\frac {9}{2}\right ) \left (2 k +6\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {3}{2} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {3}{2}}, a_{k +3}=-\frac {2 a \,k^{2} a_{k +1}+2 b \,k^{2} a_{k +2}+9 a k a_{k +1}+12 b k a_{k +2}+10 a a_{k +1}+\frac {35}{2} b a_{k +2}+\lambda a_{k}}{c \left (k +\frac {9}{2}\right ) \left (2 k +6\right )}, a_{1}=-\frac {3 a_{0} b}{10 c}, a_{2}=-\frac {3 a_{0} \left (4 a c -3 b^{2}\right )}{56 c^{2}}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y \left (x \right )=\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} x^{k +\frac {3}{2}}\right ), d_{k +3}=-\frac {2 a \,k^{2} d_{k +1}+2 b \,k^{2} d_{k +2}+3 a k d_{k +1}+6 b k d_{k +2}+a d_{k +1}+4 b d_{k +2}+\lambda d_{k}}{c \left (k +3\right ) \left (2 k +3\right )}, d_{1}=0, d_{2}=0, e_{k +3}=-\frac {2 a \,k^{2} e_{k +1}+2 b \,k^{2} e_{k +2}+9 a k e_{k +1}+12 b k e_{k +2}+10 a e_{k +1}+\frac {35}{2} b e_{k +2}+\lambda e_{k}}{c \left (k +\frac {9}{2}\right ) \left (2 k +6\right )}, e_{1}=-\frac {3 e_{0} b}{10 c}, e_{2}=-\frac {3 e_{0} \left (4 a c -3 b^{2}\right )}{56 c^{2}}\right ] \end {array} \]

2.32.20.3 ✓ Mathematica. Time used: 100.084 (sec). Leaf size: 501

ode=2*x*(a*x^2+b*x+c)*D[y[x],{x,2}]+(a*x^2-c)*D[y[x],x]+\[Lambda]*x^2*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to c_1 \cosh \left (\frac {\sqrt {\lambda } \left (\sqrt {b^2-4 a c}-b\right ) \sqrt {\sqrt {b^2-4 a c}+2 a x+b} \sqrt {\frac {2 a x}{b-\sqrt {b^2-4 a c}}+1} \left (E\left (i \text {arcsinh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x}}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )-\operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x}}{\sqrt {b+\sqrt {b^2-4 a c}}}\right ),\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )\right )}{2 a^{3/2} \sqrt {x (a x+b)+c}}\right )+i c_2 \sinh \left (\frac {\sqrt {\lambda } \left (\sqrt {b^2-4 a c}-b\right ) \sqrt {\sqrt {b^2-4 a c}+2 a x+b} \sqrt {\frac {2 a x}{b-\sqrt {b^2-4 a c}}+1} \left (E\left (i \text {arcsinh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x}}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )-\operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x}}{\sqrt {b+\sqrt {b^2-4 a c}}}\right ),\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )\right )}{2 a^{3/2} \sqrt {x (a x+b)+c}}\right ) \end{align*}

2.32.20.4 ✗ Sympy

from sympy import * 
x = symbols("x") 
a = symbols("a") 
b = symbols("b") 
c = symbols("c") 
lambda_ = symbols("lambda_") 
y = Function("y") 
ode = Eq(lambda_*x**2*y(x) + 2*x*(a*x**2 + b*x + c)*Derivative(y(x), (x, 2)) + (a*x**2 - c)*Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False
 

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0
 

classify_ode(ode,func=y(x)) 
 
('factorable', '2nd_power_series_regular')